Let $\mathcal{A}$ be an abelian category. Suppose we have objects $A$, $B$ and $C$, and morphisms $f:A\to B$, $g:B\to C$ with $g\circ f=0$ i.e. is equal to the zero morphism $A\to C$. In order to make sense of exact sequences in an arbitrary abelian category, it is presumably necessary to have a canonical morphism $\text{im}(f)\to\text{ker}(g)$ making some diagram commutative, but I can't see quite how to pin it down.
There is a morphism $A\to\text{coim}(f)\to\text{im}(f)$, and since $g\circ f=0$, there exists a unique morphism $\alpha:A\to\text{ker}(g)$ with $i\circ\alpha=f$, where $i:\text{ker}(g)\to B$ is the kernel of $g$. But how do I get the right morphism $\text{im}(f)\to\text{ker}(g)$?
Best Answer
The kernel of $g$ is defined as the pullback of $g \colon B \rightarrow C$ and the zero morphism $0 \rightarrow C$. Therefore defining a morphism into this kernel amounts to defining two "compatible" morphisms, namely one into the zero object $0$ (here we have no choice) and one into $B$. Do you see how to get a canonical morphism from $\text{im}(f)$ to $B$?