Does Exactness in Presheaves Imply Exactness in Sheaves?

Question

I am reading Rotman, Introduction to Homological Algebra, and it reads that "Exactness of sheaves means exactness of stalks, which is usually different from exactness of presheaves." I don't think I am really understanding this sentence. Suppose $\mathcal{F}$,$\mathcal{G}$, $\mathcal{H}$ are sheaves of abelian group over a topological space $X$, and the following sequence $$ \mathcal{F} \to \mathcal{G} \to \mathcal{H} $$ is exact as presheaves, which means for every $U$ open set, $$ \mathcal{F}(U) \to \mathcal{G}(U) \to \mathcal{H}(U) $$ is exact. Then is it not implying that for all $x\in X$, a sequence of stalks $$ \mathcal{F}_x \to \mathcal{G}_x \to \mathcal{H}_x $$ exact? Does it not become a exact sequence of stalks if we take a limit over all open set $U$ containing $x$?

I suppose this is not true because the image of presheaf map is not always sheaf in $\mathcal{G}$, so the equalizer condition may not word in the image of $\mathcal{F}\to\mathcal{G}$. I am not so sure.

Is there any counterexample, or did I misunderstand the sentence?

Answer 1

I like to distinguish between sheaves and presheaves. Let $i$ be the forgetful functor; that is if $F$ is a sheaf, let $iF$ denote the reinterpretation as a presheaf. Let $a$ be the associated sheaf functor.

The crucial facts to know about $i$ and $a$ are that:

• $a$ is left adjoint to $i$
  • this implies $a$ is right exact; in fact it preserves all colimits
  • this implies $i$ is left exact; in fact it preserves all limits
• $a$ is furthermore left exact
• There is a natural isomorphism $\mathcal{F} \cong ai\mathcal{F}$ for sheaves $\mathcal{F}$.

This means that if you have an exact sequence of presheaves

$$ 0 \to \mathcal{P} \to \mathcal{Q} \to \mathcal{R} \to 0 $$

you also have an exact sequence of associated sheaves

$$ 0 \to a\mathcal{P} \to a\mathcal{Q} \to a\mathcal{R} \to 0 $$

However, if you have an exact sequence of sheaves

$$ 0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0 $$

you can generally only expect a left exact sequence of the corresponding presheaves

$$ 0 \to i\mathcal{F} \to i\mathcal{G} \to i\mathcal{H} $$

Answer 2

If $\mathcal{F}$, $\mathcal{G}$ and $\mathcal{H}$ are sheaves and fitting into an exact sequence $$0\to \mathcal{F}\to \mathcal{G}\to \mathcal{H} \to 0$$ in the category of presheaves, then the sequence is in fact exact as a sequence of sheaves. This follows (for example) from the fact that sheafification preserves finite limits and colimits.

However, if you're given an exact sequence in the category of sheaves, it may not be exact in the category of presheaves -- this failure is precisely what sheaf cohomology measures. Here is a simple example. Consider the exponential sequence $$\mathbb{Z}\overset{\times 2\pi i }{\to} \mathbb{C} \overset{{\rm exp}}{\to} \mathbb{C}^{\times}\;.$$ and the corresponding sequence of sheaves on $\mathbb{C}^{\times}=\mathbb{C}-\{0\}$ $$\underline{\mathbb{Z}}\to \mathcal{O}_{\mathbb{C}^{\times}}\to \mathcal{O}^*_{\mathbb{C}^{\times}}\;.$$ This sequence is exact in the category of sheaves, but not in the category of presheaves. Indeed, if it were exact in presheaves then we would have a surjection on global sections $$\mathcal{O}_{\mathbb{C}^{\times}}(\mathbb{C}^{\times})\overset{\rm exp}{\to} \mathcal{O}^*_{\mathbb{C}^{\times}}(\mathbb{C}^{\times}).$$ In particular, this would give the existence of a global complex logarithm.