showing exact functors preserve exact sequences (abelian categories, additive functors, and kernels)

Question

I'm working through Vakil's algebraic geometry text and I've been stuck on Exercise 1.6.E (page 52 on http://math.stanford.edu/~vakil/216blog/FOAGjun1113public.pdf.)

Suppose that $F$ is an exact functor. Show that applying $F$ to an exact sequence preserves exactness. For example, if $F$ is covariant and $A' \to A \to A''$ is exact, then $FA' \to FA \to FA''$ is exact.

Here's what I've been thinking:

Let the maps be denoted $f, g$ (so we have $A' \xrightarrow{f} A \xrightarrow{g} A''$).

We know $F$ is left-exact and right-exact. To use the left-exactness of $F$, we note that $0 \to \ker f \to A \xrightarrow{g} A''$ is exact, so $0 \to F(\ker f) \to FA \xrightarrow{Fg} FA''$ is exact.

However, I'm not quite sure what to do with the $F(\ker f)$ object. (It'd be nice if $F(\ker f) = \ker Ff$ but I don't see any reason for this to actually be true.)

Answer 1

We have a diagram

with exact diagonals. Applying $F$ gives a diagram

also with exact diagonals.

Now, note that \begin{align*} \DeclareMathOperator{Im}{Im}\Im F(f) &= \Im\big(F(A^\prime)\to F(\Im f)\to F(A)\big) \\ &\overset{\ast}{=} \Im\big(F(\Im f)\to F(A)\big) \\ &= \DeclareMathOperator{Ker}{Ker}\Ker\big(F(A)\to F(\Im g)\big) \\ &\overset{\circledast}{=} \Ker\big(F(A)\to F(\Im g)\to F(A^{\prime\prime})\big) \\ &= \Ker F(g) \end{align*} where $\ast$ holds since $F(A^\prime)\to F(\Im f)$ is epi and $\circledast$ holds since $F(\Im g) \to F(A^{\prime\prime})$ is mono. Hence $$ F(A^\prime)\to F(A)\to F(A^{\prime\prime}) $$ is exact.

Of course, the above argument extends to prove that an exact functor maps acyclic complexes to acyclic complexes.