Mixing
Why are two isomorphic objects “categorically the same”?
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If I understand correctly, in category theory, two objects $X,Y$ are isomorphic if there exists a morphism $f:Xto Y$, and a morphism $f^{-1}:Yto X$ such that $fcirc f^{-1}=id_Y$ and $f^{-1}circ f=id_X$.
Why does this imply that we can treat the two objects as "equivalent" within a category?
For example, the categorical product is "unique up to isomorphism". Why does this imply that we can "pretend as if there is only one categorical product"?
category-theory
asked Jan 29 at 5:26
user56834user56834
3,37921253
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add a comment |
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If I understand correctly, in category theory, two objects $X,Y$ are isomorphic if there exists a morphism $f:Xto Y$, and a morphism $f^{-1}:Yto X$ such that $fcirc f^{-1}=id_Y$ and $f^{-1}circ f=id_X$.
Why does this imply that we can treat the two objects as "equivalent" within a category?
For example, the categorical product is "unique up to isomorphism". Why does this imply that we can "pretend as if there is only one categorical product"?
category-theory
asked Jan 29 at 5:26
user56834user56834
3,37921253
$endgroup$
add a comment |
$begingroup$
If I understand correctly, in category theory, two objects $X,Y$ are isomorphic if there exists a morphism $f:Xto Y$, and a morphism $f^{-1}:Yto X$ such that $fcirc f^{-1}=id_Y$ and $f^{-1}circ f=id_X$.
Why does this imply that we can treat the two objects as "equivalent" within a category?
For example, the categorical product is "unique up to isomorphism". Why does this imply that we can "pretend as if there is only one categorical product"?
category-theory
asked Jan 29 at 5:26
user56834user56834
3,37921253
$endgroup$
If I understand correctly, in category theory, two objects $X,Y$ are isomorphic if there exists a morphism $f:Xto Y$, and a morphism $f^{-1}:Yto X$ such that $fcirc f^{-1}=id_Y$ and $f^{-1}circ f=id_X$.
Why does this imply that we can treat the two objects as "equivalent" within a category?
For example, the categorical product is "unique up to isomorphism". Why does this imply that we can "pretend as if there is only one categorical product"?
category-theory
category-theory
asked Jan 29 at 5:26
user56834user56834
3,37921253
asked Jan 29 at 5:26
user56834user56834
3,37921253
edited Jan 29 at 5:39
user56834
asked Jan 29 at 5:26
user56834user56834
3,37921253
asked Jan 29 at 5:26
user56834user56834
3,37921253
asked Jan 29 at 5:26
user56834user56834
3,37921253
3,37921253
add a comment |
add a comment |
4 Answers
4
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oldest
votes
$begingroup$
A first point is that this process of identifying isomorphic objects comes with practise : with practise, you'll start noticing that any "categorical property" an object has is shared by every object that is isomorphic to it; and you'll start being able to see why it makes sense to identify isomorphic objects.
So the best way to understand what's happening is to take a lot of examples and work through them to see why it makes sense. Now if you want a general argument, here's sort of one :
Suppose $f:Ato B$ is an isomorphism in a category $C$.
Suppose now you have a complicated situation involving $A$, and many other morphisms $f_i :Ato C_i, iin I$, $g_j: B_jto A$, satisfying certain equations (some involving $A$, others not involving it). It's the stuff that's interesting to you about $A$.
Well now you can transport this situation to the same situation about $B$ : put $f'_i = f_icirc f^{-1} : Bto C_i$, $g'_j := fcirc g_j : B_jto B$. This situation will satisfy the same equations as the one before : those not involving $A$ will be the same by replacing every $f_i, g_j$ by $f'_i, g'_j$; those involving $A$ will be the same but replacing every occurrence of $A$ by $B$ and conjugating by $f,f^{-1}$ to make things consistent. That is because $ff^{-1} = id_B, f^{-1}f =id_A$ and identities are neutral.
What this means is that if you have an interesting situation about an object $A$ in a category $C$, this situation being all about $C$-properties, and an isomorphism $f:Ato B$, then by conjugating by $f$ or $f^{-1}$ whenever it makes sense will give the same interesting situation about the object $B$.
In other words, along $f$, you cannot distinguish between $A$ and $B$ by properties of $C$: if you cannot distinguish between them, if they have the same properties, they're "essentially the same" - that's what we mean when we say we can identify isomorphic objects.
Note, however, that the identification, the process of saying they are the same involves not just saying that $A$ and $B$ are isomorphic, but picking a specific isomorphism. It's very important to note that we can identify to isomorphic objects so long as we have a specific isomorphism in mind.
For instance, for the categorical product, there is a unique isomorphism that commutes with the projections: it's the one we're considering.
answered Jan 29 at 11:05
MaxMax
15.8k11143
$endgroup$
2
$begingroup$
I would like to be able to give you another +1 just for emphasizing that one have to keep track of the isomorphisms.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:18
add a comment |
$begingroup$
The following answer provide another, hopefully convincing, argument on why in category theory one can indentify isomorphic objects.
Categories of (multi-sorted) structures.
When dealing with categories of structures isomorphisms are bijective mapping preserving the structure. Basically this amounts saying that isomorphisms are ways to parametrize/rename the elements of one structure with names of another.
In this context changing the elements used for representing the structure does not change the structures, exactly like changing between decimal or hexadecimal basis does not change the structure of natural numbers. So isomorphic structures can and should be considered the same.
Also, when we fix the isomorphism we can identify elements of the two structures that are related by the isomorphism (but we have to keep track of the isomorphism considered, since different isomorphisms identify different pairs of elements).
With this little premise we can provide two reasons for why isomorphic objects should be considered the same in CT.
Philosophical reason.
One why to think about categories is to consider the objects as some abstract structures (like elements of a set can be thought as abstract points) and the morphisms as some abstract ways of relating them.
If we follow this point of view, and think category theory as a theory of abstract structures, then it should be natural to consider equal isomorphic objects, by abstaction of the structures' paradigm "isomorphic structures are the same".
Technical reason.
In this part we will make the previous argument formal but in order to do so we need a little premise on presheaves, also I will show an application of Yoneda Lemma which I consider particularly enlightening, so hold on and follow me.
A presheaf, on a category $mathbf C$, is a functor $P colon mathbf C^text{op} to mathbf{Set}$. You can think a presheaf as a multi-sorted algebra whose carriers are the sets $(P(c))_{c in mathbf C}$ and operations are the $(P(sigma))_{x,y in mathbf C,sigma in mathbf C[x,y]}$. Natural transformations are exactly homomorphisms for these algebras.
Now the yoneda embedding
$$ y colon mathbf C to [mathbf C^text{op},mathbf {Set}]$$
$$y(c) = mathbf C[-,c]$$
provides an isomorphism between the category $mathbf C$ (which as an algebraic structure) with a category of presheaves, namely the category of the presheaves of the representable presheaves.
From the discourse on isomorphisms of structures above we should think these two categories as being the same and we could identify (via the yoneda embedding) every object $c$ with the algebraic structure $y(c)=mathbf C[-,c]$.
We now are ready to our final claim.
Let $c_1$ and $c_2$ be two isomorphic objects of $mathbf C$. Clearly the algebras $mathbf C[-,c_1]$ and $C[-,c_2]$ must be isomorphic as well.
Now putting together what we said in the beginning,
we can identify the $c_i$'s with their algebras, the $mathbf C[-,c_i]$'s.
But since the $mathbf C[-,c_i]$'s are isomorphic structures they should be regarded as the same, and so by a transitivity argument it should be natural to consider also $c_1$ and $c_2$ to be the same.
Hopefully this more practical argument will help you strenghten the belief that isomorphic objects can and should be considered the same.
I apologize for being so long, but I do not think I could have shorten this answer.
answered Jan 31 at 10:55
Giorgio MossaGiorgio Mossa
14.3k11749
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Nice way of viewing the Yoneda lemma, though of course for large categories it becomes a bit more cumbersome to justify this view (but just move to a higher universe and this problem disappears)
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– Max
Jan 31 at 14:31
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Let universes' axiom be praised!
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– Giorgio Mossa
Jan 31 at 18:39
add a comment |
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In a category we can only talk about properties in terms of morphisms and objects and their relations (diagrams). Any object that can be talked about or defined in those terms can be replaced by an isomorphic object without changing the truth of statements. So within the framework of category theory, these objects can be interchanged everywhere they are used in statements. This is what identity "means" within a logical framework.
answered Jan 29 at 6:58
Henno BrandsmaHenno Brandsma
115k348124
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But it's not obvious to me that isomorphism induces this property that we can replace the two objects without changing the truth of statements. For example, consider a category consisting of $A,B,X$, where $A,B$ have an isomorphism, and there is a morphism from $X$ to $A$, and no other morphisms. Then the statement "there is a morphism from $X$ to $A$" is true, but if we replace $A$ with $B$, then it is no longer true.
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– user56834
Jan 29 at 7:06
1
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@user56834 what about the composition between the isomorphism from $B$ to $A$ and that morphism from $A$ to $X$? You specified an incomplete category.
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– Henno Brandsma
Jan 29 at 7:09
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of course... I see. This removes my counterargument to your claim. However, it's still not obvious to me that your claim holds. Is there a proof that "all categorical propositions conditioned on object $A$ are true iff they are true when $A$ is replaced with an object isomorphic to $A$"?
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– user56834
Jan 29 at 7:12
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@user56834 its a sort of metatheorem. To formally show it you have to define what a category theoretical statement is and show it. It can be shown I think. It gives some motivation/ intuition. Just like when we identify homeomorphic topological spaces.
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– Henno Brandsma
Jan 29 at 7:21
1
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@HennoBrandsma I thinks that's not so easy. One have to specifies exactly what are the categorical statements. As an example consider the, apparently innocuos, statement $text{src}(f)=X$, that states that $f$ is a morphism with source $X$. This statement cannot be categorical, if we replace $X$ with an isomorphic object, let say $Y$, the statement $text{src}(f)=Y$ is not true, clearly.
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– Giorgio Mossa
Jan 29 at 13:16
|
show 2 more comments
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There are some really good answers here.
Here is another one which try to approach the question from another perspective.
Two categories are considered the same if they are related by categorical equivalences. These are basically couples of functors which are inverse to each other up to natural-isomorphisms.
One of the key property is that every category $mathbf C$ is equivalent to its skeleton,
that is basically a full-subcategory containing a unique object for each isomorphism-class of objects of $mathbf C$.
This implies that when we are studying a category we could pass to one of its skeleton where isomorphic objects are the same. So it does make sense to consider two isomorphic objects as the same, because when seen through the equivalence they are really the same.
One could wonder on why we want to use the concept of equivalence of categories in the first place. But that is a long story for another time.
Edit: and that time is now.
Basically the reason why equivalence is correct notion of sameness for categories is due to the following fact.
Categorical equivalences are functors that preserve the truth of logical formulas in the categories, seen as models of the theory of categories.
This means that every closed statement that holds in a category does hold in every equivalent category.
I hope this helps.
answered Jan 29 at 13:34
Giorgio MossaGiorgio Mossa
14.3k11749
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I think it'd be hard to convince someone that equivalent categories can be treated as if they were identical, if they are not already convinced that isomorphic objects can be treated as equal... After all you need isomorphisms to define equivalence in the first place!
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– Arnaud D.
Jan 29 at 17:02
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@ArnaudD.Ar well yes and no. The point is that you need the notion of natural isomorphism for defining equivalences but you do not need to believe that isomorphic objects are the same. The small addition I have made I hope can provide the reason why equivalences are the good notion of morphisms that preserve the structure.
$endgroup$
– Giorgio Mossa
Jan 30 at 13:06
add a comment |
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4 Answers
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4 Answers
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$begingroup$
A first point is that this process of identifying isomorphic objects comes with practise : with practise, you'll start noticing that any "categorical property" an object has is shared by every object that is isomorphic to it; and you'll start being able to see why it makes sense to identify isomorphic objects.
So the best way to understand what's happening is to take a lot of examples and work through them to see why it makes sense. Now if you want a general argument, here's sort of one :
Suppose $f:Ato B$ is an isomorphism in a category $C$.
Suppose now you have a complicated situation involving $A$, and many other morphisms $f_i :Ato C_i, iin I$, $g_j: B_jto A$, satisfying certain equations (some involving $A$, others not involving it). It's the stuff that's interesting to you about $A$.
Well now you can transport this situation to the same situation about $B$ : put $f'_i = f_icirc f^{-1} : Bto C_i$, $g'_j := fcirc g_j : B_jto B$. This situation will satisfy the same equations as the one before : those not involving $A$ will be the same by replacing every $f_i, g_j$ by $f'_i, g'_j$; those involving $A$ will be the same but replacing every occurrence of $A$ by $B$ and conjugating by $f,f^{-1}$ to make things consistent. That is because $ff^{-1} = id_B, f^{-1}f =id_A$ and identities are neutral.
What this means is that if you have an interesting situation about an object $A$ in a category $C$, this situation being all about $C$-properties, and an isomorphism $f:Ato B$, then by conjugating by $f$ or $f^{-1}$ whenever it makes sense will give the same interesting situation about the object $B$.
In other words, along $f$, you cannot distinguish between $A$ and $B$ by properties of $C$: if you cannot distinguish between them, if they have the same properties, they're "essentially the same" - that's what we mean when we say we can identify isomorphic objects.
Note, however, that the identification, the process of saying they are the same involves not just saying that $A$ and $B$ are isomorphic, but picking a specific isomorphism. It's very important to note that we can identify to isomorphic objects so long as we have a specific isomorphism in mind.
For instance, for the categorical product, there is a unique isomorphism that commutes with the projections: it's the one we're considering.
answered Jan 29 at 11:05
MaxMax
15.8k11143
$endgroup$
2
$begingroup$
I would like to be able to give you another +1 just for emphasizing that one have to keep track of the isomorphisms.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:18
add a comment |
$begingroup$
A first point is that this process of identifying isomorphic objects comes with practise : with practise, you'll start noticing that any "categorical property" an object has is shared by every object that is isomorphic to it; and you'll start being able to see why it makes sense to identify isomorphic objects.
So the best way to understand what's happening is to take a lot of examples and work through them to see why it makes sense. Now if you want a general argument, here's sort of one :
Suppose $f:Ato B$ is an isomorphism in a category $C$.
Suppose now you have a complicated situation involving $A$, and many other morphisms $f_i :Ato C_i, iin I$, $g_j: B_jto A$, satisfying certain equations (some involving $A$, others not involving it). It's the stuff that's interesting to you about $A$.
Well now you can transport this situation to the same situation about $B$ : put $f'_i = f_icirc f^{-1} : Bto C_i$, $g'_j := fcirc g_j : B_jto B$. This situation will satisfy the same equations as the one before : those not involving $A$ will be the same by replacing every $f_i, g_j$ by $f'_i, g'_j$; those involving $A$ will be the same but replacing every occurrence of $A$ by $B$ and conjugating by $f,f^{-1}$ to make things consistent. That is because $ff^{-1} = id_B, f^{-1}f =id_A$ and identities are neutral.
What this means is that if you have an interesting situation about an object $A$ in a category $C$, this situation being all about $C$-properties, and an isomorphism $f:Ato B$, then by conjugating by $f$ or $f^{-1}$ whenever it makes sense will give the same interesting situation about the object $B$.
In other words, along $f$, you cannot distinguish between $A$ and $B$ by properties of $C$: if you cannot distinguish between them, if they have the same properties, they're "essentially the same" - that's what we mean when we say we can identify isomorphic objects.
Note, however, that the identification, the process of saying they are the same involves not just saying that $A$ and $B$ are isomorphic, but picking a specific isomorphism. It's very important to note that we can identify to isomorphic objects so long as we have a specific isomorphism in mind.
For instance, for the categorical product, there is a unique isomorphism that commutes with the projections: it's the one we're considering.
answered Jan 29 at 11:05
MaxMax
15.8k11143
$endgroup$
2
$begingroup$
I would like to be able to give you another +1 just for emphasizing that one have to keep track of the isomorphisms.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:18
add a comment |
$begingroup$
A first point is that this process of identifying isomorphic objects comes with practise : with practise, you'll start noticing that any "categorical property" an object has is shared by every object that is isomorphic to it; and you'll start being able to see why it makes sense to identify isomorphic objects.
So the best way to understand what's happening is to take a lot of examples and work through them to see why it makes sense. Now if you want a general argument, here's sort of one :
Suppose $f:Ato B$ is an isomorphism in a category $C$.
Suppose now you have a complicated situation involving $A$, and many other morphisms $f_i :Ato C_i, iin I$, $g_j: B_jto A$, satisfying certain equations (some involving $A$, others not involving it). It's the stuff that's interesting to you about $A$.
Well now you can transport this situation to the same situation about $B$ : put $f'_i = f_icirc f^{-1} : Bto C_i$, $g'_j := fcirc g_j : B_jto B$. This situation will satisfy the same equations as the one before : those not involving $A$ will be the same by replacing every $f_i, g_j$ by $f'_i, g'_j$; those involving $A$ will be the same but replacing every occurrence of $A$ by $B$ and conjugating by $f,f^{-1}$ to make things consistent. That is because $ff^{-1} = id_B, f^{-1}f =id_A$ and identities are neutral.
What this means is that if you have an interesting situation about an object $A$ in a category $C$, this situation being all about $C$-properties, and an isomorphism $f:Ato B$, then by conjugating by $f$ or $f^{-1}$ whenever it makes sense will give the same interesting situation about the object $B$.
In other words, along $f$, you cannot distinguish between $A$ and $B$ by properties of $C$: if you cannot distinguish between them, if they have the same properties, they're "essentially the same" - that's what we mean when we say we can identify isomorphic objects.
Note, however, that the identification, the process of saying they are the same involves not just saying that $A$ and $B$ are isomorphic, but picking a specific isomorphism. It's very important to note that we can identify to isomorphic objects so long as we have a specific isomorphism in mind.
For instance, for the categorical product, there is a unique isomorphism that commutes with the projections: it's the one we're considering.
answered Jan 29 at 11:05
MaxMax
15.8k11143
$endgroup$
A first point is that this process of identifying isomorphic objects comes with practise : with practise, you'll start noticing that any "categorical property" an object has is shared by every object that is isomorphic to it; and you'll start being able to see why it makes sense to identify isomorphic objects.
So the best way to understand what's happening is to take a lot of examples and work through them to see why it makes sense. Now if you want a general argument, here's sort of one :
Suppose $f:Ato B$ is an isomorphism in a category $C$.
Suppose now you have a complicated situation involving $A$, and many other morphisms $f_i :Ato C_i, iin I$, $g_j: B_jto A$, satisfying certain equations (some involving $A$, others not involving it). It's the stuff that's interesting to you about $A$.
Well now you can transport this situation to the same situation about $B$ : put $f'_i = f_icirc f^{-1} : Bto C_i$, $g'_j := fcirc g_j : B_jto B$. This situation will satisfy the same equations as the one before : those not involving $A$ will be the same by replacing every $f_i, g_j$ by $f'_i, g'_j$; those involving $A$ will be the same but replacing every occurrence of $A$ by $B$ and conjugating by $f,f^{-1}$ to make things consistent. That is because $ff^{-1} = id_B, f^{-1}f =id_A$ and identities are neutral.
What this means is that if you have an interesting situation about an object $A$ in a category $C$, this situation being all about $C$-properties, and an isomorphism $f:Ato B$, then by conjugating by $f$ or $f^{-1}$ whenever it makes sense will give the same interesting situation about the object $B$.
In other words, along $f$, you cannot distinguish between $A$ and $B$ by properties of $C$: if you cannot distinguish between them, if they have the same properties, they're "essentially the same" - that's what we mean when we say we can identify isomorphic objects.
Note, however, that the identification, the process of saying they are the same involves not just saying that $A$ and $B$ are isomorphic, but picking a specific isomorphism. It's very important to note that we can identify to isomorphic objects so long as we have a specific isomorphism in mind.
For instance, for the categorical product, there is a unique isomorphism that commutes with the projections: it's the one we're considering.
answered Jan 29 at 11:05
MaxMax
15.8k11143
answered Jan 29 at 11:05
MaxMax
15.8k11143
answered Jan 29 at 11:05
MaxMax
15.8k11143
answered Jan 29 at 11:05
MaxMax
15.8k11143
15.8k11143
2
$begingroup$
I would like to be able to give you another +1 just for emphasizing that one have to keep track of the isomorphisms.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:18
add a comment |
2
$begingroup$
I would like to be able to give you another +1 just for emphasizing that one have to keep track of the isomorphisms.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:18
2
2
$begingroup$
I would like to be able to give you another +1 just for emphasizing that one have to keep track of the isomorphisms.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:18
$begingroup$
I would like to be able to give you another +1 just for emphasizing that one have to keep track of the isomorphisms.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:18
add a comment |
$begingroup$
The following answer provide another, hopefully convincing, argument on why in category theory one can indentify isomorphic objects.
Categories of (multi-sorted) structures.
When dealing with categories of structures isomorphisms are bijective mapping preserving the structure. Basically this amounts saying that isomorphisms are ways to parametrize/rename the elements of one structure with names of another.
In this context changing the elements used for representing the structure does not change the structures, exactly like changing between decimal or hexadecimal basis does not change the structure of natural numbers. So isomorphic structures can and should be considered the same.
Also, when we fix the isomorphism we can identify elements of the two structures that are related by the isomorphism (but we have to keep track of the isomorphism considered, since different isomorphisms identify different pairs of elements).
With this little premise we can provide two reasons for why isomorphic objects should be considered the same in CT.
Philosophical reason.
One why to think about categories is to consider the objects as some abstract structures (like elements of a set can be thought as abstract points) and the morphisms as some abstract ways of relating them.
If we follow this point of view, and think category theory as a theory of abstract structures, then it should be natural to consider equal isomorphic objects, by abstaction of the structures' paradigm "isomorphic structures are the same".
Technical reason.
In this part we will make the previous argument formal but in order to do so we need a little premise on presheaves, also I will show an application of Yoneda Lemma which I consider particularly enlightening, so hold on and follow me.
A presheaf, on a category $mathbf C$, is a functor $P colon mathbf C^text{op} to mathbf{Set}$. You can think a presheaf as a multi-sorted algebra whose carriers are the sets $(P(c))_{c in mathbf C}$ and operations are the $(P(sigma))_{x,y in mathbf C,sigma in mathbf C[x,y]}$. Natural transformations are exactly homomorphisms for these algebras.
Now the yoneda embedding
$$ y colon mathbf C to [mathbf C^text{op},mathbf {Set}]$$
$$y(c) = mathbf C[-,c]$$
provides an isomorphism between the category $mathbf C$ (which as an algebraic structure) with a category of presheaves, namely the category of the presheaves of the representable presheaves.
From the discourse on isomorphisms of structures above we should think these two categories as being the same and we could identify (via the yoneda embedding) every object $c$ with the algebraic structure $y(c)=mathbf C[-,c]$.
We now are ready to our final claim.
Let $c_1$ and $c_2$ be two isomorphic objects of $mathbf C$. Clearly the algebras $mathbf C[-,c_1]$ and $C[-,c_2]$ must be isomorphic as well.
Now putting together what we said in the beginning,
we can identify the $c_i$'s with their algebras, the $mathbf C[-,c_i]$'s.
But since the $mathbf C[-,c_i]$'s are isomorphic structures they should be regarded as the same, and so by a transitivity argument it should be natural to consider also $c_1$ and $c_2$ to be the same.
Hopefully this more practical argument will help you strenghten the belief that isomorphic objects can and should be considered the same.
I apologize for being so long, but I do not think I could have shorten this answer.
answered Jan 31 at 10:55
Giorgio MossaGiorgio Mossa
14.3k11749
$endgroup$
$begingroup$
Nice way of viewing the Yoneda lemma, though of course for large categories it becomes a bit more cumbersome to justify this view (but just move to a higher universe and this problem disappears)
$endgroup$
– Max
Jan 31 at 14:31
$begingroup$
Let universes' axiom be praised!
$endgroup$
– Giorgio Mossa
Jan 31 at 18:39
add a comment |
$begingroup$
The following answer provide another, hopefully convincing, argument on why in category theory one can indentify isomorphic objects.
Categories of (multi-sorted) structures.
When dealing with categories of structures isomorphisms are bijective mapping preserving the structure. Basically this amounts saying that isomorphisms are ways to parametrize/rename the elements of one structure with names of another.
In this context changing the elements used for representing the structure does not change the structures, exactly like changing between decimal or hexadecimal basis does not change the structure of natural numbers. So isomorphic structures can and should be considered the same.
Also, when we fix the isomorphism we can identify elements of the two structures that are related by the isomorphism (but we have to keep track of the isomorphism considered, since different isomorphisms identify different pairs of elements).
With this little premise we can provide two reasons for why isomorphic objects should be considered the same in CT.
Philosophical reason.
One why to think about categories is to consider the objects as some abstract structures (like elements of a set can be thought as abstract points) and the morphisms as some abstract ways of relating them.
If we follow this point of view, and think category theory as a theory of abstract structures, then it should be natural to consider equal isomorphic objects, by abstaction of the structures' paradigm "isomorphic structures are the same".
Technical reason.
In this part we will make the previous argument formal but in order to do so we need a little premise on presheaves, also I will show an application of Yoneda Lemma which I consider particularly enlightening, so hold on and follow me.
A presheaf, on a category $mathbf C$, is a functor $P colon mathbf C^text{op} to mathbf{Set}$. You can think a presheaf as a multi-sorted algebra whose carriers are the sets $(P(c))_{c in mathbf C}$ and operations are the $(P(sigma))_{x,y in mathbf C,sigma in mathbf C[x,y]}$. Natural transformations are exactly homomorphisms for these algebras.
Now the yoneda embedding
$$ y colon mathbf C to [mathbf C^text{op},mathbf {Set}]$$
$$y(c) = mathbf C[-,c]$$
provides an isomorphism between the category $mathbf C$ (which as an algebraic structure) with a category of presheaves, namely the category of the presheaves of the representable presheaves.
From the discourse on isomorphisms of structures above we should think these two categories as being the same and we could identify (via the yoneda embedding) every object $c$ with the algebraic structure $y(c)=mathbf C[-,c]$.
We now are ready to our final claim.
Let $c_1$ and $c_2$ be two isomorphic objects of $mathbf C$. Clearly the algebras $mathbf C[-,c_1]$ and $C[-,c_2]$ must be isomorphic as well.
Now putting together what we said in the beginning,
we can identify the $c_i$'s with their algebras, the $mathbf C[-,c_i]$'s.
But since the $mathbf C[-,c_i]$'s are isomorphic structures they should be regarded as the same, and so by a transitivity argument it should be natural to consider also $c_1$ and $c_2$ to be the same.
Hopefully this more practical argument will help you strenghten the belief that isomorphic objects can and should be considered the same.
I apologize for being so long, but I do not think I could have shorten this answer.
answered Jan 31 at 10:55
Giorgio MossaGiorgio Mossa
14.3k11749
$endgroup$
$begingroup$
Nice way of viewing the Yoneda lemma, though of course for large categories it becomes a bit more cumbersome to justify this view (but just move to a higher universe and this problem disappears)
$endgroup$
– Max
Jan 31 at 14:31
$begingroup$
Let universes' axiom be praised!
$endgroup$
– Giorgio Mossa
Jan 31 at 18:39
add a comment |
$begingroup$
The following answer provide another, hopefully convincing, argument on why in category theory one can indentify isomorphic objects.
Categories of (multi-sorted) structures.
When dealing with categories of structures isomorphisms are bijective mapping preserving the structure. Basically this amounts saying that isomorphisms are ways to parametrize/rename the elements of one structure with names of another.
In this context changing the elements used for representing the structure does not change the structures, exactly like changing between decimal or hexadecimal basis does not change the structure of natural numbers. So isomorphic structures can and should be considered the same.
Also, when we fix the isomorphism we can identify elements of the two structures that are related by the isomorphism (but we have to keep track of the isomorphism considered, since different isomorphisms identify different pairs of elements).
With this little premise we can provide two reasons for why isomorphic objects should be considered the same in CT.
Philosophical reason.
One why to think about categories is to consider the objects as some abstract structures (like elements of a set can be thought as abstract points) and the morphisms as some abstract ways of relating them.
If we follow this point of view, and think category theory as a theory of abstract structures, then it should be natural to consider equal isomorphic objects, by abstaction of the structures' paradigm "isomorphic structures are the same".
Technical reason.
In this part we will make the previous argument formal but in order to do so we need a little premise on presheaves, also I will show an application of Yoneda Lemma which I consider particularly enlightening, so hold on and follow me.
A presheaf, on a category $mathbf C$, is a functor $P colon mathbf C^text{op} to mathbf{Set}$. You can think a presheaf as a multi-sorted algebra whose carriers are the sets $(P(c))_{c in mathbf C}$ and operations are the $(P(sigma))_{x,y in mathbf C,sigma in mathbf C[x,y]}$. Natural transformations are exactly homomorphisms for these algebras.
Now the yoneda embedding
$$ y colon mathbf C to [mathbf C^text{op},mathbf {Set}]$$
$$y(c) = mathbf C[-,c]$$
provides an isomorphism between the category $mathbf C$ (which as an algebraic structure) with a category of presheaves, namely the category of the presheaves of the representable presheaves.
From the discourse on isomorphisms of structures above we should think these two categories as being the same and we could identify (via the yoneda embedding) every object $c$ with the algebraic structure $y(c)=mathbf C[-,c]$.
We now are ready to our final claim.
Let $c_1$ and $c_2$ be two isomorphic objects of $mathbf C$. Clearly the algebras $mathbf C[-,c_1]$ and $C[-,c_2]$ must be isomorphic as well.
Now putting together what we said in the beginning,
we can identify the $c_i$'s with their algebras, the $mathbf C[-,c_i]$'s.
But since the $mathbf C[-,c_i]$'s are isomorphic structures they should be regarded as the same, and so by a transitivity argument it should be natural to consider also $c_1$ and $c_2$ to be the same.
Hopefully this more practical argument will help you strenghten the belief that isomorphic objects can and should be considered the same.
I apologize for being so long, but I do not think I could have shorten this answer.
answered Jan 31 at 10:55
Giorgio MossaGiorgio Mossa
14.3k11749
$endgroup$
The following answer provide another, hopefully convincing, argument on why in category theory one can indentify isomorphic objects.
Categories of (multi-sorted) structures.
When dealing with categories of structures isomorphisms are bijective mapping preserving the structure. Basically this amounts saying that isomorphisms are ways to parametrize/rename the elements of one structure with names of another.
In this context changing the elements used for representing the structure does not change the structures, exactly like changing between decimal or hexadecimal basis does not change the structure of natural numbers. So isomorphic structures can and should be considered the same.
Also, when we fix the isomorphism we can identify elements of the two structures that are related by the isomorphism (but we have to keep track of the isomorphism considered, since different isomorphisms identify different pairs of elements).
With this little premise we can provide two reasons for why isomorphic objects should be considered the same in CT.
Philosophical reason.
One why to think about categories is to consider the objects as some abstract structures (like elements of a set can be thought as abstract points) and the morphisms as some abstract ways of relating them.
If we follow this point of view, and think category theory as a theory of abstract structures, then it should be natural to consider equal isomorphic objects, by abstaction of the structures' paradigm "isomorphic structures are the same".
Technical reason.
In this part we will make the previous argument formal but in order to do so we need a little premise on presheaves, also I will show an application of Yoneda Lemma which I consider particularly enlightening, so hold on and follow me.
A presheaf, on a category $mathbf C$, is a functor $P colon mathbf C^text{op} to mathbf{Set}$. You can think a presheaf as a multi-sorted algebra whose carriers are the sets $(P(c))_{c in mathbf C}$ and operations are the $(P(sigma))_{x,y in mathbf C,sigma in mathbf C[x,y]}$. Natural transformations are exactly homomorphisms for these algebras.
Now the yoneda embedding
$$ y colon mathbf C to [mathbf C^text{op},mathbf {Set}]$$
$$y(c) = mathbf C[-,c]$$
provides an isomorphism between the category $mathbf C$ (which as an algebraic structure) with a category of presheaves, namely the category of the presheaves of the representable presheaves.
From the discourse on isomorphisms of structures above we should think these two categories as being the same and we could identify (via the yoneda embedding) every object $c$ with the algebraic structure $y(c)=mathbf C[-,c]$.
We now are ready to our final claim.
Let $c_1$ and $c_2$ be two isomorphic objects of $mathbf C$. Clearly the algebras $mathbf C[-,c_1]$ and $C[-,c_2]$ must be isomorphic as well.
Now putting together what we said in the beginning,
we can identify the $c_i$'s with their algebras, the $mathbf C[-,c_i]$'s.
But since the $mathbf C[-,c_i]$'s are isomorphic structures they should be regarded as the same, and so by a transitivity argument it should be natural to consider also $c_1$ and $c_2$ to be the same.
Hopefully this more practical argument will help you strenghten the belief that isomorphic objects can and should be considered the same.
I apologize for being so long, but I do not think I could have shorten this answer.
answered Jan 31 at 10:55
Giorgio MossaGiorgio Mossa
14.3k11749
answered Jan 31 at 10:55
Giorgio MossaGiorgio Mossa
14.3k11749
answered Jan 31 at 10:55
Giorgio MossaGiorgio Mossa
14.3k11749
answered Jan 31 at 10:55
Giorgio MossaGiorgio Mossa
14.3k11749
14.3k11749
$begingroup$
Nice way of viewing the Yoneda lemma, though of course for large categories it becomes a bit more cumbersome to justify this view (but just move to a higher universe and this problem disappears)
$endgroup$
– Max
Jan 31 at 14:31
$begingroup$
Let universes' axiom be praised!
$endgroup$
– Giorgio Mossa
Jan 31 at 18:39
add a comment |
$begingroup$
Nice way of viewing the Yoneda lemma, though of course for large categories it becomes a bit more cumbersome to justify this view (but just move to a higher universe and this problem disappears)
$endgroup$
– Max
Jan 31 at 14:31
$begingroup$
Let universes' axiom be praised!
$endgroup$
– Giorgio Mossa
Jan 31 at 18:39
$begingroup$
Nice way of viewing the Yoneda lemma, though of course for large categories it becomes a bit more cumbersome to justify this view (but just move to a higher universe and this problem disappears)
$endgroup$
– Max
Jan 31 at 14:31
$begingroup$
Nice way of viewing the Yoneda lemma, though of course for large categories it becomes a bit more cumbersome to justify this view (but just move to a higher universe and this problem disappears)
$endgroup$
– Max
Jan 31 at 14:31
$begingroup$
Let universes' axiom be praised!
$endgroup$
– Giorgio Mossa
Jan 31 at 18:39
$begingroup$
Let universes' axiom be praised!
$endgroup$
– Giorgio Mossa
Jan 31 at 18:39
add a comment |
$begingroup$
In a category we can only talk about properties in terms of morphisms and objects and their relations (diagrams). Any object that can be talked about or defined in those terms can be replaced by an isomorphic object without changing the truth of statements. So within the framework of category theory, these objects can be interchanged everywhere they are used in statements. This is what identity "means" within a logical framework.
answered Jan 29 at 6:58
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
$begingroup$
But it's not obvious to me that isomorphism induces this property that we can replace the two objects without changing the truth of statements. For example, consider a category consisting of $A,B,X$, where $A,B$ have an isomorphism, and there is a morphism from $X$ to $A$, and no other morphisms. Then the statement "there is a morphism from $X$ to $A$" is true, but if we replace $A$ with $B$, then it is no longer true.
$endgroup$
– user56834
Jan 29 at 7:06
1
$begingroup$
@user56834 what about the composition between the isomorphism from $B$ to $A$ and that morphism from $A$ to $X$? You specified an incomplete category.
$endgroup$
– Henno Brandsma
Jan 29 at 7:09
$begingroup$
of course... I see. This removes my counterargument to your claim. However, it's still not obvious to me that your claim holds. Is there a proof that "all categorical propositions conditioned on object $A$ are true iff they are true when $A$ is replaced with an object isomorphic to $A$"?
$endgroup$
– user56834
Jan 29 at 7:12
$begingroup$
@user56834 its a sort of metatheorem. To formally show it you have to define what a category theoretical statement is and show it. It can be shown I think. It gives some motivation/ intuition. Just like when we identify homeomorphic topological spaces.
$endgroup$
– Henno Brandsma
Jan 29 at 7:21
1
$begingroup$
@HennoBrandsma I thinks that's not so easy. One have to specifies exactly what are the categorical statements. As an example consider the, apparently innocuos, statement $text{src}(f)=X$, that states that $f$ is a morphism with source $X$. This statement cannot be categorical, if we replace $X$ with an isomorphic object, let say $Y$, the statement $text{src}(f)=Y$ is not true, clearly.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:16
|
show 2 more comments
$begingroup$
In a category we can only talk about properties in terms of morphisms and objects and their relations (diagrams). Any object that can be talked about or defined in those terms can be replaced by an isomorphic object without changing the truth of statements. So within the framework of category theory, these objects can be interchanged everywhere they are used in statements. This is what identity "means" within a logical framework.
answered Jan 29 at 6:58
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
$begingroup$
But it's not obvious to me that isomorphism induces this property that we can replace the two objects without changing the truth of statements. For example, consider a category consisting of $A,B,X$, where $A,B$ have an isomorphism, and there is a morphism from $X$ to $A$, and no other morphisms. Then the statement "there is a morphism from $X$ to $A$" is true, but if we replace $A$ with $B$, then it is no longer true.
$endgroup$
– user56834
Jan 29 at 7:06
1
$begingroup$
@user56834 what about the composition between the isomorphism from $B$ to $A$ and that morphism from $A$ to $X$? You specified an incomplete category.
$endgroup$
– Henno Brandsma
Jan 29 at 7:09
$begingroup$
of course... I see. This removes my counterargument to your claim. However, it's still not obvious to me that your claim holds. Is there a proof that "all categorical propositions conditioned on object $A$ are true iff they are true when $A$ is replaced with an object isomorphic to $A$"?
$endgroup$
– user56834
Jan 29 at 7:12
$begingroup$
@user56834 its a sort of metatheorem. To formally show it you have to define what a category theoretical statement is and show it. It can be shown I think. It gives some motivation/ intuition. Just like when we identify homeomorphic topological spaces.
$endgroup$
– Henno Brandsma
Jan 29 at 7:21
1
$begingroup$
@HennoBrandsma I thinks that's not so easy. One have to specifies exactly what are the categorical statements. As an example consider the, apparently innocuos, statement $text{src}(f)=X$, that states that $f$ is a morphism with source $X$. This statement cannot be categorical, if we replace $X$ with an isomorphic object, let say $Y$, the statement $text{src}(f)=Y$ is not true, clearly.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:16
|
show 2 more comments
$begingroup$
In a category we can only talk about properties in terms of morphisms and objects and their relations (diagrams). Any object that can be talked about or defined in those terms can be replaced by an isomorphic object without changing the truth of statements. So within the framework of category theory, these objects can be interchanged everywhere they are used in statements. This is what identity "means" within a logical framework.
answered Jan 29 at 6:58
Henno BrandsmaHenno Brandsma
115k348124
$endgroup$
In a category we can only talk about properties in terms of morphisms and objects and their relations (diagrams). Any object that can be talked about or defined in those terms can be replaced by an isomorphic object without changing the truth of statements. So within the framework of category theory, these objects can be interchanged everywhere they are used in statements. This is what identity "means" within a logical framework.
answered Jan 29 at 6:58
Henno BrandsmaHenno Brandsma
115k348124
answered Jan 29 at 6:58
Henno BrandsmaHenno Brandsma
115k348124
answered Jan 29 at 6:58
Henno BrandsmaHenno Brandsma
115k348124
answered Jan 29 at 6:58
Henno BrandsmaHenno Brandsma
115k348124
115k348124
$begingroup$
But it's not obvious to me that isomorphism induces this property that we can replace the two objects without changing the truth of statements. For example, consider a category consisting of $A,B,X$, where $A,B$ have an isomorphism, and there is a morphism from $X$ to $A$, and no other morphisms. Then the statement "there is a morphism from $X$ to $A$" is true, but if we replace $A$ with $B$, then it is no longer true.
$endgroup$
– user56834
Jan 29 at 7:06
1
$begingroup$
@user56834 what about the composition between the isomorphism from $B$ to $A$ and that morphism from $A$ to $X$? You specified an incomplete category.
$endgroup$
– Henno Brandsma
Jan 29 at 7:09
$begingroup$
of course... I see. This removes my counterargument to your claim. However, it's still not obvious to me that your claim holds. Is there a proof that "all categorical propositions conditioned on object $A$ are true iff they are true when $A$ is replaced with an object isomorphic to $A$"?
$endgroup$
– user56834
Jan 29 at 7:12
$begingroup$
@user56834 its a sort of metatheorem. To formally show it you have to define what a category theoretical statement is and show it. It can be shown I think. It gives some motivation/ intuition. Just like when we identify homeomorphic topological spaces.
$endgroup$
– Henno Brandsma
Jan 29 at 7:21
1
$begingroup$
@HennoBrandsma I thinks that's not so easy. One have to specifies exactly what are the categorical statements. As an example consider the, apparently innocuos, statement $text{src}(f)=X$, that states that $f$ is a morphism with source $X$. This statement cannot be categorical, if we replace $X$ with an isomorphic object, let say $Y$, the statement $text{src}(f)=Y$ is not true, clearly.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:16
|
show 2 more comments
$begingroup$
But it's not obvious to me that isomorphism induces this property that we can replace the two objects without changing the truth of statements. For example, consider a category consisting of $A,B,X$, where $A,B$ have an isomorphism, and there is a morphism from $X$ to $A$, and no other morphisms. Then the statement "there is a morphism from $X$ to $A$" is true, but if we replace $A$ with $B$, then it is no longer true.
$endgroup$
– user56834
Jan 29 at 7:06
1
$begingroup$
@user56834 what about the composition between the isomorphism from $B$ to $A$ and that morphism from $A$ to $X$? You specified an incomplete category.
$endgroup$
– Henno Brandsma
Jan 29 at 7:09
$begingroup$
of course... I see. This removes my counterargument to your claim. However, it's still not obvious to me that your claim holds. Is there a proof that "all categorical propositions conditioned on object $A$ are true iff they are true when $A$ is replaced with an object isomorphic to $A$"?
$endgroup$
– user56834
Jan 29 at 7:12
$begingroup$
@user56834 its a sort of metatheorem. To formally show it you have to define what a category theoretical statement is and show it. It can be shown I think. It gives some motivation/ intuition. Just like when we identify homeomorphic topological spaces.
$endgroup$
– Henno Brandsma
Jan 29 at 7:21
1
$begingroup$
@HennoBrandsma I thinks that's not so easy. One have to specifies exactly what are the categorical statements. As an example consider the, apparently innocuos, statement $text{src}(f)=X$, that states that $f$ is a morphism with source $X$. This statement cannot be categorical, if we replace $X$ with an isomorphic object, let say $Y$, the statement $text{src}(f)=Y$ is not true, clearly.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:16
$begingroup$
But it's not obvious to me that isomorphism induces this property that we can replace the two objects without changing the truth of statements. For example, consider a category consisting of $A,B,X$, where $A,B$ have an isomorphism, and there is a morphism from $X$ to $A$, and no other morphisms. Then the statement "there is a morphism from $X$ to $A$" is true, but if we replace $A$ with $B$, then it is no longer true.
$endgroup$
– user56834
Jan 29 at 7:06
$begingroup$
But it's not obvious to me that isomorphism induces this property that we can replace the two objects without changing the truth of statements. For example, consider a category consisting of $A,B,X$, where $A,B$ have an isomorphism, and there is a morphism from $X$ to $A$, and no other morphisms. Then the statement "there is a morphism from $X$ to $A$" is true, but if we replace $A$ with $B$, then it is no longer true.
$endgroup$
– user56834
Jan 29 at 7:06
1
1
$begingroup$
@user56834 what about the composition between the isomorphism from $B$ to $A$ and that morphism from $A$ to $X$? You specified an incomplete category.
$endgroup$
– Henno Brandsma
Jan 29 at 7:09
$begingroup$
@user56834 what about the composition between the isomorphism from $B$ to $A$ and that morphism from $A$ to $X$? You specified an incomplete category.
$endgroup$
– Henno Brandsma
Jan 29 at 7:09
$begingroup$
of course... I see. This removes my counterargument to your claim. However, it's still not obvious to me that your claim holds. Is there a proof that "all categorical propositions conditioned on object $A$ are true iff they are true when $A$ is replaced with an object isomorphic to $A$"?
$endgroup$
– user56834
Jan 29 at 7:12
$begingroup$
of course... I see. This removes my counterargument to your claim. However, it's still not obvious to me that your claim holds. Is there a proof that "all categorical propositions conditioned on object $A$ are true iff they are true when $A$ is replaced with an object isomorphic to $A$"?
$endgroup$
– user56834
Jan 29 at 7:12
$begingroup$
@user56834 its a sort of metatheorem. To formally show it you have to define what a category theoretical statement is and show it. It can be shown I think. It gives some motivation/ intuition. Just like when we identify homeomorphic topological spaces.
$endgroup$
– Henno Brandsma
Jan 29 at 7:21
$begingroup$
@user56834 its a sort of metatheorem. To formally show it you have to define what a category theoretical statement is and show it. It can be shown I think. It gives some motivation/ intuition. Just like when we identify homeomorphic topological spaces.
$endgroup$
– Henno Brandsma
Jan 29 at 7:21
1
1
$begingroup$
@HennoBrandsma I thinks that's not so easy. One have to specifies exactly what are the categorical statements. As an example consider the, apparently innocuos, statement $text{src}(f)=X$, that states that $f$ is a morphism with source $X$. This statement cannot be categorical, if we replace $X$ with an isomorphic object, let say $Y$, the statement $text{src}(f)=Y$ is not true, clearly.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:16
$begingroup$
@HennoBrandsma I thinks that's not so easy. One have to specifies exactly what are the categorical statements. As an example consider the, apparently innocuos, statement $text{src}(f)=X$, that states that $f$ is a morphism with source $X$. This statement cannot be categorical, if we replace $X$ with an isomorphic object, let say $Y$, the statement $text{src}(f)=Y$ is not true, clearly.
$endgroup$
– Giorgio Mossa
Jan 29 at 13:16
|
show 2 more comments
$begingroup$
There are some really good answers here.
Here is another one which try to approach the question from another perspective.
Two categories are considered the same if they are related by categorical equivalences. These are basically couples of functors which are inverse to each other up to natural-isomorphisms.
One of the key property is that every category $mathbf C$ is equivalent to its skeleton,
that is basically a full-subcategory containing a unique object for each isomorphism-class of objects of $mathbf C$.
This implies that when we are studying a category we could pass to one of its skeleton where isomorphic objects are the same. So it does make sense to consider two isomorphic objects as the same, because when seen through the equivalence they are really the same.
One could wonder on why we want to use the concept of equivalence of categories in the first place. But that is a long story for another time.
Edit: and that time is now.
Basically the reason why equivalence is correct notion of sameness for categories is due to the following fact.
Categorical equivalences are functors that preserve the truth of logical formulas in the categories, seen as models of the theory of categories.
This means that every closed statement that holds in a category does hold in every equivalent category.
I hope this helps.
answered Jan 29 at 13:34
Giorgio MossaGiorgio Mossa
14.3k11749
$endgroup$
$begingroup$
I think it'd be hard to convince someone that equivalent categories can be treated as if they were identical, if they are not already convinced that isomorphic objects can be treated as equal... After all you need isomorphisms to define equivalence in the first place!
$endgroup$
– Arnaud D.
Jan 29 at 17:02
$begingroup$
@ArnaudD.Ar well yes and no. The point is that you need the notion of natural isomorphism for defining equivalences but you do not need to believe that isomorphic objects are the same. The small addition I have made I hope can provide the reason why equivalences are the good notion of morphisms that preserve the structure.
$endgroup$
– Giorgio Mossa
Jan 30 at 13:06
add a comment |
$begingroup$
There are some really good answers here.
Here is another one which try to approach the question from another perspective.
Two categories are considered the same if they are related by categorical equivalences. These are basically couples of functors which are inverse to each other up to natural-isomorphisms.
One of the key property is that every category $mathbf C$ is equivalent to its skeleton,
that is basically a full-subcategory containing a unique object for each isomorphism-class of objects of $mathbf C$.
This implies that when we are studying a category we could pass to one of its skeleton where isomorphic objects are the same. So it does make sense to consider two isomorphic objects as the same, because when seen through the equivalence they are really the same.
One could wonder on why we want to use the concept of equivalence of categories in the first place. But that is a long story for another time.
Edit: and that time is now.
Basically the reason why equivalence is correct notion of sameness for categories is due to the following fact.
Categorical equivalences are functors that preserve the truth of logical formulas in the categories, seen as models of the theory of categories.
This means that every closed statement that holds in a category does hold in every equivalent category.
I hope this helps.
answered Jan 29 at 13:34
Giorgio MossaGiorgio Mossa
14.3k11749
$endgroup$
$begingroup$
I think it'd be hard to convince someone that equivalent categories can be treated as if they were identical, if they are not already convinced that isomorphic objects can be treated as equal... After all you need isomorphisms to define equivalence in the first place!
$endgroup$
– Arnaud D.
Jan 29 at 17:02
$begingroup$
@ArnaudD.Ar well yes and no. The point is that you need the notion of natural isomorphism for defining equivalences but you do not need to believe that isomorphic objects are the same. The small addition I have made I hope can provide the reason why equivalences are the good notion of morphisms that preserve the structure.
$endgroup$
– Giorgio Mossa
Jan 30 at 13:06
add a comment |
$begingroup$
There are some really good answers here.
Here is another one which try to approach the question from another perspective.
Two categories are considered the same if they are related by categorical equivalences. These are basically couples of functors which are inverse to each other up to natural-isomorphisms.
One of the key property is that every category $mathbf C$ is equivalent to its skeleton,
that is basically a full-subcategory containing a unique object for each isomorphism-class of objects of $mathbf C$.
This implies that when we are studying a category we could pass to one of its skeleton where isomorphic objects are the same. So it does make sense to consider two isomorphic objects as the same, because when seen through the equivalence they are really the same.
One could wonder on why we want to use the concept of equivalence of categories in the first place. But that is a long story for another time.
Edit: and that time is now.
Basically the reason why equivalence is correct notion of sameness for categories is due to the following fact.
Categorical equivalences are functors that preserve the truth of logical formulas in the categories, seen as models of the theory of categories.
This means that every closed statement that holds in a category does hold in every equivalent category.
I hope this helps.
answered Jan 29 at 13:34
Giorgio MossaGiorgio Mossa
14.3k11749
$endgroup$
There are some really good answers here.
Here is another one which try to approach the question from another perspective.
Two categories are considered the same if they are related by categorical equivalences. These are basically couples of functors which are inverse to each other up to natural-isomorphisms.
One of the key property is that every category $mathbf C$ is equivalent to its skeleton,
that is basically a full-subcategory containing a unique object for each isomorphism-class of objects of $mathbf C$.
This implies that when we are studying a category we could pass to one of its skeleton where isomorphic objects are the same. So it does make sense to consider two isomorphic objects as the same, because when seen through the equivalence they are really the same.
One could wonder on why we want to use the concept of equivalence of categories in the first place. But that is a long story for another time.
Edit: and that time is now.
Basically the reason why equivalence is correct notion of sameness for categories is due to the following fact.
Categorical equivalences are functors that preserve the truth of logical formulas in the categories, seen as models of the theory of categories.
This means that every closed statement that holds in a category does hold in every equivalent category.
I hope this helps.
answered Jan 29 at 13:34
Giorgio MossaGiorgio Mossa
14.3k11749
edited Jan 30 at 13:04
answered Jan 29 at 13:34
Giorgio MossaGiorgio Mossa
14.3k11749
answered Jan 29 at 13:34
Giorgio MossaGiorgio Mossa
14.3k11749
answered Jan 29 at 13:34
Giorgio MossaGiorgio Mossa
14.3k11749
14.3k11749
$begingroup$
I think it'd be hard to convince someone that equivalent categories can be treated as if they were identical, if they are not already convinced that isomorphic objects can be treated as equal... After all you need isomorphisms to define equivalence in the first place!
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– Arnaud D.
Jan 29 at 17:02
$begingroup$
@ArnaudD.Ar well yes and no. The point is that you need the notion of natural isomorphism for defining equivalences but you do not need to believe that isomorphic objects are the same. The small addition I have made I hope can provide the reason why equivalences are the good notion of morphisms that preserve the structure.
$endgroup$
– Giorgio Mossa
Jan 30 at 13:06
add a comment |
$begingroup$
I think it'd be hard to convince someone that equivalent categories can be treated as if they were identical, if they are not already convinced that isomorphic objects can be treated as equal... After all you need isomorphisms to define equivalence in the first place!
$endgroup$
– Arnaud D.
Jan 29 at 17:02
$begingroup$
@ArnaudD.Ar well yes and no. The point is that you need the notion of natural isomorphism for defining equivalences but you do not need to believe that isomorphic objects are the same. The small addition I have made I hope can provide the reason why equivalences are the good notion of morphisms that preserve the structure.
$endgroup$
– Giorgio Mossa
Jan 30 at 13:06
$begingroup$
I think it'd be hard to convince someone that equivalent categories can be treated as if they were identical, if they are not already convinced that isomorphic objects can be treated as equal... After all you need isomorphisms to define equivalence in the first place!
$endgroup$
– Arnaud D.
Jan 29 at 17:02
$begingroup$
I think it'd be hard to convince someone that equivalent categories can be treated as if they were identical, if they are not already convinced that isomorphic objects can be treated as equal... After all you need isomorphisms to define equivalence in the first place!
$endgroup$
– Arnaud D.
Jan 29 at 17:02
$begingroup$
@ArnaudD.Ar well yes and no. The point is that you need the notion of natural isomorphism for defining equivalences but you do not need to believe that isomorphic objects are the same. The small addition I have made I hope can provide the reason why equivalences are the good notion of morphisms that preserve the structure.
$endgroup$
– Giorgio Mossa
Jan 30 at 13:06
$begingroup$
@ArnaudD.Ar well yes and no. The point is that you need the notion of natural isomorphism for defining equivalences but you do not need to believe that isomorphic objects are the same. The small addition I have made I hope can provide the reason why equivalences are the good notion of morphisms that preserve the structure.
$endgroup$
– Giorgio Mossa
Jan 30 at 13:06
add a comment |
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