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.)