Exact sequence of sheaves in Serre's "Faisceaux algebriques coherents"

Question

I hope this is not a duplicate but I could not find any reference to what I'm about to ask in the other questions. I am studying Serre's article, and in particular the 1st chapter, in view of an exam, and I am stuck at $\S 24$, where the author states the following:

let $\mathfrak U$ be a cover of a topological space $X$ and assume the sequence $0\rightarrow \mathcal F\xrightarrow{\alpha} \mathcal G\xrightarrow{\beta} \mathcal H\rightarrow 0$ of sheaves is exact; then the following sequence (where by $C(\mathfrak U,\mathcal F)$ we mean the cochain complex associated to the cover $\mathfrak U$ and the sheaf $\mathcal F$) is exact $$ 0\rightarrow C(\mathfrak U,\mathcal F)\xrightarrow{\alpha} C(\mathfrak U,\mathcal G) \xrightarrow{\beta} C(\mathfrak U,\mathcal H),$$ but the map $\beta$ need not be surjective.

Now, I have a couple of questions about this, which I hope aren't too trivial (I am definitely neither an expert in homological algebra nor in sheaf theory, much less algebraic geometry):

1. why is that the sequence of cochain complexes is exact in the first two complexes and not in the third, that is, why is $\alpha$ necessarily injective and the sequence exact in $C(\mathfrak U,\mathcal G)$, while $\beta$ is not surjective? By this I mean: where's the error in the following proof for the surjectivity of $\beta$? (I have an idea but I want some feedback)

Let $h\in C(\mathfrak U,\mathcal H)$ and assume $h$ is mapped to zero by the map induced by the map $\mathcal H\rightarrow 0$ (I denote thi map by $\phi$). Then for each $s\in I^{q+1}\quad (\phi h)_s=0$. That is $ \phi(h_s)=0 $ by definition. By surjectivity of $\beta$ at sheaf level I can find a section $g\in\mathcal G$ such that $\beta(g_s)=h_s$, that is $(\beta g)_s=h_s$, which is the same as $\beta g=h$ by the sheaf axioms. Hence $\beta$ is surjective at cochain level.

2. can anybody provide an example of an exact sequence of sheaves where $\beta$ is surjective in the induced sequence of complexes and one in which $\beta$ is not?

Thanks in advance for all help.

Answer 1

I think you might be misunderstanding what it means for the map $\beta\colon \mathcal{G} \to \mathcal{H}$ to be surjective. It does not mean that for each $U \subseteq X$ open the induced map $\mathcal{G}(U) \to \mathcal{H}(U)$ is surjective! Rather, it means that the quotient sheaf (i.e., the sheafification of the quotient presheaf) is zero, or equivalently, that for each $x \in X$ the stalk map $\mathcal{G}_x \to \mathcal{H}_x$ is surjective.

So the error in your reasoning is in the line "By surjectivity of $\beta$ at sheaf level ...".

There is a natural example of such an exact sequence in complex analysis. Consider $X = \mathbb{C} \setminus\{0\}$ with the usual topology, and let $\mathfrak{U} = \{X\}$ (other open covers will work as well, but this one is particularly easy to work with). We let $\mathcal{O}_X$ denote the sheaf of analytic functions on $X$, and $\mathbb{C}_X$ the sheaf of locally constant fucntions on $X$. Then we have a sequence $$ 0 \to \mathbb{C}_X \overset{\alpha}\longrightarrow \mathcal{O}_X \overset{\beta}\longrightarrow \mathcal{O}_X \to 0,$$

where $\alpha$ is the inclusion, and $\beta$ sends a function $f$ to its derivative $\mathrm{d}f/\mathrm{d}z$. You should think a bit about why this sequence is exact: injectivity of $\alpha$ is obvious, and exactness in the middle just says that the functions whose derivative are zero are exactly those who are locally constant. The surjectivity of $\beta$ says that an analytic function defined in an open neighborhood of $x \in X$ has a primitive in a (possibly smaller) open neighborhood of $x$.

But if you now take global sections, you see that the map $\mathcal{O}_X(X) \to \mathcal{O}_X(X)$ is not surjective: the function $1/z$ is not in the image. Indeed, while $1/z$ locally admits primitives around each point, of the form $\log z + C$, there is no way to choose choose these primitives in such a way that they glue to an analytic function defined on all of $X$.

Answer 2

The injectivity of $\alpha$ and exactness at the middle is just brutal checking and elementary. I suggest you do it yourself, no thinking necessary.

One easy case where $\beta$ is surjective is when the sequence splits. In other words, take $\mathcal{G}=\mathcal{F}\oplus\mathcal{H}$ with the obvious maps, where $\mathfrak{U}$ can be any cover.

For $\beta$ non-surjective, take the Euler sequence $0\to \mathcal{O}(-2)\to\mathcal{O}(-1)^{\oplus 2}\to \mathcal{O}\to 0$ on the projective line and take $\mathfrak{U}=\{\mathbb{P}^1\}$