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I'm working through Vakil's algebraic geometry text and I've been stuck on Exercise 1.6.E (page 52 on

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.)

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Dear Alan, What does it mean to you that $0 \to F(\ker f) \to FA \xrightarrow{Fg} FA''$ is exact? – Bruno Joyal May 21 '14 at 1:22
Hmm, exactness at $F(\ker f)$ means that $F(\ker f) \to FA$ is a monomorphism. Our goal is to show that $\ker Fg = \operatorname{im} Ff$, so we need to extract some information about $\ker Fg$ from the monomorphism $F(\ker f) \to FA$, somehow? – Alan C May 21 '14 at 1:44
Indeed, it's a monomorphism. What about exactness at $FA$? – Bruno Joyal May 21 '14 at 1:46
Let $\phi$ denote the monomorphism. Then exactness at $FA$ tells us $\ker Fg = \operatorname{im} \phi$. And then maybe it helps if we identify $\operatorname{im} \phi$ with $F(\ker f)$? (Maybe part of my confusion is due to the fact that I'm still used to thinking of kernels and images as objects?) – Alan C May 21 '14 at 1:55
That's right. In an abelian category, the image of a monomorphism is the monomorphism itself. As a rule of thumb, you can always just pretend that you are working in a category of modules over a general ring. – Bruno Joyal May 21 '14 at 2:02
up vote 7 down vote accepted

We have a diagram

enter image description here

with exact diagonals. Applying $F$ gives a diagram

enter image description here

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.

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+1. This old answer of mine comes to mind. – Bruno Stonek May 21 '14 at 8:39

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