Is it true that $C=operatornamecoker(f)$?
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In an abelian category, is it true that
$$0to Astackrel fto Bstackrel gto Cto 0$$
being exact means that $C=operatornamecoker(f)$?
I checked it in the module category but am having trouble in an arbitrary abelian category.
First isomorphism theorem $B/A=B/operatornameim(f)cong B/ker(g)=C$
where $B/A:=operatornamecoker(f)$ for $f:Ahookrightarrow B$ is the definition of a quotient object I think
category-theory abelian-categories
add a comment |Â
up vote
0
down vote
favorite
In an abelian category, is it true that
$$0to Astackrel fto Bstackrel gto Cto 0$$
being exact means that $C=operatornamecoker(f)$?
I checked it in the module category but am having trouble in an arbitrary abelian category.
First isomorphism theorem $B/A=B/operatornameim(f)cong B/ker(g)=C$
where $B/A:=operatornamecoker(f)$ for $f:Ahookrightarrow B$ is the definition of a quotient object I think
category-theory abelian-categories
2
Yes, or more strictly the arrow $g$ is the cokernel.
â Lord Shark the Unknown
Aug 26 at 6:48
See, for instance: (1) Freyd - Abelian Categories (2003) â Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) â page 200-201
â Steenis
Aug 29 at 16:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In an abelian category, is it true that
$$0to Astackrel fto Bstackrel gto Cto 0$$
being exact means that $C=operatornamecoker(f)$?
I checked it in the module category but am having trouble in an arbitrary abelian category.
First isomorphism theorem $B/A=B/operatornameim(f)cong B/ker(g)=C$
where $B/A:=operatornamecoker(f)$ for $f:Ahookrightarrow B$ is the definition of a quotient object I think
category-theory abelian-categories
In an abelian category, is it true that
$$0to Astackrel fto Bstackrel gto Cto 0$$
being exact means that $C=operatornamecoker(f)$?
I checked it in the module category but am having trouble in an arbitrary abelian category.
First isomorphism theorem $B/A=B/operatornameim(f)cong B/ker(g)=C$
where $B/A:=operatornamecoker(f)$ for $f:Ahookrightarrow B$ is the definition of a quotient object I think
category-theory abelian-categories
edited Aug 26 at 9:25
Bernard
111k635103
111k635103
asked Aug 26 at 6:16
user587226
61
61
2
Yes, or more strictly the arrow $g$ is the cokernel.
â Lord Shark the Unknown
Aug 26 at 6:48
See, for instance: (1) Freyd - Abelian Categories (2003) â Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) â page 200-201
â Steenis
Aug 29 at 16:07
add a comment |Â
2
Yes, or more strictly the arrow $g$ is the cokernel.
â Lord Shark the Unknown
Aug 26 at 6:48
See, for instance: (1) Freyd - Abelian Categories (2003) â Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) â page 200-201
â Steenis
Aug 29 at 16:07
2
2
Yes, or more strictly the arrow $g$ is the cokernel.
â Lord Shark the Unknown
Aug 26 at 6:48
Yes, or more strictly the arrow $g$ is the cokernel.
â Lord Shark the Unknown
Aug 26 at 6:48
See, for instance: (1) Freyd - Abelian Categories (2003) â Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) â page 200-201
â Steenis
Aug 29 at 16:07
See, for instance: (1) Freyd - Abelian Categories (2003) â Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) â page 200-201
â Steenis
Aug 29 at 16:07
add a comment |Â
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2
Yes, or more strictly the arrow $g$ is the cokernel.
â Lord Shark the Unknown
Aug 26 at 6:48
See, for instance: (1) Freyd - Abelian Categories (2003) â Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) â page 200-201
â Steenis
Aug 29 at 16:07