Mixing
Two questions regarding the convention concerning concrete categories
$begingroup$
Definition of Concrete Categories
Let $mathbf{X}$ be a category. A concrete category over $mathbf{X}$ is a pair $(mathbf{A}, U)$, where $mathbf{A}$ is a category and $U : mathbf{A} to mathbf{X}$ is a faithful functor. Sometimes $U$ is called the
forgetful (or underlying) functor of the concrete category and $mathbf{X}$ is called the base category for $(mathbf{A}, U)$.
This definition above is given only for the sake of completeness and is copied verbatim from Definition 5.1(1) of The Joy of Cats.
While I was reading The Joy of Cats and in pp. 59-60, I found that shortly after giving the examples of concrete categories (after the above definition) it is written (see Remark 5.3 (1)),
Since faithful functors are injective on $text{hom}$-sets, we usually assume...that $text{hom}_{mathbf{A}}(A,B)$ is a subset of $text{hom}_{mathbf{X}}(UA,UB)$ for each pair $(A,B)$ of $mathbf{A}$-objects. This familiar convention allows one to express the property that
"for $mathbf{A}$-objects $A$ and $B$ and an $mathbf{X}$-morphism $f:UAto UB$ there exists a (necessarily unique) $mathbf{A}$-morphism $require{AMScd}begin{CD}Arightarrow Bend{CD}$ with, $U(A→B)=f:UAto UB$"
much more succinctly, by stating,
"$f:UAto UB$ is an $mathbf{A}$-morphism (from $A$ to $B$)"
However, I find this convention to be very much confusing. My questions are,
Why would we assume that $text{hom}_{mathbf{A}}(A,B)$ is a subset of $text{hom}_{mathbf{X}}(UA,UB)$ even when it may not be the case? I agree that a convention is a convention (and I have no problem with that) but I don't think that a convention should be such that it creates more confusion than lessening it. So, I think that there is something "categorical way" of thinking that I am missing. I am interested to know it. Hence this question.
After the sentence "$f:UAto UB$ is an $mathbf{A}$-morphism (from $A$ to $B$)" as written above the authors write in a footnote the following,
"Observe that analogues of this expression (i.e. $f:UA→UX$ is an $mathbf{A}$-morphism) are frequently used in “concrete situations”, e.g., in saying
that a certain function between vector spaces “is linear” or that a certain function between topological spaces “is continuous”, etc. "
However I don't understand how this remark relates to the convention. Can anyone explain that to me?
category-theory big-picture
asked Jan 23 at 14:06
user 170039user 170039
10.5k42466
$endgroup$
|
show 3 more comments
$begingroup$
Definition of Concrete Categories
Let $mathbf{X}$ be a category. A concrete category over $mathbf{X}$ is a pair $(mathbf{A}, U)$, where $mathbf{A}$ is a category and $U : mathbf{A} to mathbf{X}$ is a faithful functor. Sometimes $U$ is called the
forgetful (or underlying) functor of the concrete category and $mathbf{X}$ is called the base category for $(mathbf{A}, U)$.
This definition above is given only for the sake of completeness and is copied verbatim from Definition 5.1(1) of The Joy of Cats.
While I was reading The Joy of Cats and in pp. 59-60, I found that shortly after giving the examples of concrete categories (after the above definition) it is written (see Remark 5.3 (1)),
Since faithful functors are injective on $text{hom}$-sets, we usually assume...that $text{hom}_{mathbf{A}}(A,B)$ is a subset of $text{hom}_{mathbf{X}}(UA,UB)$ for each pair $(A,B)$ of $mathbf{A}$-objects. This familiar convention allows one to express the property that
"for $mathbf{A}$-objects $A$ and $B$ and an $mathbf{X}$-morphism $f:UAto UB$ there exists a (necessarily unique) $mathbf{A}$-morphism $require{AMScd}begin{CD}Arightarrow Bend{CD}$ with, $U(A→B)=f:UAto UB$"
much more succinctly, by stating,
"$f:UAto UB$ is an $mathbf{A}$-morphism (from $A$ to $B$)"
However, I find this convention to be very much confusing. My questions are,
Why would we assume that $text{hom}_{mathbf{A}}(A,B)$ is a subset of $text{hom}_{mathbf{X}}(UA,UB)$ even when it may not be the case? I agree that a convention is a convention (and I have no problem with that) but I don't think that a convention should be such that it creates more confusion than lessening it. So, I think that there is something "categorical way" of thinking that I am missing. I am interested to know it. Hence this question.
After the sentence "$f:UAto UB$ is an $mathbf{A}$-morphism (from $A$ to $B$)" as written above the authors write in a footnote the following,
"Observe that analogues of this expression (i.e. $f:UA→UX$ is an $mathbf{A}$-morphism) are frequently used in “concrete situations”, e.g., in saying
that a certain function between vector spaces “is linear” or that a certain function between topological spaces “is continuous”, etc. "
However I don't understand how this remark relates to the convention. Can anyone explain that to me?
category-theory big-picture
asked Jan 23 at 14:06
user 170039user 170039
10.5k42466
$endgroup$
1
$begingroup$
It is not more than a convention practicized in JoC and you can just neglect it if it causes confusion (as long as you keep in mind of course what the authors mean by it). What helped me (I encountered it too) in understanding the underlying motivation was mainly the look on functions $Xto Y$ where $X$ and $Y$ are topological spaces. It is quite common to involve functions $Xto Y$ that are not necessarily continuous while we are practicizing topology. The statement "$f$ is continuous" is then reformed into "$f$ is a $mathbf{Top}$-morphism".
$endgroup$
– drhab
Jan 23 at 14:24
$begingroup$
@drhab: Thank you very much for the comment. Indeed that would be a good way to look at the situation if $mathbf{X}$ is $mathbf{Set}$ itself. Inspired by this, for the general case would it be ok to view the $mathbf{X}$-morphism $f:UAto UB$ is essentially same as the unique $mathbf{A}$-morphism $g:Ato B$ (say) such that $U(g)=f$ if you ignore the structure that $g$ has? But does there exist more formal way to say this notion of "essential sameness"?
$endgroup$
– user 170039
Jan 23 at 14:50
$begingroup$
Although since we are allowed to take arbitrary categories, if the base category $mathbf{X}$ has "more structure" this viewpoint becomes problematic.
$endgroup$
– user 170039
Jan 23 at 14:55
1
$begingroup$
One way to defend the convention is to observe that every concrete category is concretely isomorphic to one with the same objects, in which the morphisms sets really are subsets of the morphism sets in $X$.
$endgroup$
– Kevin Carlson
Jan 23 at 16:06
1
$begingroup$
@KevinCarlson : but your point is a good one too, in the sense that we can make the hom-sets "as close to subsets as they could be", i.e. there is a conrectely isomorphic category where $hom(A,B)$ is a subset of $hom(UA,UB)times {(A,B)}$
$endgroup$
– Max
Jan 23 at 21:22
|
show 3 more comments
$begingroup$
Definition of Concrete Categories
Let $mathbf{X}$ be a category. A concrete category over $mathbf{X}$ is a pair $(mathbf{A}, U)$, where $mathbf{A}$ is a category and $U : mathbf{A} to mathbf{X}$ is a faithful functor. Sometimes $U$ is called the
forgetful (or underlying) functor of the concrete category and $mathbf{X}$ is called the base category for $(mathbf{A}, U)$.
This definition above is given only for the sake of completeness and is copied verbatim from Definition 5.1(1) of The Joy of Cats.
While I was reading The Joy of Cats and in pp. 59-60, I found that shortly after giving the examples of concrete categories (after the above definition) it is written (see Remark 5.3 (1)),
Since faithful functors are injective on $text{hom}$-sets, we usually assume...that $text{hom}_{mathbf{A}}(A,B)$ is a subset of $text{hom}_{mathbf{X}}(UA,UB)$ for each pair $(A,B)$ of $mathbf{A}$-objects. This familiar convention allows one to express the property that
"for $mathbf{A}$-objects $A$ and $B$ and an $mathbf{X}$-morphism $f:UAto UB$ there exists a (necessarily unique) $mathbf{A}$-morphism $require{AMScd}begin{CD}Arightarrow Bend{CD}$ with, $U(A→B)=f:UAto UB$"
much more succinctly, by stating,
"$f:UAto UB$ is an $mathbf{A}$-morphism (from $A$ to $B$)"
However, I find this convention to be very much confusing. My questions are,
Why would we assume that $text{hom}_{mathbf{A}}(A,B)$ is a subset of $text{hom}_{mathbf{X}}(UA,UB)$ even when it may not be the case? I agree that a convention is a convention (and I have no problem with that) but I don't think that a convention should be such that it creates more confusion than lessening it. So, I think that there is something "categorical way" of thinking that I am missing. I am interested to know it. Hence this question.
After the sentence "$f:UAto UB$ is an $mathbf{A}$-morphism (from $A$ to $B$)" as written above the authors write in a footnote the following,
"Observe that analogues of this expression (i.e. $f:UA→UX$ is an $mathbf{A}$-morphism) are frequently used in “concrete situations”, e.g., in saying
that a certain function between vector spaces “is linear” or that a certain function between topological spaces “is continuous”, etc. "
However I don't understand how this remark relates to the convention. Can anyone explain that to me?
category-theory big-picture
asked Jan 23 at 14:06
user 170039user 170039
10.5k42466
$endgroup$
Definition of Concrete Categories
Let $mathbf{X}$ be a category. A concrete category over $mathbf{X}$ is a pair $(mathbf{A}, U)$, where $mathbf{A}$ is a category and $U : mathbf{A} to mathbf{X}$ is a faithful functor. Sometimes $U$ is called the
forgetful (or underlying) functor of the concrete category and $mathbf{X}$ is called the base category for $(mathbf{A}, U)$.
This definition above is given only for the sake of completeness and is copied verbatim from Definition 5.1(1) of The Joy of Cats.
While I was reading The Joy of Cats and in pp. 59-60, I found that shortly after giving the examples of concrete categories (after the above definition) it is written (see Remark 5.3 (1)),
Since faithful functors are injective on $text{hom}$-sets, we usually assume...that $text{hom}_{mathbf{A}}(A,B)$ is a subset of $text{hom}_{mathbf{X}}(UA,UB)$ for each pair $(A,B)$ of $mathbf{A}$-objects. This familiar convention allows one to express the property that
"for $mathbf{A}$-objects $A$ and $B$ and an $mathbf{X}$-morphism $f:UAto UB$ there exists a (necessarily unique) $mathbf{A}$-morphism $require{AMScd}begin{CD}Arightarrow Bend{CD}$ with, $U(A→B)=f:UAto UB$"
much more succinctly, by stating,
"$f:UAto UB$ is an $mathbf{A}$-morphism (from $A$ to $B$)"
However, I find this convention to be very much confusing. My questions are,
Why would we assume that $text{hom}_{mathbf{A}}(A,B)$ is a subset of $text{hom}_{mathbf{X}}(UA,UB)$ even when it may not be the case? I agree that a convention is a convention (and I have no problem with that) but I don't think that a convention should be such that it creates more confusion than lessening it. So, I think that there is something "categorical way" of thinking that I am missing. I am interested to know it. Hence this question.
After the sentence "$f:UAto UB$ is an $mathbf{A}$-morphism (from $A$ to $B$)" as written above the authors write in a footnote the following,
"Observe that analogues of this expression (i.e. $f:UA→UX$ is an $mathbf{A}$-morphism) are frequently used in “concrete situations”, e.g., in saying
that a certain function between vector spaces “is linear” or that a certain function between topological spaces “is continuous”, etc. "
However I don't understand how this remark relates to the convention. Can anyone explain that to me?
category-theory big-picture
category-theory big-picture
asked Jan 23 at 14:06
user 170039user 170039
10.5k42466
asked Jan 23 at 14:06
user 170039user 170039
10.5k42466
edited Jan 23 at 14:37
user 170039
asked Jan 23 at 14:06
user 170039user 170039
10.5k42466
asked Jan 23 at 14:06
user 170039user 170039
10.5k42466
asked Jan 23 at 14:06
user 170039user 170039
10.5k42466
10.5k42466
1
$begingroup$
It is not more than a convention practicized in JoC and you can just neglect it if it causes confusion (as long as you keep in mind of course what the authors mean by it). What helped me (I encountered it too) in understanding the underlying motivation was mainly the look on functions $Xto Y$ where $X$ and $Y$ are topological spaces. It is quite common to involve functions $Xto Y$ that are not necessarily continuous while we are practicizing topology. The statement "$f$ is continuous" is then reformed into "$f$ is a $mathbf{Top}$-morphism".
$endgroup$
– drhab
Jan 23 at 14:24
$begingroup$
@drhab: Thank you very much for the comment. Indeed that would be a good way to look at the situation if $mathbf{X}$ is $mathbf{Set}$ itself. Inspired by this, for the general case would it be ok to view the $mathbf{X}$-morphism $f:UAto UB$ is essentially same as the unique $mathbf{A}$-morphism $g:Ato B$ (say) such that $U(g)=f$ if you ignore the structure that $g$ has? But does there exist more formal way to say this notion of "essential sameness"?
$endgroup$
– user 170039
Jan 23 at 14:50
$begingroup$
Although since we are allowed to take arbitrary categories, if the base category $mathbf{X}$ has "more structure" this viewpoint becomes problematic.
$endgroup$
– user 170039
Jan 23 at 14:55
1
$begingroup$
One way to defend the convention is to observe that every concrete category is concretely isomorphic to one with the same objects, in which the morphisms sets really are subsets of the morphism sets in $X$.
$endgroup$
– Kevin Carlson
Jan 23 at 16:06
1
$begingroup$
@KevinCarlson : but your point is a good one too, in the sense that we can make the hom-sets "as close to subsets as they could be", i.e. there is a conrectely isomorphic category where $hom(A,B)$ is a subset of $hom(UA,UB)times {(A,B)}$
$endgroup$
– Max
Jan 23 at 21:22
|
show 3 more comments
1
$begingroup$
It is not more than a convention practicized in JoC and you can just neglect it if it causes confusion (as long as you keep in mind of course what the authors mean by it). What helped me (I encountered it too) in understanding the underlying motivation was mainly the look on functions $Xto Y$ where $X$ and $Y$ are topological spaces. It is quite common to involve functions $Xto Y$ that are not necessarily continuous while we are practicizing topology. The statement "$f$ is continuous" is then reformed into "$f$ is a $mathbf{Top}$-morphism".
$endgroup$
– drhab
Jan 23 at 14:24
$begingroup$
@drhab: Thank you very much for the comment. Indeed that would be a good way to look at the situation if $mathbf{X}$ is $mathbf{Set}$ itself. Inspired by this, for the general case would it be ok to view the $mathbf{X}$-morphism $f:UAto UB$ is essentially same as the unique $mathbf{A}$-morphism $g:Ato B$ (say) such that $U(g)=f$ if you ignore the structure that $g$ has? But does there exist more formal way to say this notion of "essential sameness"?
$endgroup$
– user 170039
Jan 23 at 14:50
$begingroup$
Although since we are allowed to take arbitrary categories, if the base category $mathbf{X}$ has "more structure" this viewpoint becomes problematic.
$endgroup$
– user 170039
Jan 23 at 14:55
1
$begingroup$
One way to defend the convention is to observe that every concrete category is concretely isomorphic to one with the same objects, in which the morphisms sets really are subsets of the morphism sets in $X$.
$endgroup$
– Kevin Carlson
Jan 23 at 16:06
1
$begingroup$
@KevinCarlson : but your point is a good one too, in the sense that we can make the hom-sets "as close to subsets as they could be", i.e. there is a conrectely isomorphic category where $hom(A,B)$ is a subset of $hom(UA,UB)times {(A,B)}$
$endgroup$
– Max
Jan 23 at 21:22
1
1
$begingroup$
It is not more than a convention practicized in JoC and you can just neglect it if it causes confusion (as long as you keep in mind of course what the authors mean by it). What helped me (I encountered it too) in understanding the underlying motivation was mainly the look on functions $Xto Y$ where $X$ and $Y$ are topological spaces. It is quite common to involve functions $Xto Y$ that are not necessarily continuous while we are practicizing topology. The statement "$f$ is continuous" is then reformed into "$f$ is a $mathbf{Top}$-morphism".
$endgroup$
– drhab
Jan 23 at 14:24
$begingroup$
It is not more than a convention practicized in JoC and you can just neglect it if it causes confusion (as long as you keep in mind of course what the authors mean by it). What helped me (I encountered it too) in understanding the underlying motivation was mainly the look on functions $Xto Y$ where $X$ and $Y$ are topological spaces. It is quite common to involve functions $Xto Y$ that are not necessarily continuous while we are practicizing topology. The statement "$f$ is continuous" is then reformed into "$f$ is a $mathbf{Top}$-morphism".
$endgroup$
– drhab
Jan 23 at 14:24
$begingroup$
@drhab: Thank you very much for the comment. Indeed that would be a good way to look at the situation if $mathbf{X}$ is $mathbf{Set}$ itself. Inspired by this, for the general case would it be ok to view the $mathbf{X}$-morphism $f:UAto UB$ is essentially same as the unique $mathbf{A}$-morphism $g:Ato B$ (say) such that $U(g)=f$ if you ignore the structure that $g$ has? But does there exist more formal way to say this notion of "essential sameness"?
$endgroup$
– user 170039
Jan 23 at 14:50
$begingroup$
@drhab: Thank you very much for the comment. Indeed that would be a good way to look at the situation if $mathbf{X}$ is $mathbf{Set}$ itself. Inspired by this, for the general case would it be ok to view the $mathbf{X}$-morphism $f:UAto UB$ is essentially same as the unique $mathbf{A}$-morphism $g:Ato B$ (say) such that $U(g)=f$ if you ignore the structure that $g$ has? But does there exist more formal way to say this notion of "essential sameness"?
$endgroup$
– user 170039
Jan 23 at 14:50
$begingroup$
Although since we are allowed to take arbitrary categories, if the base category $mathbf{X}$ has "more structure" this viewpoint becomes problematic.
$endgroup$
– user 170039
Jan 23 at 14:55
$begingroup$
Although since we are allowed to take arbitrary categories, if the base category $mathbf{X}$ has "more structure" this viewpoint becomes problematic.
$endgroup$
– user 170039
Jan 23 at 14:55
1
1
$begingroup$
One way to defend the convention is to observe that every concrete category is concretely isomorphic to one with the same objects, in which the morphisms sets really are subsets of the morphism sets in $X$.
$endgroup$
– Kevin Carlson
Jan 23 at 16:06
$begingroup$
One way to defend the convention is to observe that every concrete category is concretely isomorphic to one with the same objects, in which the morphisms sets really are subsets of the morphism sets in $X$.
$endgroup$
– Kevin Carlson
Jan 23 at 16:06
1
1
$begingroup$
@KevinCarlson : but your point is a good one too, in the sense that we can make the hom-sets "as close to subsets as they could be", i.e. there is a conrectely isomorphic category where $hom(A,B)$ is a subset of $hom(UA,UB)times {(A,B)}$
$endgroup$
– Max
Jan 23 at 21:22
$begingroup$
@KevinCarlson : but your point is a good one too, in the sense that we can make the hom-sets "as close to subsets as they could be", i.e. there is a conrectely isomorphic category where $hom(A,B)$ is a subset of $hom(UA,UB)times {(A,B)}$
$endgroup$
– Max
Jan 23 at 21:22
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Say $mathbf{X}$ is the category of sets, and $mathbf{A}$ is the category of groups with the usual forgetful functor (which is faithful).
Then given two groups $underline{G}=(G,times, e, i)$ and $underline{H}=(H, *, 1, iota)$, it's not uncommon (or rather it's the most used convention/notation/way of writing things) to say, given a map of sets $f:Gto H$ that "it is a group morphism". But $G,H$ aren't groups, they're sets. It just so happens that $G=U(G,times ,e,i)$ and similarly for $H$, and that we know which group structure we're referring to from context; but in a pure categorical sense, we should be saying something like (depending on your definition of group morphisms, but it's only a technicality) "$f=U(f,underline{G},underline{H})= Uunderline{f}$ for some group morphism $underline{f}: underline{G}to underline{H}$"
Obviously this is not tractable, and we'd rather have a small (tiny in practice) ambiguity as to what we mean when we say "$f:Gto H$ is a group morphism" than have to write this. What happens, is that from context it is clear which group structure (read "antecedent under $U$") we are referring to on $G$ and on $H$, and $f$ (because a morphism in $mathbf{A}$ contains the data of which groups it is a morphism of, and in principle there are plenty of group structures on $G,H$ for which $f$ could be a group morphism)
Well here this convention is the same ! You see $mathbf{A}$ as a category of $mathbf{X}$'s with some extra structure and the maps are $mathbf{X}$-maps preserving this extra structure (this is the point of having a faithful functor !!). So you see an $mathbf{A}$-object $A$ as $UA$ but with extra structure. Then it only makes sense to say that a map $UAto UB$ "is an $mathbf{A}$-map" : it's an $mathbf{X}$-map that preserves the extra structure, i.e. is in $mathbf{A}$.
Seeing the inlcusion by $U$ as an actual subset inclusion allows you to do what we do every day with groups and topological spaces: identify the morphism (group morphism, continuous map, linear map) with the underlying set-theoretic map, and just say "it's a group morphism" instead of "it's in the image of the forgetful functor".
I think the main point here (I don't remember if JoC stresses that point) is that a faithful functor can be thought of as "we have a category $mathbf{X}$, its objects may (or may not) be furnished with some extra structure (whatever that means), and we get a new category $mathbf{A}$ by putting as objects : $mathbf{X}$-objects furnished with that extra structure and as maps : $mathbf{X}$-maps that preserve said structure".
You can make this precise in a pointless way (I'll leave that to you, it's really uninteresting), or try to see why it makes sense by looking at tons of faithful functors and seeing what makes them different from non faithful ones; and convincing yourself that faithful functors are "always" of this shape.
answered Jan 23 at 20:40
MaxMax
15.3k11143
$endgroup$
$begingroup$
So, instead of thinking $text{hom}_{mathbf{A}}(A,B)$ to be subset of $text{hom}_{mathbf{X}}(UA,UB)$ in the literal sense it is better to think of it as an "essential subset" of $text{hom}_{mathbf{X}}(UA,UB)$ in the sense that the $mathbf{A}$-objects are essentially the $mathbf{X}$-objects furnished with some "extra structure" and the $mathbf{A}$-morphisms are essentially the $mathbf{X}$-morphisms that preserve said structure. Does this way of viewing things sound ok to you?
$endgroup$
– user 170039
Jan 24 at 4:09
$begingroup$
Yes, but I don't see how that would, in practice, differ from their convention. Note that in practice, for instance for groups or topological spaces, we do act as if it were a subset
$endgroup$
– Max
Jan 24 at 9:13
$begingroup$
Indeed. But I think there is a crucial difference in these cases. For example if we consider the forgetful functor $U:mathbf{Grp}to mathbf{Set}$ then $hom_{mathbf{Grp}}(A,B)$ is indeed a subset of $hom_{mathbf{Set}}(UA,UB)$.
$endgroup$
– user 170039
Jan 24 at 13:40
$begingroup$
This can be clarified by the saying that a group homomorphisms (i.e., $mathbf{Grp}$-morphisms) for two given groups $(G,circ_G)$ and $(H,circ_H)$ is precisely those functions (i.e., $mathbf{Set}$-morphisms) $f:Gto H$ such that the following diagram commutes, $$require{AMScd}begin{CD} Gtimes G @>{ftimes f}>> Htimes H\ @V{circ_G}VV @VV{circ_H}V \ G @>{f}>> H end{CD}$$
$endgroup$
– user 170039
Jan 24 at 13:41
1
$begingroup$
No, that's not true, it isn't (and it can't be !), that's what I've been trying to explain. Moreover even if it were the case, I don't see why that would be a crucial difference
$endgroup$
– Max
Jan 24 at 13:41
|
show 3 more comments
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$begingroup$
Say $mathbf{X}$ is the category of sets, and $mathbf{A}$ is the category of groups with the usual forgetful functor (which is faithful).
Then given two groups $underline{G}=(G,times, e, i)$ and $underline{H}=(H, *, 1, iota)$, it's not uncommon (or rather it's the most used convention/notation/way of writing things) to say, given a map of sets $f:Gto H$ that "it is a group morphism". But $G,H$ aren't groups, they're sets. It just so happens that $G=U(G,times ,e,i)$ and similarly for $H$, and that we know which group structure we're referring to from context; but in a pure categorical sense, we should be saying something like (depending on your definition of group morphisms, but it's only a technicality) "$f=U(f,underline{G},underline{H})= Uunderline{f}$ for some group morphism $underline{f}: underline{G}to underline{H}$"
Obviously this is not tractable, and we'd rather have a small (tiny in practice) ambiguity as to what we mean when we say "$f:Gto H$ is a group morphism" than have to write this. What happens, is that from context it is clear which group structure (read "antecedent under $U$") we are referring to on $G$ and on $H$, and $f$ (because a morphism in $mathbf{A}$ contains the data of which groups it is a morphism of, and in principle there are plenty of group structures on $G,H$ for which $f$ could be a group morphism)
Well here this convention is the same ! You see $mathbf{A}$ as a category of $mathbf{X}$'s with some extra structure and the maps are $mathbf{X}$-maps preserving this extra structure (this is the point of having a faithful functor !!). So you see an $mathbf{A}$-object $A$ as $UA$ but with extra structure. Then it only makes sense to say that a map $UAto UB$ "is an $mathbf{A}$-map" : it's an $mathbf{X}$-map that preserves the extra structure, i.e. is in $mathbf{A}$.
Seeing the inlcusion by $U$ as an actual subset inclusion allows you to do what we do every day with groups and topological spaces: identify the morphism (group morphism, continuous map, linear map) with the underlying set-theoretic map, and just say "it's a group morphism" instead of "it's in the image of the forgetful functor".
I think the main point here (I don't remember if JoC stresses that point) is that a faithful functor can be thought of as "we have a category $mathbf{X}$, its objects may (or may not) be furnished with some extra structure (whatever that means), and we get a new category $mathbf{A}$ by putting as objects : $mathbf{X}$-objects furnished with that extra structure and as maps : $mathbf{X}$-maps that preserve said structure".
You can make this precise in a pointless way (I'll leave that to you, it's really uninteresting), or try to see why it makes sense by looking at tons of faithful functors and seeing what makes them different from non faithful ones; and convincing yourself that faithful functors are "always" of this shape.
answered Jan 23 at 20:40
MaxMax
15.3k11143
$endgroup$
$begingroup$
So, instead of thinking $text{hom}_{mathbf{A}}(A,B)$ to be subset of $text{hom}_{mathbf{X}}(UA,UB)$ in the literal sense it is better to think of it as an "essential subset" of $text{hom}_{mathbf{X}}(UA,UB)$ in the sense that the $mathbf{A}$-objects are essentially the $mathbf{X}$-objects furnished with some "extra structure" and the $mathbf{A}$-morphisms are essentially the $mathbf{X}$-morphisms that preserve said structure. Does this way of viewing things sound ok to you?
$endgroup$
– user 170039
Jan 24 at 4:09
$begingroup$
Yes, but I don't see how that would, in practice, differ from their convention. Note that in practice, for instance for groups or topological spaces, we do act as if it were a subset
$endgroup$
– Max
Jan 24 at 9:13
$begingroup$
Indeed. But I think there is a crucial difference in these cases. For example if we consider the forgetful functor $U:mathbf{Grp}to mathbf{Set}$ then $hom_{mathbf{Grp}}(A,B)$ is indeed a subset of $hom_{mathbf{Set}}(UA,UB)$.
$endgroup$
– user 170039
Jan 24 at 13:40
$begingroup$
This can be clarified by the saying that a group homomorphisms (i.e., $mathbf{Grp}$-morphisms) for two given groups $(G,circ_G)$ and $(H,circ_H)$ is precisely those functions (i.e., $mathbf{Set}$-morphisms) $f:Gto H$ such that the following diagram commutes, $$require{AMScd}begin{CD} Gtimes G @>{ftimes f}>> Htimes H\ @V{circ_G}VV @VV{circ_H}V \ G @>{f}>> H end{CD}$$
$endgroup$
– user 170039
Jan 24 at 13:41
1
$begingroup$
No, that's not true, it isn't (and it can't be !), that's what I've been trying to explain. Moreover even if it were the case, I don't see why that would be a crucial difference
$endgroup$
– Max
Jan 24 at 13:41
|
show 3 more comments
$begingroup$
Say $mathbf{X}$ is the category of sets, and $mathbf{A}$ is the category of groups with the usual forgetful functor (which is faithful).
Then given two groups $underline{G}=(G,times, e, i)$ and $underline{H}=(H, *, 1, iota)$, it's not uncommon (or rather it's the most used convention/notation/way of writing things) to say, given a map of sets $f:Gto H$ that "it is a group morphism". But $G,H$ aren't groups, they're sets. It just so happens that $G=U(G,times ,e,i)$ and similarly for $H$, and that we know which group structure we're referring to from context; but in a pure categorical sense, we should be saying something like (depending on your definition of group morphisms, but it's only a technicality) "$f=U(f,underline{G},underline{H})= Uunderline{f}$ for some group morphism $underline{f}: underline{G}to underline{H}$"
Obviously this is not tractable, and we'd rather have a small (tiny in practice) ambiguity as to what we mean when we say "$f:Gto H$ is a group morphism" than have to write this. What happens, is that from context it is clear which group structure (read "antecedent under $U$") we are referring to on $G$ and on $H$, and $f$ (because a morphism in $mathbf{A}$ contains the data of which groups it is a morphism of, and in principle there are plenty of group structures on $G,H$ for which $f$ could be a group morphism)
Well here this convention is the same ! You see $mathbf{A}$ as a category of $mathbf{X}$'s with some extra structure and the maps are $mathbf{X}$-maps preserving this extra structure (this is the point of having a faithful functor !!). So you see an $mathbf{A}$-object $A$ as $UA$ but with extra structure. Then it only makes sense to say that a map $UAto UB$ "is an $mathbf{A}$-map" : it's an $mathbf{X}$-map that preserves the extra structure, i.e. is in $mathbf{A}$.
Seeing the inlcusion by $U$ as an actual subset inclusion allows you to do what we do every day with groups and topological spaces: identify the morphism (group morphism, continuous map, linear map) with the underlying set-theoretic map, and just say "it's a group morphism" instead of "it's in the image of the forgetful functor".
I think the main point here (I don't remember if JoC stresses that point) is that a faithful functor can be thought of as "we have a category $mathbf{X}$, its objects may (or may not) be furnished with some extra structure (whatever that means), and we get a new category $mathbf{A}$ by putting as objects : $mathbf{X}$-objects furnished with that extra structure and as maps : $mathbf{X}$-maps that preserve said structure".
You can make this precise in a pointless way (I'll leave that to you, it's really uninteresting), or try to see why it makes sense by looking at tons of faithful functors and seeing what makes them different from non faithful ones; and convincing yourself that faithful functors are "always" of this shape.
answered Jan 23 at 20:40
MaxMax
15.3k11143
$endgroup$
$begingroup$
So, instead of thinking $text{hom}_{mathbf{A}}(A,B)$ to be subset of $text{hom}_{mathbf{X}}(UA,UB)$ in the literal sense it is better to think of it as an "essential subset" of $text{hom}_{mathbf{X}}(UA,UB)$ in the sense that the $mathbf{A}$-objects are essentially the $mathbf{X}$-objects furnished with some "extra structure" and the $mathbf{A}$-morphisms are essentially the $mathbf{X}$-morphisms that preserve said structure. Does this way of viewing things sound ok to you?
$endgroup$
– user 170039
Jan 24 at 4:09
$begingroup$
Yes, but I don't see how that would, in practice, differ from their convention. Note that in practice, for instance for groups or topological spaces, we do act as if it were a subset
$endgroup$
– Max
Jan 24 at 9:13
$begingroup$
Indeed. But I think there is a crucial difference in these cases. For example if we consider the forgetful functor $U:mathbf{Grp}to mathbf{Set}$ then $hom_{mathbf{Grp}}(A,B)$ is indeed a subset of $hom_{mathbf{Set}}(UA,UB)$.
$endgroup$
– user 170039
Jan 24 at 13:40
$begingroup$
This can be clarified by the saying that a group homomorphisms (i.e., $mathbf{Grp}$-morphisms) for two given groups $(G,circ_G)$ and $(H,circ_H)$ is precisely those functions (i.e., $mathbf{Set}$-morphisms) $f:Gto H$ such that the following diagram commutes, $$require{AMScd}begin{CD} Gtimes G @>{ftimes f}>> Htimes H\ @V{circ_G}VV @VV{circ_H}V \ G @>{f}>> H end{CD}$$
$endgroup$
– user 170039
Jan 24 at 13:41
1
$begingroup$
No, that's not true, it isn't (and it can't be !), that's what I've been trying to explain. Moreover even if it were the case, I don't see why that would be a crucial difference
$endgroup$
– Max
Jan 24 at 13:41
|
show 3 more comments
$begingroup$
Say $mathbf{X}$ is the category of sets, and $mathbf{A}$ is the category of groups with the usual forgetful functor (which is faithful).
Then given two groups $underline{G}=(G,times, e, i)$ and $underline{H}=(H, *, 1, iota)$, it's not uncommon (or rather it's the most used convention/notation/way of writing things) to say, given a map of sets $f:Gto H$ that "it is a group morphism". But $G,H$ aren't groups, they're sets. It just so happens that $G=U(G,times ,e,i)$ and similarly for $H$, and that we know which group structure we're referring to from context; but in a pure categorical sense, we should be saying something like (depending on your definition of group morphisms, but it's only a technicality) "$f=U(f,underline{G},underline{H})= Uunderline{f}$ for some group morphism $underline{f}: underline{G}to underline{H}$"
Obviously this is not tractable, and we'd rather have a small (tiny in practice) ambiguity as to what we mean when we say "$f:Gto H$ is a group morphism" than have to write this. What happens, is that from context it is clear which group structure (read "antecedent under $U$") we are referring to on $G$ and on $H$, and $f$ (because a morphism in $mathbf{A}$ contains the data of which groups it is a morphism of, and in principle there are plenty of group structures on $G,H$ for which $f$ could be a group morphism)
Well here this convention is the same ! You see $mathbf{A}$ as a category of $mathbf{X}$'s with some extra structure and the maps are $mathbf{X}$-maps preserving this extra structure (this is the point of having a faithful functor !!). So you see an $mathbf{A}$-object $A$ as $UA$ but with extra structure. Then it only makes sense to say that a map $UAto UB$ "is an $mathbf{A}$-map" : it's an $mathbf{X}$-map that preserves the extra structure, i.e. is in $mathbf{A}$.
Seeing the inlcusion by $U$ as an actual subset inclusion allows you to do what we do every day with groups and topological spaces: identify the morphism (group morphism, continuous map, linear map) with the underlying set-theoretic map, and just say "it's a group morphism" instead of "it's in the image of the forgetful functor".
I think the main point here (I don't remember if JoC stresses that point) is that a faithful functor can be thought of as "we have a category $mathbf{X}$, its objects may (or may not) be furnished with some extra structure (whatever that means), and we get a new category $mathbf{A}$ by putting as objects : $mathbf{X}$-objects furnished with that extra structure and as maps : $mathbf{X}$-maps that preserve said structure".
You can make this precise in a pointless way (I'll leave that to you, it's really uninteresting), or try to see why it makes sense by looking at tons of faithful functors and seeing what makes them different from non faithful ones; and convincing yourself that faithful functors are "always" of this shape.
answered Jan 23 at 20:40
MaxMax
15.3k11143
$endgroup$
Say $mathbf{X}$ is the category of sets, and $mathbf{A}$ is the category of groups with the usual forgetful functor (which is faithful).
Then given two groups $underline{G}=(G,times, e, i)$ and $underline{H}=(H, *, 1, iota)$, it's not uncommon (or rather it's the most used convention/notation/way of writing things) to say, given a map of sets $f:Gto H$ that "it is a group morphism". But $G,H$ aren't groups, they're sets. It just so happens that $G=U(G,times ,e,i)$ and similarly for $H$, and that we know which group structure we're referring to from context; but in a pure categorical sense, we should be saying something like (depending on your definition of group morphisms, but it's only a technicality) "$f=U(f,underline{G},underline{H})= Uunderline{f}$ for some group morphism $underline{f}: underline{G}to underline{H}$"
Obviously this is not tractable, and we'd rather have a small (tiny in practice) ambiguity as to what we mean when we say "$f:Gto H$ is a group morphism" than have to write this. What happens, is that from context it is clear which group structure (read "antecedent under $U$") we are referring to on $G$ and on $H$, and $f$ (because a morphism in $mathbf{A}$ contains the data of which groups it is a morphism of, and in principle there are plenty of group structures on $G,H$ for which $f$ could be a group morphism)
Well here this convention is the same ! You see $mathbf{A}$ as a category of $mathbf{X}$'s with some extra structure and the maps are $mathbf{X}$-maps preserving this extra structure (this is the point of having a faithful functor !!). So you see an $mathbf{A}$-object $A$ as $UA$ but with extra structure. Then it only makes sense to say that a map $UAto UB$ "is an $mathbf{A}$-map" : it's an $mathbf{X}$-map that preserves the extra structure, i.e. is in $mathbf{A}$.
Seeing the inlcusion by $U$ as an actual subset inclusion allows you to do what we do every day with groups and topological spaces: identify the morphism (group morphism, continuous map, linear map) with the underlying set-theoretic map, and just say "it's a group morphism" instead of "it's in the image of the forgetful functor".
I think the main point here (I don't remember if JoC stresses that point) is that a faithful functor can be thought of as "we have a category $mathbf{X}$, its objects may (or may not) be furnished with some extra structure (whatever that means), and we get a new category $mathbf{A}$ by putting as objects : $mathbf{X}$-objects furnished with that extra structure and as maps : $mathbf{X}$-maps that preserve said structure".
You can make this precise in a pointless way (I'll leave that to you, it's really uninteresting), or try to see why it makes sense by looking at tons of faithful functors and seeing what makes them different from non faithful ones; and convincing yourself that faithful functors are "always" of this shape.
answered Jan 23 at 20:40
MaxMax
15.3k11143
answered Jan 23 at 20:40
MaxMax
15.3k11143
answered Jan 23 at 20:40
MaxMax
15.3k11143
answered Jan 23 at 20:40
MaxMax
15.3k11143
15.3k11143
$begingroup$
So, instead of thinking $text{hom}_{mathbf{A}}(A,B)$ to be subset of $text{hom}_{mathbf{X}}(UA,UB)$ in the literal sense it is better to think of it as an "essential subset" of $text{hom}_{mathbf{X}}(UA,UB)$ in the sense that the $mathbf{A}$-objects are essentially the $mathbf{X}$-objects furnished with some "extra structure" and the $mathbf{A}$-morphisms are essentially the $mathbf{X}$-morphisms that preserve said structure. Does this way of viewing things sound ok to you?
$endgroup$
– user 170039
Jan 24 at 4:09
$begingroup$
Yes, but I don't see how that would, in practice, differ from their convention. Note that in practice, for instance for groups or topological spaces, we do act as if it were a subset
$endgroup$
– Max
Jan 24 at 9:13
$begingroup$
Indeed. But I think there is a crucial difference in these cases. For example if we consider the forgetful functor $U:mathbf{Grp}to mathbf{Set}$ then $hom_{mathbf{Grp}}(A,B)$ is indeed a subset of $hom_{mathbf{Set}}(UA,UB)$.
$endgroup$
– user 170039
Jan 24 at 13:40
$begingroup$
This can be clarified by the saying that a group homomorphisms (i.e., $mathbf{Grp}$-morphisms) for two given groups $(G,circ_G)$ and $(H,circ_H)$ is precisely those functions (i.e., $mathbf{Set}$-morphisms) $f:Gto H$ such that the following diagram commutes, $$require{AMScd}begin{CD} Gtimes G @>{ftimes f}>> Htimes H\ @V{circ_G}VV @VV{circ_H}V \ G @>{f}>> H end{CD}$$
$endgroup$
– user 170039
Jan 24 at 13:41
1
$begingroup$
No, that's not true, it isn't (and it can't be !), that's what I've been trying to explain. Moreover even if it were the case, I don't see why that would be a crucial difference
$endgroup$
– Max
Jan 24 at 13:41
|
show 3 more comments
$begingroup$
So, instead of thinking $text{hom}_{mathbf{A}}(A,B)$ to be subset of $text{hom}_{mathbf{X}}(UA,UB)$ in the literal sense it is better to think of it as an "essential subset" of $text{hom}_{mathbf{X}}(UA,UB)$ in the sense that the $mathbf{A}$-objects are essentially the $mathbf{X}$-objects furnished with some "extra structure" and the $mathbf{A}$-morphisms are essentially the $mathbf{X}$-morphisms that preserve said structure. Does this way of viewing things sound ok to you?
$endgroup$
– user 170039
Jan 24 at 4:09
$begingroup$
Yes, but I don't see how that would, in practice, differ from their convention. Note that in practice, for instance for groups or topological spaces, we do act as if it were a subset
$endgroup$
– Max
Jan 24 at 9:13
$begingroup$
Indeed. But I think there is a crucial difference in these cases. For example if we consider the forgetful functor $U:mathbf{Grp}to mathbf{Set}$ then $hom_{mathbf{Grp}}(A,B)$ is indeed a subset of $hom_{mathbf{Set}}(UA,UB)$.
$endgroup$
– user 170039
Jan 24 at 13:40
$begingroup$
This can be clarified by the saying that a group homomorphisms (i.e., $mathbf{Grp}$-morphisms) for two given groups $(G,circ_G)$ and $(H,circ_H)$ is precisely those functions (i.e., $mathbf{Set}$-morphisms) $f:Gto H$ such that the following diagram commutes, $$require{AMScd}begin{CD} Gtimes G @>{ftimes f}>> Htimes H\ @V{circ_G}VV @VV{circ_H}V \ G @>{f}>> H end{CD}$$
$endgroup$
– user 170039
Jan 24 at 13:41
1
$begingroup$
No, that's not true, it isn't (and it can't be !), that's what I've been trying to explain. Moreover even if it were the case, I don't see why that would be a crucial difference
$endgroup$
– Max
Jan 24 at 13:41
$begingroup$
So, instead of thinking $text{hom}_{mathbf{A}}(A,B)$ to be subset of $text{hom}_{mathbf{X}}(UA,UB)$ in the literal sense it is better to think of it as an "essential subset" of $text{hom}_{mathbf{X}}(UA,UB)$ in the sense that the $mathbf{A}$-objects are essentially the $mathbf{X}$-objects furnished with some "extra structure" and the $mathbf{A}$-morphisms are essentially the $mathbf{X}$-morphisms that preserve said structure. Does this way of viewing things sound ok to you?
$endgroup$
– user 170039
Jan 24 at 4:09
$begingroup$
So, instead of thinking $text{hom}_{mathbf{A}}(A,B)$ to be subset of $text{hom}_{mathbf{X}}(UA,UB)$ in the literal sense it is better to think of it as an "essential subset" of $text{hom}_{mathbf{X}}(UA,UB)$ in the sense that the $mathbf{A}$-objects are essentially the $mathbf{X}$-objects furnished with some "extra structure" and the $mathbf{A}$-morphisms are essentially the $mathbf{X}$-morphisms that preserve said structure. Does this way of viewing things sound ok to you?
$endgroup$
– user 170039
Jan 24 at 4:09
$begingroup$
Yes, but I don't see how that would, in practice, differ from their convention. Note that in practice, for instance for groups or topological spaces, we do act as if it were a subset
$endgroup$
– Max
Jan 24 at 9:13
$begingroup$
Yes, but I don't see how that would, in practice, differ from their convention. Note that in practice, for instance for groups or topological spaces, we do act as if it were a subset
$endgroup$
– Max
Jan 24 at 9:13
$begingroup$
Indeed. But I think there is a crucial difference in these cases. For example if we consider the forgetful functor $U:mathbf{Grp}to mathbf{Set}$ then $hom_{mathbf{Grp}}(A,B)$ is indeed a subset of $hom_{mathbf{Set}}(UA,UB)$.
$endgroup$
– user 170039
Jan 24 at 13:40
$begingroup$
Indeed. But I think there is a crucial difference in these cases. For example if we consider the forgetful functor $U:mathbf{Grp}to mathbf{Set}$ then $hom_{mathbf{Grp}}(A,B)$ is indeed a subset of $hom_{mathbf{Set}}(UA,UB)$.
$endgroup$
– user 170039
Jan 24 at 13:40
$begingroup$
This can be clarified by the saying that a group homomorphisms (i.e., $mathbf{Grp}$-morphisms) for two given groups $(G,circ_G)$ and $(H,circ_H)$ is precisely those functions (i.e., $mathbf{Set}$-morphisms) $f:Gto H$ such that the following diagram commutes, $$require{AMScd}begin{CD} Gtimes G @>{ftimes f}>> Htimes H\ @V{circ_G}VV @VV{circ_H}V \ G @>{f}>> H end{CD}$$
$endgroup$
– user 170039
Jan 24 at 13:41
$begingroup$
This can be clarified by the saying that a group homomorphisms (i.e., $mathbf{Grp}$-morphisms) for two given groups $(G,circ_G)$ and $(H,circ_H)$ is precisely those functions (i.e., $mathbf{Set}$-morphisms) $f:Gto H$ such that the following diagram commutes, $$require{AMScd}begin{CD} Gtimes G @>{ftimes f}>> Htimes H\ @V{circ_G}VV @VV{circ_H}V \ G @>{f}>> H end{CD}$$
$endgroup$
– user 170039
Jan 24 at 13:41
1
1
$begingroup$
No, that's not true, it isn't (and it can't be !), that's what I've been trying to explain. Moreover even if it were the case, I don't see why that would be a crucial difference
$endgroup$
– Max
Jan 24 at 13:41
$begingroup$
No, that's not true, it isn't (and it can't be !), that's what I've been trying to explain. Moreover even if it were the case, I don't see why that would be a crucial difference
$endgroup$
– Max
Jan 24 at 13:41
|
show 3 more comments
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$begingroup$
It is not more than a convention practicized in JoC and you can just neglect it if it causes confusion (as long as you keep in mind of course what the authors mean by it). What helped me (I encountered it too) in understanding the underlying motivation was mainly the look on functions $Xto Y$ where $X$ and $Y$ are topological spaces. It is quite common to involve functions $Xto Y$ that are not necessarily continuous while we are practicizing topology. The statement "$f$ is continuous" is then reformed into "$f$ is a $mathbf{Top}$-morphism".
$endgroup$
– drhab
Jan 23 at 14:24
$begingroup$
@drhab: Thank you very much for the comment. Indeed that would be a good way to look at the situation if $mathbf{X}$ is $mathbf{Set}$ itself. Inspired by this, for the general case would it be ok to view the $mathbf{X}$-morphism $f:UAto UB$ is essentially same as the unique $mathbf{A}$-morphism $g:Ato B$ (say) such that $U(g)=f$ if you ignore the structure that $g$ has? But does there exist more formal way to say this notion of "essential sameness"?
$endgroup$
– user 170039
Jan 23 at 14:50
$begingroup$
Although since we are allowed to take arbitrary categories, if the base category $mathbf{X}$ has "more structure" this viewpoint becomes problematic.
$endgroup$
– user 170039
Jan 23 at 14:55
1
$begingroup$
One way to defend the convention is to observe that every concrete category is concretely isomorphic to one with the same objects, in which the morphisms sets really are subsets of the morphism sets in $X$.
$endgroup$
– Kevin Carlson
Jan 23 at 16:06
1
$begingroup$
@KevinCarlson : but your point is a good one too, in the sense that we can make the hom-sets "as close to subsets as they could be", i.e. there is a conrectely isomorphic category where $hom(A,B)$ is a subset of $hom(UA,UB)times {(A,B)}$
$endgroup$
– Max
Jan 23 at 21:22