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







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    Yes, or more strictly the arrow $g$ is the cokernel.
    – Lord Shark the Unknown
    Aug 26 at 6:48










  • See, for instance: (1) Freyd - Abelian Categories (2003) – Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) – page 200-201
    – Steenis
    Aug 29 at 16:07














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







share|cite|improve this question


















  • 2




    Yes, or more strictly the arrow $g$ is the cokernel.
    – Lord Shark the Unknown
    Aug 26 at 6:48










  • See, for instance: (1) Freyd - Abelian Categories (2003) – Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) – page 200-201
    – Steenis
    Aug 29 at 16:07












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







share|cite|improve this question














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









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 26 at 9:25









Bernard

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asked Aug 26 at 6:16









user587226

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  • 2




    Yes, or more strictly the arrow $g$ is the cokernel.
    – Lord Shark the Unknown
    Aug 26 at 6:48










  • See, for instance: (1) Freyd - Abelian Categories (2003) – Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) – page 200-201
    – Steenis
    Aug 29 at 16:07












  • 2




    Yes, or more strictly the arrow $g$ is the cokernel.
    – Lord Shark the Unknown
    Aug 26 at 6:48










  • See, for instance: (1) Freyd - Abelian Categories (2003) – Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) – page 200-201
    – Steenis
    Aug 29 at 16:07







2




2




Yes, or more strictly the arrow $g$ is the cokernel.
– Lord Shark the Unknown
Aug 26 at 6:48




Yes, or more strictly the arrow $g$ is the cokernel.
– Lord Shark the Unknown
Aug 26 at 6:48












See, for instance: (1) Freyd - Abelian Categories (2003) – Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) – page 200-201
– Steenis
Aug 29 at 16:07




See, for instance: (1) Freyd - Abelian Categories (2003) – Section 2.2, page 45 (2) Mac Lane - Categories for the Working Mathematician (2nd edition,1998) – page 200-201
– Steenis
Aug 29 at 16:07















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