If $A$ abelian category and $X$ a small category (a set) then $Fun(X,A)$ is abelian category.
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I managed to show that $Fun(X,A)={functors:X->A}$ is a category and i am trying to show that it is abelian.It is easy to show that is pre-additive but i can't define the zero object of the category,to prove that every homomorphism has a kernel and cokernel and finally that the induced homomorphism $f^{-}:CoImf->Imf$ is isomorphism for every $f$ homomorphism.I actually have a problem on how to use the zero object of A to define the zero object of $Fun(X,A)$.I would appreciate any help.
category-theory
asked Nov 22 at 20:12
T.Karawolf
235
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I managed to show that $Fun(X,A)={functors:X->A}$ is a category and i am trying to show that it is abelian.It is easy to show that is pre-additive but i can't define the zero object of the category,to prove that every homomorphism has a kernel and cokernel and finally that the induced homomorphism $f^{-}:CoImf->Imf$ is isomorphism for every $f$ homomorphism.I actually have a problem on how to use the zero object of A to define the zero object of $Fun(X,A)$.I would appreciate any help.
category-theory
asked Nov 22 at 20:12
T.Karawolf
235
1
Limits and colimits are always computed pointwise in functor categories, so the terminal object is the pointwise terminal object, kernels are pointwise, etc.
– Qiaochu Yuan
Nov 22 at 20:22
To define a zero object of $Fun(X,A)$,shouldn't i define an object $o$ of $Fun(X,A)$,a Functor,such that for every other functor F the sets $Hom(F,o)$ and $Hom(o,F)$ have only one element?I can't understand you @Qiaochu Yuan,maybe because i am a new student in category theory and i don't know all the terminology.
– T.Karawolf
Nov 22 at 20:42
Yes, that's the definition. The zero object is both a terminal object and an initial object, so once you've computed both of these and verified that they're the same you've found the zero object. And computing terminal and initial objects is a special case of computing limits and colimits, and those are always computed pointwise. What this all adds up to is that the zero object in $text{Fun}(X, A)$ is the constant functor with constant value the zero object in $A$; check this.
– Qiaochu Yuan
Nov 22 at 20:44
oh that was so simple anyway thank you
– T.Karawolf
Nov 22 at 21:06
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I managed to show that $Fun(X,A)={functors:X->A}$ is a category and i am trying to show that it is abelian.It is easy to show that is pre-additive but i can't define the zero object of the category,to prove that every homomorphism has a kernel and cokernel and finally that the induced homomorphism $f^{-}:CoImf->Imf$ is isomorphism for every $f$ homomorphism.I actually have a problem on how to use the zero object of A to define the zero object of $Fun(X,A)$.I would appreciate any help.
category-theory
asked Nov 22 at 20:12
T.Karawolf
235
I managed to show that $Fun(X,A)={functors:X->A}$ is a category and i am trying to show that it is abelian.It is easy to show that is pre-additive but i can't define the zero object of the category,to prove that every homomorphism has a kernel and cokernel and finally that the induced homomorphism $f^{-}:CoImf->Imf$ is isomorphism for every $f$ homomorphism.I actually have a problem on how to use the zero object of A to define the zero object of $Fun(X,A)$.I would appreciate any help.
category-theory
category-theory
asked Nov 22 at 20:12
T.Karawolf
235
asked Nov 22 at 20:12
T.Karawolf
235
asked Nov 22 at 20:12
T.Karawolf
235
asked Nov 22 at 20:12
T.Karawolf
235
asked Nov 22 at 20:12
T.Karawolf
235
235
1
Limits and colimits are always computed pointwise in functor categories, so the terminal object is the pointwise terminal object, kernels are pointwise, etc.
– Qiaochu Yuan
Nov 22 at 20:22
To define a zero object of $Fun(X,A)$,shouldn't i define an object $o$ of $Fun(X,A)$,a Functor,such that for every other functor F the sets $Hom(F,o)$ and $Hom(o,F)$ have only one element?I can't understand you @Qiaochu Yuan,maybe because i am a new student in category theory and i don't know all the terminology.
– T.Karawolf
Nov 22 at 20:42
Yes, that's the definition. The zero object is both a terminal object and an initial object, so once you've computed both of these and verified that they're the same you've found the zero object. And computing terminal and initial objects is a special case of computing limits and colimits, and those are always computed pointwise. What this all adds up to is that the zero object in $text{Fun}(X, A)$ is the constant functor with constant value the zero object in $A$; check this.
– Qiaochu Yuan
Nov 22 at 20:44
oh that was so simple anyway thank you
– T.Karawolf
Nov 22 at 21:06
add a comment |
1
Limits and colimits are always computed pointwise in functor categories, so the terminal object is the pointwise terminal object, kernels are pointwise, etc.
– Qiaochu Yuan
Nov 22 at 20:22
To define a zero object of $Fun(X,A)$,shouldn't i define an object $o$ of $Fun(X,A)$,a Functor,such that for every other functor F the sets $Hom(F,o)$ and $Hom(o,F)$ have only one element?I can't understand you @Qiaochu Yuan,maybe because i am a new student in category theory and i don't know all the terminology.
– T.Karawolf
Nov 22 at 20:42
Yes, that's the definition. The zero object is both a terminal object and an initial object, so once you've computed both of these and verified that they're the same you've found the zero object. And computing terminal and initial objects is a special case of computing limits and colimits, and those are always computed pointwise. What this all adds up to is that the zero object in $text{Fun}(X, A)$ is the constant functor with constant value the zero object in $A$; check this.
– Qiaochu Yuan
Nov 22 at 20:44
oh that was so simple anyway thank you
– T.Karawolf
Nov 22 at 21:06
1
1
Limits and colimits are always computed pointwise in functor categories, so the terminal object is the pointwise terminal object, kernels are pointwise, etc.
– Qiaochu Yuan
Nov 22 at 20:22
Limits and colimits are always computed pointwise in functor categories, so the terminal object is the pointwise terminal object, kernels are pointwise, etc.
– Qiaochu Yuan
Nov 22 at 20:22
To define a zero object of $Fun(X,A)$,shouldn't i define an object $o$ of $Fun(X,A)$,a Functor,such that for every other functor F the sets $Hom(F,o)$ and $Hom(o,F)$ have only one element?I can't understand you @Qiaochu Yuan,maybe because i am a new student in category theory and i don't know all the terminology.
– T.Karawolf
Nov 22 at 20:42
To define a zero object of $Fun(X,A)$,shouldn't i define an object $o$ of $Fun(X,A)$,a Functor,such that for every other functor F the sets $Hom(F,o)$ and $Hom(o,F)$ have only one element?I can't understand you @Qiaochu Yuan,maybe because i am a new student in category theory and i don't know all the terminology.
– T.Karawolf
Nov 22 at 20:42
Yes, that's the definition. The zero object is both a terminal object and an initial object, so once you've computed both of these and verified that they're the same you've found the zero object. And computing terminal and initial objects is a special case of computing limits and colimits, and those are always computed pointwise. What this all adds up to is that the zero object in $text{Fun}(X, A)$ is the constant functor with constant value the zero object in $A$; check this.
– Qiaochu Yuan
Nov 22 at 20:44
Yes, that's the definition. The zero object is both a terminal object and an initial object, so once you've computed both of these and verified that they're the same you've found the zero object. And computing terminal and initial objects is a special case of computing limits and colimits, and those are always computed pointwise. What this all adds up to is that the zero object in $text{Fun}(X, A)$ is the constant functor with constant value the zero object in $A$; check this.
– Qiaochu Yuan
Nov 22 at 20:44
oh that was so simple anyway thank you
– T.Karawolf
Nov 22 at 21:06
oh that was so simple anyway thank you
– T.Karawolf
Nov 22 at 21:06
add a comment |
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1
Limits and colimits are always computed pointwise in functor categories, so the terminal object is the pointwise terminal object, kernels are pointwise, etc.
– Qiaochu Yuan
Nov 22 at 20:22
To define a zero object of $Fun(X,A)$,shouldn't i define an object $o$ of $Fun(X,A)$,a Functor,such that for every other functor F the sets $Hom(F,o)$ and $Hom(o,F)$ have only one element?I can't understand you @Qiaochu Yuan,maybe because i am a new student in category theory and i don't know all the terminology.
– T.Karawolf
Nov 22 at 20:42
Yes, that's the definition. The zero object is both a terminal object and an initial object, so once you've computed both of these and verified that they're the same you've found the zero object. And computing terminal and initial objects is a special case of computing limits and colimits, and those are always computed pointwise. What this all adds up to is that the zero object in $text{Fun}(X, A)$ is the constant functor with constant value the zero object in $A$; check this.
– Qiaochu Yuan
Nov 22 at 20:44
oh that was so simple anyway thank you
– T.Karawolf
Nov 22 at 21:06