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.)
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.