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.

  1. 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.


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 $X$.

  • $\begingroup$ Ty for the explaination, this clarifies much. Just to check my line of thought, were I to prove that such a map is surjective, I should verify that stalk-level surjectivity implies cochain-level surjectivity, which is the step that fails in this particular case, am I right? On the other hand, stalk-level injectivity always implies cochain-level injectivity, and this is essentially why the sequence is exact in the first complex? $\endgroup$ – Alessandro Contini Jun 20 '18 at 19:58
  • $\begingroup$ In regard to the example, the sheaf (stalk)-level surjectivity happens because locally I can always take the Taylor series of $f$ in a neighbourhood of $x\in X$ and integrate it term-wise to obtain the Taylor series of the primitive, which is again analytic? This should give sheaf-level surjectivity since the stalks of $\mathcal O_X$ are just germs of analytic functions, hence local power series. Then the global sections sequence is not exact since different representations of an analytic function may not glue, as with the $1\diagup z$ function you mentioned. $\endgroup$ – Alessandro Contini Jun 20 '18 at 20:09
  • $\begingroup$ @AlessandroContini What you mean by "different representations of an analytic function may not glue"? The sheaf of analytic functions is indeed a sheaf, so its sections glue uniquely whenever they satisfy the gluing condition (i.e. on each intersection the two restrictions coincide). The failure of surjectivity at the global level comes from the fact that primitives are unique only up to constants, and there is no choice of these constants such that the gluing condition is satisfied. $\endgroup$ – Marc Paul Jun 20 '18 at 20:51
  • $\begingroup$ I meant exactly the part "primitives are unique only... " in your comment, altough I admit I expressed myself very poorly up there. Thank you for your time and precision $\endgroup$ – Alessandro Contini Jun 20 '18 at 21:19

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 $\mathfrak{U}=\{\mathbb{P}^1\}$

  • $\begingroup$ Thank you for the answer. Of course, as you suggested, the injectivity and the exactness at the second complex can be derived from a simple check instead of looking for 'abstract' reasons. My question, however, was more related to (and this might not be clear form the text of the question, which I shall edit in order to clarify the point) the reason why I cannot apply the same line of thought of these two checks to prove the surjectivity. Any hints in this direction? $\endgroup$ – Alessandro Contini Jun 20 '18 at 13:53
  • $\begingroup$ In regard to the surjectivity example, I can see that in this case the thesis holds. Indeed given $h\in C(\mathfrak U,\mathcal H)$, since $\mathcal H$ is the quotient sheaf $\mathcal F\oplus\mathcal H\diagup\mathcal F$, I can consider the section $f+h\in\mathcal F\oplus\mathcal H$ for some $f\in C(\mathfrak U,\mathcal F)$ and this projects to the desired section $h$ in view of the exactness at shaf level. Is this correct? $\endgroup$ – Alessandro Contini Jun 20 '18 at 14:11

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