# Is it true that $C=\operatorname{coker}(f)$?

In an abelian category, is it true that $$0\to A\stackrel f\to B\stackrel g\to C\to 0$$ being exact means that $C=\operatorname{coker}(f)$?

I checked it in the module category but am having trouble in an arbitrary abelian category.

First isomorphism theorem $B/A=B/\operatorname{im}(f)\cong B/\ker(g)=C$

where $B/A:=\operatorname{coker}(f)$ for $f:A\hookrightarrow B$ is the definition of a quotient object I think

• Yes, or more strictly the arrow $g$ is the cokernel. – Lord Shark the Unknown Aug 26 '18 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 '18 at 16:07