How can a hom-set have a group structure?
$begingroup$
I'm trying to understand the definition of an Ab-enriched category, but I don't get how a hom-set can have a group structure. Doesn't $Hom_C(a, b)$ consist only of morphisms of the form $f: a rightarrow b$? How does then composition of morphisms in the group work when the domain and codomain don't match? Is the group operation not morphism composition or is there something I am missing here?
category-theory
asked Dec 16 '18 at 21:17
npubemnpubem
32
$endgroup$
add a comment |
$begingroup$
I'm trying to understand the definition of an Ab-enriched category, but I don't get how a hom-set can have a group structure. Doesn't $Hom_C(a, b)$ consist only of morphisms of the form $f: a rightarrow b$? How does then composition of morphisms in the group work when the domain and codomain don't match? Is the group operation not morphism composition or is there something I am missing here?
category-theory
asked Dec 16 '18 at 21:17
npubemnpubem
32
$endgroup$
$begingroup$
Think about the example of the category of Abelian groups.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 21:19
6
$begingroup$
The group structure on each hom set is extra structure when one describes an $Ab$-enriched category. It is not a result of a category structure.
$endgroup$
– Ittay Weiss
Dec 16 '18 at 21:23
$begingroup$
As an example, the abelian group structure on morphisms between abelian groups is point-wise addition of morphisms, not composition. In general, the group structure on morphisms will be something other than composition (indeed, composition is instead a group homomorphism $operatorname{Hom}_C(b,c)otimesoperatorname{Hom}_C(a,b)tooperatorname{Hom}_C(a,c).$)
$endgroup$
– Stahl
Dec 16 '18 at 21:31
add a comment |
$begingroup$
I'm trying to understand the definition of an Ab-enriched category, but I don't get how a hom-set can have a group structure. Doesn't $Hom_C(a, b)$ consist only of morphisms of the form $f: a rightarrow b$? How does then composition of morphisms in the group work when the domain and codomain don't match? Is the group operation not morphism composition or is there something I am missing here?
category-theory
asked Dec 16 '18 at 21:17
npubemnpubem
32
$endgroup$
I'm trying to understand the definition of an Ab-enriched category, but I don't get how a hom-set can have a group structure. Doesn't $Hom_C(a, b)$ consist only of morphisms of the form $f: a rightarrow b$? How does then composition of morphisms in the group work when the domain and codomain don't match? Is the group operation not morphism composition or is there something I am missing here?
category-theory
category-theory
asked Dec 16 '18 at 21:17
npubemnpubem
32
asked Dec 16 '18 at 21:17
npubemnpubem
32
asked Dec 16 '18 at 21:17
npubemnpubem
32
asked Dec 16 '18 at 21:17
npubemnpubem
32
asked Dec 16 '18 at 21:17
npubemnpubem
32
32
$begingroup$
Think about the example of the category of Abelian groups.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 21:19
6
$begingroup$
The group structure on each hom set is extra structure when one describes an $Ab$-enriched category. It is not a result of a category structure.
$endgroup$
– Ittay Weiss
Dec 16 '18 at 21:23
$begingroup$
As an example, the abelian group structure on morphisms between abelian groups is point-wise addition of morphisms, not composition. In general, the group structure on morphisms will be something other than composition (indeed, composition is instead a group homomorphism $operatorname{Hom}_C(b,c)otimesoperatorname{Hom}_C(a,b)tooperatorname{Hom}_C(a,c).$)
$endgroup$
– Stahl
Dec 16 '18 at 21:31
add a comment |
$begingroup$
Think about the example of the category of Abelian groups.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 21:19
6
$begingroup$
The group structure on each hom set is extra structure when one describes an $Ab$-enriched category. It is not a result of a category structure.
$endgroup$
– Ittay Weiss
Dec 16 '18 at 21:23
$begingroup$
As an example, the abelian group structure on morphisms between abelian groups is point-wise addition of morphisms, not composition. In general, the group structure on morphisms will be something other than composition (indeed, composition is instead a group homomorphism $operatorname{Hom}_C(b,c)otimesoperatorname{Hom}_C(a,b)tooperatorname{Hom}_C(a,c).$)
$endgroup$
– Stahl
Dec 16 '18 at 21:31
$begingroup$
Think about the example of the category of Abelian groups.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 21:19
$begingroup$
Think about the example of the category of Abelian groups.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 21:19
6
6
$begingroup$
The group structure on each hom set is extra structure when one describes an $Ab$-enriched category. It is not a result of a category structure.
$endgroup$
– Ittay Weiss
Dec 16 '18 at 21:23
$begingroup$
The group structure on each hom set is extra structure when one describes an $Ab$-enriched category. It is not a result of a category structure.
$endgroup$
– Ittay Weiss
Dec 16 '18 at 21:23
$begingroup$
As an example, the abelian group structure on morphisms between abelian groups is point-wise addition of morphisms, not composition. In general, the group structure on morphisms will be something other than composition (indeed, composition is instead a group homomorphism $operatorname{Hom}_C(b,c)otimesoperatorname{Hom}_C(a,b)tooperatorname{Hom}_C(a,c).$)
$endgroup$
– Stahl
Dec 16 '18 at 21:31
$begingroup$
As an example, the abelian group structure on morphisms between abelian groups is point-wise addition of morphisms, not composition. In general, the group structure on morphisms will be something other than composition (indeed, composition is instead a group homomorphism $operatorname{Hom}_C(b,c)otimesoperatorname{Hom}_C(a,b)tooperatorname{Hom}_C(a,c).$)
$endgroup$
– Stahl
Dec 16 '18 at 21:31
add a comment |
2 Answers
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$begingroup$
Indeed, the group operation is not composition of morphisms. It is just some group operation, which is specified as part of the $Ab$-enrichment. In typical examples, it is the operation of "pointwise addition of homomorphisms". For instance, if $A$ and $B$ are abelian groups (so our category is $Ab$ itself), then the set $operatorname{Hom}_{Ab}(A,B)$ of homomorphisms $Ato B$ has a natural abelian group structure: if $f,g:Ato B$ are homomorphisms, then the function $f+g$ defined by $(f+g)(x)=f(x)+g(x)$ is also a homomorphism. This group structure on $operatorname{Hom}_{Ab}(A,B)$ for each $A,B$ makes $Ab$ into an $Ab$-enriched category.
answered Dec 16 '18 at 21:39
Eric WofseyEric Wofsey
186k14215342
$endgroup$
add a comment |
$begingroup$
I believe you are studying the notion of group object in a category $C$; $G$ is a group object if $Hom(X,G)$ has a structure of group and $Xrightarrow Hom(X,G)$ is a contravariant functor to the category of groups.
Suppose for example $C$ is the category of sets, and $G$ is a group, you endow $Hom(X,G)$ with the structure defined by $(f.g)(x)=f(x).g(x)$. The neutral is the constant map $f(x)=1_G$ and the inverse of $f$ is $(f^{-1})(x)=f(x)^{-1}$.
answered Dec 16 '18 at 21:23
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
3
$begingroup$
I don't think this is in fact what the OP is asking, although it's of course related.
$endgroup$
– Noah Schweber
Dec 16 '18 at 21:41
add a comment |
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$begingroup$
Indeed, the group operation is not composition of morphisms. It is just some group operation, which is specified as part of the $Ab$-enrichment. In typical examples, it is the operation of "pointwise addition of homomorphisms". For instance, if $A$ and $B$ are abelian groups (so our category is $Ab$ itself), then the set $operatorname{Hom}_{Ab}(A,B)$ of homomorphisms $Ato B$ has a natural abelian group structure: if $f,g:Ato B$ are homomorphisms, then the function $f+g$ defined by $(f+g)(x)=f(x)+g(x)$ is also a homomorphism. This group structure on $operatorname{Hom}_{Ab}(A,B)$ for each $A,B$ makes $Ab$ into an $Ab$-enriched category.
answered Dec 16 '18 at 21:39
Eric WofseyEric Wofsey
186k14215342
$endgroup$
add a comment |
$begingroup$
Indeed, the group operation is not composition of morphisms. It is just some group operation, which is specified as part of the $Ab$-enrichment. In typical examples, it is the operation of "pointwise addition of homomorphisms". For instance, if $A$ and $B$ are abelian groups (so our category is $Ab$ itself), then the set $operatorname{Hom}_{Ab}(A,B)$ of homomorphisms $Ato B$ has a natural abelian group structure: if $f,g:Ato B$ are homomorphisms, then the function $f+g$ defined by $(f+g)(x)=f(x)+g(x)$ is also a homomorphism. This group structure on $operatorname{Hom}_{Ab}(A,B)$ for each $A,B$ makes $Ab$ into an $Ab$-enriched category.
answered Dec 16 '18 at 21:39
Eric WofseyEric Wofsey
186k14215342
$endgroup$
add a comment |
$begingroup$
Indeed, the group operation is not composition of morphisms. It is just some group operation, which is specified as part of the $Ab$-enrichment. In typical examples, it is the operation of "pointwise addition of homomorphisms". For instance, if $A$ and $B$ are abelian groups (so our category is $Ab$ itself), then the set $operatorname{Hom}_{Ab}(A,B)$ of homomorphisms $Ato B$ has a natural abelian group structure: if $f,g:Ato B$ are homomorphisms, then the function $f+g$ defined by $(f+g)(x)=f(x)+g(x)$ is also a homomorphism. This group structure on $operatorname{Hom}_{Ab}(A,B)$ for each $A,B$ makes $Ab$ into an $Ab$-enriched category.
answered Dec 16 '18 at 21:39
Eric WofseyEric Wofsey
186k14215342
$endgroup$
Indeed, the group operation is not composition of morphisms. It is just some group operation, which is specified as part of the $Ab$-enrichment. In typical examples, it is the operation of "pointwise addition of homomorphisms". For instance, if $A$ and $B$ are abelian groups (so our category is $Ab$ itself), then the set $operatorname{Hom}_{Ab}(A,B)$ of homomorphisms $Ato B$ has a natural abelian group structure: if $f,g:Ato B$ are homomorphisms, then the function $f+g$ defined by $(f+g)(x)=f(x)+g(x)$ is also a homomorphism. This group structure on $operatorname{Hom}_{Ab}(A,B)$ for each $A,B$ makes $Ab$ into an $Ab$-enriched category.
answered Dec 16 '18 at 21:39
Eric WofseyEric Wofsey
186k14215342
answered Dec 16 '18 at 21:39
Eric WofseyEric Wofsey
186k14215342
answered Dec 16 '18 at 21:39
Eric WofseyEric Wofsey
186k14215342
answered Dec 16 '18 at 21:39
Eric WofseyEric Wofsey
186k14215342
186k14215342
add a comment |
add a comment |
$begingroup$
I believe you are studying the notion of group object in a category $C$; $G$ is a group object if $Hom(X,G)$ has a structure of group and $Xrightarrow Hom(X,G)$ is a contravariant functor to the category of groups.
Suppose for example $C$ is the category of sets, and $G$ is a group, you endow $Hom(X,G)$ with the structure defined by $(f.g)(x)=f(x).g(x)$. The neutral is the constant map $f(x)=1_G$ and the inverse of $f$ is $(f^{-1})(x)=f(x)^{-1}$.
answered Dec 16 '18 at 21:23
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
3
$begingroup$
I don't think this is in fact what the OP is asking, although it's of course related.
$endgroup$
– Noah Schweber
Dec 16 '18 at 21:41
add a comment |
$begingroup$
I believe you are studying the notion of group object in a category $C$; $G$ is a group object if $Hom(X,G)$ has a structure of group and $Xrightarrow Hom(X,G)$ is a contravariant functor to the category of groups.
Suppose for example $C$ is the category of sets, and $G$ is a group, you endow $Hom(X,G)$ with the structure defined by $(f.g)(x)=f(x).g(x)$. The neutral is the constant map $f(x)=1_G$ and the inverse of $f$ is $(f^{-1})(x)=f(x)^{-1}$.
answered Dec 16 '18 at 21:23
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
3
$begingroup$
I don't think this is in fact what the OP is asking, although it's of course related.
$endgroup$
– Noah Schweber
Dec 16 '18 at 21:41
add a comment |
$begingroup$
I believe you are studying the notion of group object in a category $C$; $G$ is a group object if $Hom(X,G)$ has a structure of group and $Xrightarrow Hom(X,G)$ is a contravariant functor to the category of groups.
Suppose for example $C$ is the category of sets, and $G$ is a group, you endow $Hom(X,G)$ with the structure defined by $(f.g)(x)=f(x).g(x)$. The neutral is the constant map $f(x)=1_G$ and the inverse of $f$ is $(f^{-1})(x)=f(x)^{-1}$.
answered Dec 16 '18 at 21:23
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
I believe you are studying the notion of group object in a category $C$; $G$ is a group object if $Hom(X,G)$ has a structure of group and $Xrightarrow Hom(X,G)$ is a contravariant functor to the category of groups.
Suppose for example $C$ is the category of sets, and $G$ is a group, you endow $Hom(X,G)$ with the structure defined by $(f.g)(x)=f(x).g(x)$. The neutral is the constant map $f(x)=1_G$ and the inverse of $f$ is $(f^{-1})(x)=f(x)^{-1}$.
answered Dec 16 '18 at 21:23
Tsemo AristideTsemo Aristide
58.3k11445
answered Dec 16 '18 at 21:23
Tsemo AristideTsemo Aristide
58.3k11445
answered Dec 16 '18 at 21:23
Tsemo AristideTsemo Aristide
58.3k11445
answered Dec 16 '18 at 21:23
Tsemo AristideTsemo Aristide
58.3k11445
58.3k11445
3
$begingroup$
I don't think this is in fact what the OP is asking, although it's of course related.
$endgroup$
– Noah Schweber
Dec 16 '18 at 21:41
add a comment |
3
$begingroup$
I don't think this is in fact what the OP is asking, although it's of course related.
$endgroup$
– Noah Schweber
Dec 16 '18 at 21:41
3
3
$begingroup$
I don't think this is in fact what the OP is asking, although it's of course related.
$endgroup$
– Noah Schweber
Dec 16 '18 at 21:41
$begingroup$
I don't think this is in fact what the OP is asking, although it's of course related.
$endgroup$
– Noah Schweber
Dec 16 '18 at 21:41
add a comment |
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$begingroup$
Think about the example of the category of Abelian groups.
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 21:19
6
$begingroup$
The group structure on each hom set is extra structure when one describes an $Ab$-enriched category. It is not a result of a category structure.
$endgroup$
– Ittay Weiss
Dec 16 '18 at 21:23
$begingroup$
As an example, the abelian group structure on morphisms between abelian groups is point-wise addition of morphisms, not composition. In general, the group structure on morphisms will be something other than composition (indeed, composition is instead a group homomorphism $operatorname{Hom}_C(b,c)otimesoperatorname{Hom}_C(a,b)tooperatorname{Hom}_C(a,c).$)
$endgroup$
– Stahl
Dec 16 '18 at 21:31