Mixing
Are equivalence classes of subobjects of some X in Set just equivalence classes of subsets of X with a...
$begingroup$
I'm trying to understand what would be the subobjects of ${0, 1}$.
Would they be ${emptyset, {0}, {1}, {0, 1} }$?
Or are ${0}$ and ${1}$ somehow identified together? Because I can map from 0 to 1 and backwards - which tells me they're a part of the same equivalence class.
And what's stopping me from picking ${7}$ as a subobject, based on this definition?
There was a similar answer from a while ago, but I fail to see how it answers the questions I'm posing above.
category-theory equivalence-relations
asked Jan 13 at 15:13
Bruno GavranovicBruno Gavranovic
434
$endgroup$
add a comment |
$begingroup$
I'm trying to understand what would be the subobjects of ${0, 1}$.
Would they be ${emptyset, {0}, {1}, {0, 1} }$?
Or are ${0}$ and ${1}$ somehow identified together? Because I can map from 0 to 1 and backwards - which tells me they're a part of the same equivalence class.
And what's stopping me from picking ${7}$ as a subobject, based on this definition?
There was a similar answer from a while ago, but I fail to see how it answers the questions I'm posing above.
category-theory equivalence-relations
asked Jan 13 at 15:13
Bruno GavranovicBruno Gavranovic
434
$endgroup$
add a comment |
$begingroup$
I'm trying to understand what would be the subobjects of ${0, 1}$.
Would they be ${emptyset, {0}, {1}, {0, 1} }$?
Or are ${0}$ and ${1}$ somehow identified together? Because I can map from 0 to 1 and backwards - which tells me they're a part of the same equivalence class.
And what's stopping me from picking ${7}$ as a subobject, based on this definition?
There was a similar answer from a while ago, but I fail to see how it answers the questions I'm posing above.
category-theory equivalence-relations
asked Jan 13 at 15:13
Bruno GavranovicBruno Gavranovic
434
$endgroup$
I'm trying to understand what would be the subobjects of ${0, 1}$.
Would they be ${emptyset, {0}, {1}, {0, 1} }$?
Or are ${0}$ and ${1}$ somehow identified together? Because I can map from 0 to 1 and backwards - which tells me they're a part of the same equivalence class.
And what's stopping me from picking ${7}$ as a subobject, based on this definition?
There was a similar answer from a while ago, but I fail to see how it answers the questions I'm posing above.
category-theory equivalence-relations
category-theory equivalence-relations
asked Jan 13 at 15:13
Bruno GavranovicBruno Gavranovic
434
asked Jan 13 at 15:13
Bruno GavranovicBruno Gavranovic
434
asked Jan 13 at 15:13
Bruno GavranovicBruno Gavranovic
434
asked Jan 13 at 15:13
Bruno GavranovicBruno Gavranovic
434
asked Jan 13 at 15:13
Bruno GavranovicBruno Gavranovic
434
434
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I preassume that you are looking at set ${0,1}$ as an object of the category of $mathbf{Sets}$ here.
Object ${0,1}$ has $4$ subobjects.
Each of them is a class of of injective functions that all have ${0,1}$ as codomain.
If $m$ denotes the unique arrow $varnothingtovarnothing$ then ${m}$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $0$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $1$ is a subobject of ${0,1}$.
The class of injective functions ${a,b}to{0,1}$ with $aneq b$ is a subobject of ${0,1}$.
answered Jan 13 at 15:40
drhabdrhab
101k545136
$endgroup$
$begingroup$
Ah I see. So in a sense, we don't care about the input, we just care about the "shape" of the image of the monomorphism. In the case of sets, any one element set can be identified with a subobject of ${0, 1}$ in two different ways. I assume that as we add more structure to the set (a monoid, a group), fewer one element sets will be able to be subobjects (as all the morphisms need to be homomorphisms for the respective type of structure).
$endgroup$
– Bruno Gavranovic
Jan 13 at 16:15
add a comment |
$begingroup$
In $Set$, at least it has exactly the four subobjects you wrote up.
Note that a subobject of some object $X$ is not only a mere object $A$ in the same category, but it is understood to be together with a monomorphism $i:Ahookrightarrow X$ which plays the role of the inclusion.
Said that, indeed nothing prevents us to take ${7}$ as a subobject of ${0,1}$, moreover there are exactly two ways to do that: $i$ either sends $7mapsto 0$, or $7mapsto 1$.
So the first one represents the same subobject as ${0}hookrightarrow{0,1}$, while the second one represents the same subobject as ${1}hookrightarrow{0,1}$, but these two are distinct.
answered Jan 13 at 15:28
BerciBerci
60.8k23673
$endgroup$
$begingroup$
Thank you for the answer! According to this answer (link above),there is a "good definition" where {2} isn't a subobject of {1}. --- "If you were to define "subobjects of A∈C" to mean "monomorphisms with codomain A", then the notion of a subobject would not generalize all of these classical notions, because you get too many different subobjects that correspond to the same "sub- something". For instance, in Set, the monomorphisms ∅→{1}, {1}→{1} and {2}→{1} would be three different subobjects of the object {1}, but there are only two subsets of the set {1}. So this would be a bad definition."
$endgroup$
– Bruno Gavranovic
Jan 13 at 15:33
$begingroup$
Yes, we need to consider certain monomorphisms to represent the same subobject, i.e. the subobjects form a quotient set (class) of the monomorphisms.
$endgroup$
– Berci
Jan 13 at 15:43
add a comment |
$begingroup$
A subobject of an object of a category is not an object of the same category, it is an equivalence class of monomorphisms to this object (see definition from your link). So $varnothing$, ${0}$, ${1}$, ${0,1}$ are not subobjects of ${0,1}$, they are its subsets. But for every set $X$ the power set $mathcal{P}(X)$ is isomorphic to the set of subobjects of $X$ in the category of sets (and they are isomorphic as ordered sets). For example, the subset ${0}subset{0,1}$ corresponds to the subobject $[i_{{0}}]$, where $i_{{0}}colon{0}to{0,1}$ is the canonical inclusion of subset and $[-]$ is taking of equivalence class of a monomorphism (as in the definition by the link). It is easy to see that if $X$, $Y$, $Z$ are sets and $fcolon Yto X$ and $gcolon Zto X$ are monomorphisms (injections), then $[f]=[g]$ as subobjects of $X$ in the category of sets if and only if their set-theoretic images are equal: $f(Y)=g(Z)$. So, for example, if $i_{{0}}colon{0}to{0,1}$ and $i_{{1}}colon{1}to{0,1}$ are canonical inclusions, then $[i_{{0}}]$ and $[i_{{1}}]$ are not equal as subobjects of ${0,1}$ in the category of sets.
answered Jan 13 at 15:34
OskarOskar
3,1431719
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
I preassume that you are looking at set ${0,1}$ as an object of the category of $mathbf{Sets}$ here.
Object ${0,1}$ has $4$ subobjects.
Each of them is a class of of injective functions that all have ${0,1}$ as codomain.
If $m$ denotes the unique arrow $varnothingtovarnothing$ then ${m}$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $0$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $1$ is a subobject of ${0,1}$.
The class of injective functions ${a,b}to{0,1}$ with $aneq b$ is a subobject of ${0,1}$.
answered Jan 13 at 15:40
drhabdrhab
101k545136
$endgroup$
$begingroup$
Ah I see. So in a sense, we don't care about the input, we just care about the "shape" of the image of the monomorphism. In the case of sets, any one element set can be identified with a subobject of ${0, 1}$ in two different ways. I assume that as we add more structure to the set (a monoid, a group), fewer one element sets will be able to be subobjects (as all the morphisms need to be homomorphisms for the respective type of structure).
$endgroup$
– Bruno Gavranovic
Jan 13 at 16:15
add a comment |
$begingroup$
I preassume that you are looking at set ${0,1}$ as an object of the category of $mathbf{Sets}$ here.
Object ${0,1}$ has $4$ subobjects.
Each of them is a class of of injective functions that all have ${0,1}$ as codomain.
If $m$ denotes the unique arrow $varnothingtovarnothing$ then ${m}$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $0$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $1$ is a subobject of ${0,1}$.
The class of injective functions ${a,b}to{0,1}$ with $aneq b$ is a subobject of ${0,1}$.
answered Jan 13 at 15:40
drhabdrhab
101k545136
$endgroup$
$begingroup$
Ah I see. So in a sense, we don't care about the input, we just care about the "shape" of the image of the monomorphism. In the case of sets, any one element set can be identified with a subobject of ${0, 1}$ in two different ways. I assume that as we add more structure to the set (a monoid, a group), fewer one element sets will be able to be subobjects (as all the morphisms need to be homomorphisms for the respective type of structure).
$endgroup$
– Bruno Gavranovic
Jan 13 at 16:15
add a comment |
$begingroup$
I preassume that you are looking at set ${0,1}$ as an object of the category of $mathbf{Sets}$ here.
Object ${0,1}$ has $4$ subobjects.
Each of them is a class of of injective functions that all have ${0,1}$ as codomain.
If $m$ denotes the unique arrow $varnothingtovarnothing$ then ${m}$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $0$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $1$ is a subobject of ${0,1}$.
The class of injective functions ${a,b}to{0,1}$ with $aneq b$ is a subobject of ${0,1}$.
answered Jan 13 at 15:40
drhabdrhab
101k545136
$endgroup$
I preassume that you are looking at set ${0,1}$ as an object of the category of $mathbf{Sets}$ here.
Object ${0,1}$ has $4$ subobjects.
Each of them is a class of of injective functions that all have ${0,1}$ as codomain.
If $m$ denotes the unique arrow $varnothingtovarnothing$ then ${m}$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $0$ is a subobject of ${0,1}$.
The class of functions ${a}to{0,1}$ where $a$ is sent to $1$ is a subobject of ${0,1}$.
The class of injective functions ${a,b}to{0,1}$ with $aneq b$ is a subobject of ${0,1}$.
answered Jan 13 at 15:40
drhabdrhab
101k545136
answered Jan 13 at 15:40
drhabdrhab
101k545136
answered Jan 13 at 15:40
drhabdrhab
101k545136
answered Jan 13 at 15:40
drhabdrhab
101k545136
101k545136
$begingroup$
Ah I see. So in a sense, we don't care about the input, we just care about the "shape" of the image of the monomorphism. In the case of sets, any one element set can be identified with a subobject of ${0, 1}$ in two different ways. I assume that as we add more structure to the set (a monoid, a group), fewer one element sets will be able to be subobjects (as all the morphisms need to be homomorphisms for the respective type of structure).
$endgroup$
– Bruno Gavranovic
Jan 13 at 16:15
add a comment |
$begingroup$
Ah I see. So in a sense, we don't care about the input, we just care about the "shape" of the image of the monomorphism. In the case of sets, any one element set can be identified with a subobject of ${0, 1}$ in two different ways. I assume that as we add more structure to the set (a monoid, a group), fewer one element sets will be able to be subobjects (as all the morphisms need to be homomorphisms for the respective type of structure).
$endgroup$
– Bruno Gavranovic
Jan 13 at 16:15
$begingroup$
Ah I see. So in a sense, we don't care about the input, we just care about the "shape" of the image of the monomorphism. In the case of sets, any one element set can be identified with a subobject of ${0, 1}$ in two different ways. I assume that as we add more structure to the set (a monoid, a group), fewer one element sets will be able to be subobjects (as all the morphisms need to be homomorphisms for the respective type of structure).
$endgroup$
– Bruno Gavranovic
Jan 13 at 16:15
$begingroup$
Ah I see. So in a sense, we don't care about the input, we just care about the "shape" of the image of the monomorphism. In the case of sets, any one element set can be identified with a subobject of ${0, 1}$ in two different ways. I assume that as we add more structure to the set (a monoid, a group), fewer one element sets will be able to be subobjects (as all the morphisms need to be homomorphisms for the respective type of structure).
$endgroup$
– Bruno Gavranovic
Jan 13 at 16:15
add a comment |
$begingroup$
In $Set$, at least it has exactly the four subobjects you wrote up.
Note that a subobject of some object $X$ is not only a mere object $A$ in the same category, but it is understood to be together with a monomorphism $i:Ahookrightarrow X$ which plays the role of the inclusion.
Said that, indeed nothing prevents us to take ${7}$ as a subobject of ${0,1}$, moreover there are exactly two ways to do that: $i$ either sends $7mapsto 0$, or $7mapsto 1$.
So the first one represents the same subobject as ${0}hookrightarrow{0,1}$, while the second one represents the same subobject as ${1}hookrightarrow{0,1}$, but these two are distinct.
answered Jan 13 at 15:28
BerciBerci
60.8k23673
$endgroup$
$begingroup$
Thank you for the answer! According to this answer (link above),there is a "good definition" where {2} isn't a subobject of {1}. --- "If you were to define "subobjects of A∈C" to mean "monomorphisms with codomain A", then the notion of a subobject would not generalize all of these classical notions, because you get too many different subobjects that correspond to the same "sub- something". For instance, in Set, the monomorphisms ∅→{1}, {1}→{1} and {2}→{1} would be three different subobjects of the object {1}, but there are only two subsets of the set {1}. So this would be a bad definition."
$endgroup$
– Bruno Gavranovic
Jan 13 at 15:33
$begingroup$
Yes, we need to consider certain monomorphisms to represent the same subobject, i.e. the subobjects form a quotient set (class) of the monomorphisms.
$endgroup$
– Berci
Jan 13 at 15:43
add a comment |
$begingroup$
In $Set$, at least it has exactly the four subobjects you wrote up.
Note that a subobject of some object $X$ is not only a mere object $A$ in the same category, but it is understood to be together with a monomorphism $i:Ahookrightarrow X$ which plays the role of the inclusion.
Said that, indeed nothing prevents us to take ${7}$ as a subobject of ${0,1}$, moreover there are exactly two ways to do that: $i$ either sends $7mapsto 0$, or $7mapsto 1$.
So the first one represents the same subobject as ${0}hookrightarrow{0,1}$, while the second one represents the same subobject as ${1}hookrightarrow{0,1}$, but these two are distinct.
answered Jan 13 at 15:28
BerciBerci
60.8k23673
$endgroup$
$begingroup$
Thank you for the answer! According to this answer (link above),there is a "good definition" where {2} isn't a subobject of {1}. --- "If you were to define "subobjects of A∈C" to mean "monomorphisms with codomain A", then the notion of a subobject would not generalize all of these classical notions, because you get too many different subobjects that correspond to the same "sub- something". For instance, in Set, the monomorphisms ∅→{1}, {1}→{1} and {2}→{1} would be three different subobjects of the object {1}, but there are only two subsets of the set {1}. So this would be a bad definition."
$endgroup$
– Bruno Gavranovic
Jan 13 at 15:33
$begingroup$
Yes, we need to consider certain monomorphisms to represent the same subobject, i.e. the subobjects form a quotient set (class) of the monomorphisms.
$endgroup$
– Berci
Jan 13 at 15:43
add a comment |
$begingroup$
In $Set$, at least it has exactly the four subobjects you wrote up.
Note that a subobject of some object $X$ is not only a mere object $A$ in the same category, but it is understood to be together with a monomorphism $i:Ahookrightarrow X$ which plays the role of the inclusion.
Said that, indeed nothing prevents us to take ${7}$ as a subobject of ${0,1}$, moreover there are exactly two ways to do that: $i$ either sends $7mapsto 0$, or $7mapsto 1$.
So the first one represents the same subobject as ${0}hookrightarrow{0,1}$, while the second one represents the same subobject as ${1}hookrightarrow{0,1}$, but these two are distinct.
answered Jan 13 at 15:28
BerciBerci
60.8k23673
$endgroup$
In $Set$, at least it has exactly the four subobjects you wrote up.
Note that a subobject of some object $X$ is not only a mere object $A$ in the same category, but it is understood to be together with a monomorphism $i:Ahookrightarrow X$ which plays the role of the inclusion.
Said that, indeed nothing prevents us to take ${7}$ as a subobject of ${0,1}$, moreover there are exactly two ways to do that: $i$ either sends $7mapsto 0$, or $7mapsto 1$.
So the first one represents the same subobject as ${0}hookrightarrow{0,1}$, while the second one represents the same subobject as ${1}hookrightarrow{0,1}$, but these two are distinct.
answered Jan 13 at 15:28
BerciBerci
60.8k23673
answered Jan 13 at 15:28
BerciBerci
60.8k23673
answered Jan 13 at 15:28
BerciBerci
60.8k23673
answered Jan 13 at 15:28
BerciBerci
60.8k23673
60.8k23673
$begingroup$
Thank you for the answer! According to this answer (link above),there is a "good definition" where {2} isn't a subobject of {1}. --- "If you were to define "subobjects of A∈C" to mean "monomorphisms with codomain A", then the notion of a subobject would not generalize all of these classical notions, because you get too many different subobjects that correspond to the same "sub- something". For instance, in Set, the monomorphisms ∅→{1}, {1}→{1} and {2}→{1} would be three different subobjects of the object {1}, but there are only two subsets of the set {1}. So this would be a bad definition."
$endgroup$
– Bruno Gavranovic
Jan 13 at 15:33
$begingroup$
Yes, we need to consider certain monomorphisms to represent the same subobject, i.e. the subobjects form a quotient set (class) of the monomorphisms.
$endgroup$
– Berci
Jan 13 at 15:43
add a comment |
$begingroup$
Thank you for the answer! According to this answer (link above),there is a "good definition" where {2} isn't a subobject of {1}. --- "If you were to define "subobjects of A∈C" to mean "monomorphisms with codomain A", then the notion of a subobject would not generalize all of these classical notions, because you get too many different subobjects that correspond to the same "sub- something". For instance, in Set, the monomorphisms ∅→{1}, {1}→{1} and {2}→{1} would be three different subobjects of the object {1}, but there are only two subsets of the set {1}. So this would be a bad definition."
$endgroup$
– Bruno Gavranovic
Jan 13 at 15:33
$begingroup$
Yes, we need to consider certain monomorphisms to represent the same subobject, i.e. the subobjects form a quotient set (class) of the monomorphisms.
$endgroup$
– Berci
Jan 13 at 15:43
$begingroup$
Thank you for the answer! According to this answer (link above),there is a "good definition" where {2} isn't a subobject of {1}. --- "If you were to define "subobjects of A∈C" to mean "monomorphisms with codomain A", then the notion of a subobject would not generalize all of these classical notions, because you get too many different subobjects that correspond to the same "sub- something". For instance, in Set, the monomorphisms ∅→{1}, {1}→{1} and {2}→{1} would be three different subobjects of the object {1}, but there are only two subsets of the set {1}. So this would be a bad definition."
$endgroup$
– Bruno Gavranovic
Jan 13 at 15:33
$begingroup$
Thank you for the answer! According to this answer (link above),there is a "good definition" where {2} isn't a subobject of {1}. --- "If you were to define "subobjects of A∈C" to mean "monomorphisms with codomain A", then the notion of a subobject would not generalize all of these classical notions, because you get too many different subobjects that correspond to the same "sub- something". For instance, in Set, the monomorphisms ∅→{1}, {1}→{1} and {2}→{1} would be three different subobjects of the object {1}, but there are only two subsets of the set {1}. So this would be a bad definition."
$endgroup$
– Bruno Gavranovic
Jan 13 at 15:33
$begingroup$
Yes, we need to consider certain monomorphisms to represent the same subobject, i.e. the subobjects form a quotient set (class) of the monomorphisms.
$endgroup$
– Berci
Jan 13 at 15:43
$begingroup$
Yes, we need to consider certain monomorphisms to represent the same subobject, i.e. the subobjects form a quotient set (class) of the monomorphisms.
$endgroup$
– Berci
Jan 13 at 15:43
add a comment |
$begingroup$
A subobject of an object of a category is not an object of the same category, it is an equivalence class of monomorphisms to this object (see definition from your link). So $varnothing$, ${0}$, ${1}$, ${0,1}$ are not subobjects of ${0,1}$, they are its subsets. But for every set $X$ the power set $mathcal{P}(X)$ is isomorphic to the set of subobjects of $X$ in the category of sets (and they are isomorphic as ordered sets). For example, the subset ${0}subset{0,1}$ corresponds to the subobject $[i_{{0}}]$, where $i_{{0}}colon{0}to{0,1}$ is the canonical inclusion of subset and $[-]$ is taking of equivalence class of a monomorphism (as in the definition by the link). It is easy to see that if $X$, $Y$, $Z$ are sets and $fcolon Yto X$ and $gcolon Zto X$ are monomorphisms (injections), then $[f]=[g]$ as subobjects of $X$ in the category of sets if and only if their set-theoretic images are equal: $f(Y)=g(Z)$. So, for example, if $i_{{0}}colon{0}to{0,1}$ and $i_{{1}}colon{1}to{0,1}$ are canonical inclusions, then $[i_{{0}}]$ and $[i_{{1}}]$ are not equal as subobjects of ${0,1}$ in the category of sets.
answered Jan 13 at 15:34
OskarOskar
3,1431719
$endgroup$
add a comment |
$begingroup$
A subobject of an object of a category is not an object of the same category, it is an equivalence class of monomorphisms to this object (see definition from your link). So $varnothing$, ${0}$, ${1}$, ${0,1}$ are not subobjects of ${0,1}$, they are its subsets. But for every set $X$ the power set $mathcal{P}(X)$ is isomorphic to the set of subobjects of $X$ in the category of sets (and they are isomorphic as ordered sets). For example, the subset ${0}subset{0,1}$ corresponds to the subobject $[i_{{0}}]$, where $i_{{0}}colon{0}to{0,1}$ is the canonical inclusion of subset and $[-]$ is taking of equivalence class of a monomorphism (as in the definition by the link). It is easy to see that if $X$, $Y$, $Z$ are sets and $fcolon Yto X$ and $gcolon Zto X$ are monomorphisms (injections), then $[f]=[g]$ as subobjects of $X$ in the category of sets if and only if their set-theoretic images are equal: $f(Y)=g(Z)$. So, for example, if $i_{{0}}colon{0}to{0,1}$ and $i_{{1}}colon{1}to{0,1}$ are canonical inclusions, then $[i_{{0}}]$ and $[i_{{1}}]$ are not equal as subobjects of ${0,1}$ in the category of sets.
answered Jan 13 at 15:34
OskarOskar
3,1431719
$endgroup$
add a comment |
$begingroup$
A subobject of an object of a category is not an object of the same category, it is an equivalence class of monomorphisms to this object (see definition from your link). So $varnothing$, ${0}$, ${1}$, ${0,1}$ are not subobjects of ${0,1}$, they are its subsets. But for every set $X$ the power set $mathcal{P}(X)$ is isomorphic to the set of subobjects of $X$ in the category of sets (and they are isomorphic as ordered sets). For example, the subset ${0}subset{0,1}$ corresponds to the subobject $[i_{{0}}]$, where $i_{{0}}colon{0}to{0,1}$ is the canonical inclusion of subset and $[-]$ is taking of equivalence class of a monomorphism (as in the definition by the link). It is easy to see that if $X$, $Y$, $Z$ are sets and $fcolon Yto X$ and $gcolon Zto X$ are monomorphisms (injections), then $[f]=[g]$ as subobjects of $X$ in the category of sets if and only if their set-theoretic images are equal: $f(Y)=g(Z)$. So, for example, if $i_{{0}}colon{0}to{0,1}$ and $i_{{1}}colon{1}to{0,1}$ are canonical inclusions, then $[i_{{0}}]$ and $[i_{{1}}]$ are not equal as subobjects of ${0,1}$ in the category of sets.
answered Jan 13 at 15:34
OskarOskar
3,1431719
$endgroup$
A subobject of an object of a category is not an object of the same category, it is an equivalence class of monomorphisms to this object (see definition from your link). So $varnothing$, ${0}$, ${1}$, ${0,1}$ are not subobjects of ${0,1}$, they are its subsets. But for every set $X$ the power set $mathcal{P}(X)$ is isomorphic to the set of subobjects of $X$ in the category of sets (and they are isomorphic as ordered sets). For example, the subset ${0}subset{0,1}$ corresponds to the subobject $[i_{{0}}]$, where $i_{{0}}colon{0}to{0,1}$ is the canonical inclusion of subset and $[-]$ is taking of equivalence class of a monomorphism (as in the definition by the link). It is easy to see that if $X$, $Y$, $Z$ are sets and $fcolon Yto X$ and $gcolon Zto X$ are monomorphisms (injections), then $[f]=[g]$ as subobjects of $X$ in the category of sets if and only if their set-theoretic images are equal: $f(Y)=g(Z)$. So, for example, if $i_{{0}}colon{0}to{0,1}$ and $i_{{1}}colon{1}to{0,1}$ are canonical inclusions, then $[i_{{0}}]$ and $[i_{{1}}]$ are not equal as subobjects of ${0,1}$ in the category of sets.
answered Jan 13 at 15:34
OskarOskar
3,1431719
edited Jan 13 at 16:38
answered Jan 13 at 15:34
OskarOskar
3,1431719
answered Jan 13 at 15:34
OskarOskar
3,1431719
answered Jan 13 at 15:34
OskarOskar
3,1431719
3,1431719
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