Definition of generator in an abelian category.
$begingroup$
Let $mathcal A$ be an abelian category. Let an object $G$ in $mathcal A$ be such that $Homleft(G,unicode{x2013} right)$ is a faithful functor from $mathcal A$ to the category of sets.(I assume that I am working with a category where any family of objects has their coproduct existing inside the category)
Why is the above equivalent to the fact that every object $X$ in $mathcal A$ admits an epimorphism $G^I$ to $X$? (where $I$ is the index category and is arbitrary)(where,$G^I$ is coproduct of copies of $G$ which exists in that category)
I do not see how injectivity of maps getting translated to epimorphism and vice versa? I guess I should start with appropriate short exact sequence and apply appropriate functor to make injectivity translated to epimorphism but I have no clue how to proceed practically, i.e how to make it precise.
Any help from anyone is welcome.
category-theory abelian-categories
asked Dec 3 '18 at 4:52
HARRYHARRY
889
$endgroup$
add a comment |
$begingroup$
Let $mathcal A$ be an abelian category. Let an object $G$ in $mathcal A$ be such that $Homleft(G,unicode{x2013} right)$ is a faithful functor from $mathcal A$ to the category of sets.(I assume that I am working with a category where any family of objects has their coproduct existing inside the category)
Why is the above equivalent to the fact that every object $X$ in $mathcal A$ admits an epimorphism $G^I$ to $X$? (where $I$ is the index category and is arbitrary)(where,$G^I$ is coproduct of copies of $G$ which exists in that category)
I do not see how injectivity of maps getting translated to epimorphism and vice versa? I guess I should start with appropriate short exact sequence and apply appropriate functor to make injectivity translated to epimorphism but I have no clue how to proceed practically, i.e how to make it precise.
Any help from anyone is welcome.
category-theory abelian-categories
asked Dec 3 '18 at 4:52
HARRYHARRY
889
$endgroup$
$begingroup$
I'm not sure it is. In general Abelian categories one cannot always take arbitrary powers of objects.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 5:01
$begingroup$
@lord-shark-the-unknown,sorry I should have mentioned that I am assuming in that category every family of objects(even possibly infinite) has its coproduct in that category.
$endgroup$
– HARRY
Dec 3 '18 at 5:11
1
$begingroup$
$G^I$ is a weird notation for a possibly infinite coproduct
$endgroup$
– Max
Dec 3 '18 at 8:11
add a comment |
$begingroup$
Let $mathcal A$ be an abelian category. Let an object $G$ in $mathcal A$ be such that $Homleft(G,unicode{x2013} right)$ is a faithful functor from $mathcal A$ to the category of sets.(I assume that I am working with a category where any family of objects has their coproduct existing inside the category)
Why is the above equivalent to the fact that every object $X$ in $mathcal A$ admits an epimorphism $G^I$ to $X$? (where $I$ is the index category and is arbitrary)(where,$G^I$ is coproduct of copies of $G$ which exists in that category)
I do not see how injectivity of maps getting translated to epimorphism and vice versa? I guess I should start with appropriate short exact sequence and apply appropriate functor to make injectivity translated to epimorphism but I have no clue how to proceed practically, i.e how to make it precise.
Any help from anyone is welcome.
category-theory abelian-categories
asked Dec 3 '18 at 4:52
HARRYHARRY
889
$endgroup$
Let $mathcal A$ be an abelian category. Let an object $G$ in $mathcal A$ be such that $Homleft(G,unicode{x2013} right)$ is a faithful functor from $mathcal A$ to the category of sets.(I assume that I am working with a category where any family of objects has their coproduct existing inside the category)
Why is the above equivalent to the fact that every object $X$ in $mathcal A$ admits an epimorphism $G^I$ to $X$? (where $I$ is the index category and is arbitrary)(where,$G^I$ is coproduct of copies of $G$ which exists in that category)
I do not see how injectivity of maps getting translated to epimorphism and vice versa? I guess I should start with appropriate short exact sequence and apply appropriate functor to make injectivity translated to epimorphism but I have no clue how to proceed practically, i.e how to make it precise.
Any help from anyone is welcome.
category-theory abelian-categories
category-theory abelian-categories
asked Dec 3 '18 at 4:52
HARRYHARRY
889
asked Dec 3 '18 at 4:52
HARRYHARRY
889
edited Dec 3 '18 at 5:14
HARRY
asked Dec 3 '18 at 4:52
HARRYHARRY
889
asked Dec 3 '18 at 4:52
HARRYHARRY
889
asked Dec 3 '18 at 4:52
HARRYHARRY
889
889
$begingroup$
I'm not sure it is. In general Abelian categories one cannot always take arbitrary powers of objects.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 5:01
$begingroup$
@lord-shark-the-unknown,sorry I should have mentioned that I am assuming in that category every family of objects(even possibly infinite) has its coproduct in that category.
$endgroup$
– HARRY
Dec 3 '18 at 5:11
1
$begingroup$
$G^I$ is a weird notation for a possibly infinite coproduct
$endgroup$
– Max
Dec 3 '18 at 8:11
add a comment |
$begingroup$
I'm not sure it is. In general Abelian categories one cannot always take arbitrary powers of objects.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 5:01
$begingroup$
@lord-shark-the-unknown,sorry I should have mentioned that I am assuming in that category every family of objects(even possibly infinite) has its coproduct in that category.
$endgroup$
– HARRY
Dec 3 '18 at 5:11
1
$begingroup$
$G^I$ is a weird notation for a possibly infinite coproduct
$endgroup$
– Max
Dec 3 '18 at 8:11
$begingroup$
I'm not sure it is. In general Abelian categories one cannot always take arbitrary powers of objects.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 5:01
$begingroup$
I'm not sure it is. In general Abelian categories one cannot always take arbitrary powers of objects.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 5:01
$begingroup$
@lord-shark-the-unknown,sorry I should have mentioned that I am assuming in that category every family of objects(even possibly infinite) has its coproduct in that category.
$endgroup$
– HARRY
Dec 3 '18 at 5:11
$begingroup$
@lord-shark-the-unknown,sorry I should have mentioned that I am assuming in that category every family of objects(even possibly infinite) has its coproduct in that category.
$endgroup$
– HARRY
Dec 3 '18 at 5:11
1
1
$begingroup$
$G^I$ is a weird notation for a possibly infinite coproduct
$endgroup$
– Max
Dec 3 '18 at 8:11
$begingroup$
$G^I$ is a weird notation for a possibly infinite coproduct
$endgroup$
– Max
Dec 3 '18 at 8:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose there is an epimorphism $p:G^Ito X$ and suppose $f:Xto Y$ is a morphism which becomes $0$ after applying the functor $operatorname{Hom}(G,-)$. This means that for every morphism $g:Gto X$, $fg=0$. In particular, by taking $g$ to be each of the inclusion maps $Gto G^I$ composed with $p$, this implies $fp=0$. Since $p$ is an epimorphism, this implies $f=0$. Thus if such an epimorphism $p$ exists for every $X$, $operatorname{Hom}(G,-)$ is faithful.
Conversely, suppose $operatorname{Hom}(G,-)$ is faithful and let $X$ be an object. Let $I=operatorname{Hom}(G,X)$ and let $p:G^Ito X$ be the unique morphism such that for each $iin I$, the composition of $p$ with the $i$th inclusion map $Gto G^I$ is $i:Gto X$. I claim $p$ is an epimorphism. To show this, it suffices to show that if $f:Xto Y$ is a morphism such that $fp=0$, then $f=0$. But given any such $f$, by composing with the inclusion maps $Gto G^I$ we see that $fi=0$ for all $i:Gto X$. This means that $operatorname{Hom}(G,-)$ sends $f$ to $0$ and thus $f=0$ since $operatorname{Hom}(G,-)$ is faithful.
answered Dec 3 '18 at 5:36
Eric WofseyEric Wofsey
182k13209337
$endgroup$
$begingroup$
I first thought your answer was wrong because of the notation $G^I$ which seems weird for a (possibly infinite) coproduct. Perhaps you could write a word about it ?
$endgroup$
– Max
Dec 3 '18 at 8:12
$begingroup$
@Max,I tried to put $(I)$ there but somehow it did not work.sorry for the notation,but later I explained what it stands for.
$endgroup$
– HARRY
Dec 3 '18 at 8:40
$begingroup$
@HARRY : yes of course, that's how I realized that the answer was actually correct (and having seen Eric's answers over time it seemed weird that he'd make this sort of mistake)
$endgroup$
– Max
Dec 3 '18 at 8:50
add a comment |
Your Answer
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$begingroup$
Suppose there is an epimorphism $p:G^Ito X$ and suppose $f:Xto Y$ is a morphism which becomes $0$ after applying the functor $operatorname{Hom}(G,-)$. This means that for every morphism $g:Gto X$, $fg=0$. In particular, by taking $g$ to be each of the inclusion maps $Gto G^I$ composed with $p$, this implies $fp=0$. Since $p$ is an epimorphism, this implies $f=0$. Thus if such an epimorphism $p$ exists for every $X$, $operatorname{Hom}(G,-)$ is faithful.
Conversely, suppose $operatorname{Hom}(G,-)$ is faithful and let $X$ be an object. Let $I=operatorname{Hom}(G,X)$ and let $p:G^Ito X$ be the unique morphism such that for each $iin I$, the composition of $p$ with the $i$th inclusion map $Gto G^I$ is $i:Gto X$. I claim $p$ is an epimorphism. To show this, it suffices to show that if $f:Xto Y$ is a morphism such that $fp=0$, then $f=0$. But given any such $f$, by composing with the inclusion maps $Gto G^I$ we see that $fi=0$ for all $i:Gto X$. This means that $operatorname{Hom}(G,-)$ sends $f$ to $0$ and thus $f=0$ since $operatorname{Hom}(G,-)$ is faithful.
answered Dec 3 '18 at 5:36
Eric WofseyEric Wofsey
182k13209337
$endgroup$
$begingroup$
I first thought your answer was wrong because of the notation $G^I$ which seems weird for a (possibly infinite) coproduct. Perhaps you could write a word about it ?
$endgroup$
– Max
Dec 3 '18 at 8:12
$begingroup$
@Max,I tried to put $(I)$ there but somehow it did not work.sorry for the notation,but later I explained what it stands for.
$endgroup$
– HARRY
Dec 3 '18 at 8:40
$begingroup$
@HARRY : yes of course, that's how I realized that the answer was actually correct (and having seen Eric's answers over time it seemed weird that he'd make this sort of mistake)
$endgroup$
– Max
Dec 3 '18 at 8:50
add a comment |
$begingroup$
Suppose there is an epimorphism $p:G^Ito X$ and suppose $f:Xto Y$ is a morphism which becomes $0$ after applying the functor $operatorname{Hom}(G,-)$. This means that for every morphism $g:Gto X$, $fg=0$. In particular, by taking $g$ to be each of the inclusion maps $Gto G^I$ composed with $p$, this implies $fp=0$. Since $p$ is an epimorphism, this implies $f=0$. Thus if such an epimorphism $p$ exists for every $X$, $operatorname{Hom}(G,-)$ is faithful.
Conversely, suppose $operatorname{Hom}(G,-)$ is faithful and let $X$ be an object. Let $I=operatorname{Hom}(G,X)$ and let $p:G^Ito X$ be the unique morphism such that for each $iin I$, the composition of $p$ with the $i$th inclusion map $Gto G^I$ is $i:Gto X$. I claim $p$ is an epimorphism. To show this, it suffices to show that if $f:Xto Y$ is a morphism such that $fp=0$, then $f=0$. But given any such $f$, by composing with the inclusion maps $Gto G^I$ we see that $fi=0$ for all $i:Gto X$. This means that $operatorname{Hom}(G,-)$ sends $f$ to $0$ and thus $f=0$ since $operatorname{Hom}(G,-)$ is faithful.
answered Dec 3 '18 at 5:36
Eric WofseyEric Wofsey
182k13209337
$endgroup$
$begingroup$
I first thought your answer was wrong because of the notation $G^I$ which seems weird for a (possibly infinite) coproduct. Perhaps you could write a word about it ?
$endgroup$
– Max
Dec 3 '18 at 8:12
$begingroup$
@Max,I tried to put $(I)$ there but somehow it did not work.sorry for the notation,but later I explained what it stands for.
$endgroup$
– HARRY
Dec 3 '18 at 8:40
$begingroup$
@HARRY : yes of course, that's how I realized that the answer was actually correct (and having seen Eric's answers over time it seemed weird that he'd make this sort of mistake)
$endgroup$
– Max
Dec 3 '18 at 8:50
add a comment |
$begingroup$
Suppose there is an epimorphism $p:G^Ito X$ and suppose $f:Xto Y$ is a morphism which becomes $0$ after applying the functor $operatorname{Hom}(G,-)$. This means that for every morphism $g:Gto X$, $fg=0$. In particular, by taking $g$ to be each of the inclusion maps $Gto G^I$ composed with $p$, this implies $fp=0$. Since $p$ is an epimorphism, this implies $f=0$. Thus if such an epimorphism $p$ exists for every $X$, $operatorname{Hom}(G,-)$ is faithful.
Conversely, suppose $operatorname{Hom}(G,-)$ is faithful and let $X$ be an object. Let $I=operatorname{Hom}(G,X)$ and let $p:G^Ito X$ be the unique morphism such that for each $iin I$, the composition of $p$ with the $i$th inclusion map $Gto G^I$ is $i:Gto X$. I claim $p$ is an epimorphism. To show this, it suffices to show that if $f:Xto Y$ is a morphism such that $fp=0$, then $f=0$. But given any such $f$, by composing with the inclusion maps $Gto G^I$ we see that $fi=0$ for all $i:Gto X$. This means that $operatorname{Hom}(G,-)$ sends $f$ to $0$ and thus $f=0$ since $operatorname{Hom}(G,-)$ is faithful.
answered Dec 3 '18 at 5:36
Eric WofseyEric Wofsey
182k13209337
$endgroup$
Suppose there is an epimorphism $p:G^Ito X$ and suppose $f:Xto Y$ is a morphism which becomes $0$ after applying the functor $operatorname{Hom}(G,-)$. This means that for every morphism $g:Gto X$, $fg=0$. In particular, by taking $g$ to be each of the inclusion maps $Gto G^I$ composed with $p$, this implies $fp=0$. Since $p$ is an epimorphism, this implies $f=0$. Thus if such an epimorphism $p$ exists for every $X$, $operatorname{Hom}(G,-)$ is faithful.
Conversely, suppose $operatorname{Hom}(G,-)$ is faithful and let $X$ be an object. Let $I=operatorname{Hom}(G,X)$ and let $p:G^Ito X$ be the unique morphism such that for each $iin I$, the composition of $p$ with the $i$th inclusion map $Gto G^I$ is $i:Gto X$. I claim $p$ is an epimorphism. To show this, it suffices to show that if $f:Xto Y$ is a morphism such that $fp=0$, then $f=0$. But given any such $f$, by composing with the inclusion maps $Gto G^I$ we see that $fi=0$ for all $i:Gto X$. This means that $operatorname{Hom}(G,-)$ sends $f$ to $0$ and thus $f=0$ since $operatorname{Hom}(G,-)$ is faithful.
answered Dec 3 '18 at 5:36
Eric WofseyEric Wofsey
182k13209337
answered Dec 3 '18 at 5:36
Eric WofseyEric Wofsey
182k13209337
answered Dec 3 '18 at 5:36
Eric WofseyEric Wofsey
182k13209337
answered Dec 3 '18 at 5:36
Eric WofseyEric Wofsey
182k13209337
182k13209337
$begingroup$
I first thought your answer was wrong because of the notation $G^I$ which seems weird for a (possibly infinite) coproduct. Perhaps you could write a word about it ?
$endgroup$
– Max
Dec 3 '18 at 8:12
$begingroup$
@Max,I tried to put $(I)$ there but somehow it did not work.sorry for the notation,but later I explained what it stands for.
$endgroup$
– HARRY
Dec 3 '18 at 8:40
$begingroup$
@HARRY : yes of course, that's how I realized that the answer was actually correct (and having seen Eric's answers over time it seemed weird that he'd make this sort of mistake)
$endgroup$
– Max
Dec 3 '18 at 8:50
add a comment |
$begingroup$
I first thought your answer was wrong because of the notation $G^I$ which seems weird for a (possibly infinite) coproduct. Perhaps you could write a word about it ?
$endgroup$
– Max
Dec 3 '18 at 8:12
$begingroup$
@Max,I tried to put $(I)$ there but somehow it did not work.sorry for the notation,but later I explained what it stands for.
$endgroup$
– HARRY
Dec 3 '18 at 8:40
$begingroup$
@HARRY : yes of course, that's how I realized that the answer was actually correct (and having seen Eric's answers over time it seemed weird that he'd make this sort of mistake)
$endgroup$
– Max
Dec 3 '18 at 8:50
$begingroup$
I first thought your answer was wrong because of the notation $G^I$ which seems weird for a (possibly infinite) coproduct. Perhaps you could write a word about it ?
$endgroup$
– Max
Dec 3 '18 at 8:12
$begingroup$
I first thought your answer was wrong because of the notation $G^I$ which seems weird for a (possibly infinite) coproduct. Perhaps you could write a word about it ?
$endgroup$
– Max
Dec 3 '18 at 8:12
$begingroup$
@Max,I tried to put $(I)$ there but somehow it did not work.sorry for the notation,but later I explained what it stands for.
$endgroup$
– HARRY
Dec 3 '18 at 8:40
$begingroup$
@Max,I tried to put $(I)$ there but somehow it did not work.sorry for the notation,but later I explained what it stands for.
$endgroup$
– HARRY
Dec 3 '18 at 8:40
$begingroup$
@HARRY : yes of course, that's how I realized that the answer was actually correct (and having seen Eric's answers over time it seemed weird that he'd make this sort of mistake)
$endgroup$
– Max
Dec 3 '18 at 8:50
$begingroup$
@HARRY : yes of course, that's how I realized that the answer was actually correct (and having seen Eric's answers over time it seemed weird that he'd make this sort of mistake)
$endgroup$
– Max
Dec 3 '18 at 8:50
add a comment |
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$begingroup$
I'm not sure it is. In general Abelian categories one cannot always take arbitrary powers of objects.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 5:01
$begingroup$
@lord-shark-the-unknown,sorry I should have mentioned that I am assuming in that category every family of objects(even possibly infinite) has its coproduct in that category.
$endgroup$
– HARRY
Dec 3 '18 at 5:11
1
$begingroup$
$G^I$ is a weird notation for a possibly infinite coproduct
$endgroup$
– Max
Dec 3 '18 at 8:11