In abelian category, $0\rightarrow A\xrightarrow{f} B\rightarrow 0$ is exact if and only if $f$ is an isomorphism.

Question

In Vakil's notes, Section 1.6.5, he mentioned that

$0\rightarrow A\xrightarrow{f} B\rightarrow 0$ is exact if and only if $f$ is an isomorphism.

I can prove that if $f$ is an isomorphism, then the sequence is exact. I have trouble proving the other direction.

Suppose the sequence is exact. Then $f$ is a monomorphism and an epimorphism. But this does not imply it is an isomorphism. To show it, I think I have to construct a map $g$ from $B$ to $A$, then show that $g\circ f$ and $f\circ g$ are the identity maps. But since all the arrows I have are from left to right, how can I even know there exists a map, other than $0$, from $B$ to $A$?

Thank you for any help!

Answer 1

The identity map $B \to B$ factors as $B \xrightarrow{g} A \xrightarrow{f} B$ by the exactness of the sequence.

The identity map $A \to A$ factors as $A \xrightarrow{f} B \xrightarrow{h} A$ by the exactness of the sequence.

Now we have $g = (h \circ f) \circ g = h \circ (f \circ g) = h$, hence the result.