Exercise III.2.7 in Hartshorne's Algebraic Geometry is the following.

Let $S^1$ be the circle (with its usual topology) and let $\mathbb Z$ be the constant sheaf $\mathbb Z$.

(a) Show that $H^1(S^1, \mathbb Z)\cong \mathbb Z$, using our definition of cohomology [derived functor cohomology]. (b) Now let $\mathcal R$ be the sheaf of germs of continuous real-valued functions on $S^1$. Show that $H^1(S^1, \mathcal R)=0$.

I have two questions. The first is just me missing something obvious, but the second seems more serious.

For part (a), it seems like the most straightforward thing to do is to build an injective resolution and take cohomology. In proposition III.2.2, Hartshorne gives us a recipe for constructing injectives: stick together a bunch of skyscraper sheaves. Let $i_p(A)$ denote the skyscraper sheaf at a point $p$ with group $A$. Then I get the resolution $$\mathbb Z \rightarrow \prod_{p\in S^1} i_p(\mathbb Q) \rightarrow \prod_{p\in S^1} i_p(\mathbb Q/\mathbb Z) \rightarrow 0.$$ But taking global sections and then cohomology gives an answer that is clearly wrong. What am I misunderstanding here?

For (b), it seems the claim is false. For example, if we compute Cech comology using the standard two-piece cover of the circle, mimicking example 4.0.4 in the text, we get that $H^1$ is nonzero. But this seems odd, since Cech cohomology should agree with derived functor cohomology for a good cover. What's going on here?

  • 1
    $\begingroup$ Is your cover in B) good? $\endgroup$ Nov 18, 2015 at 2:27
  • $\begingroup$ @MarianoSuárez-Alvarez I can't prove anything, but it sure seems like it. If the cohomology vanishes on $S^1$, then surely it should also vanish on an interval, right? (I realize the problem only claims $H^1(S^1,\mathcal R)=0$, but this seems suggestive...) $\endgroup$
    – Potato
    Nov 18, 2015 at 2:44
  • $\begingroup$ The issue is not with the cover, though. $\endgroup$ May 7, 2017 at 22:45

1 Answer 1


For (a), your resolution is not right, because $(\prod_{p\in S^1}i_p(\mathbb{Q}))/\mathbb{Z}$ is not $\prod_{p\in S^1}i_p(\mathbb{Q}/\mathbb{Z})$. Or in other words, the kernel of $\prod_{p\in S^1}i_p(\mathbb{Q})\rightarrow \prod_{p\in S^1}i_p(\mathbb{Q}/\mathbb{Z})$ is not $\mathbb{Z}$ (this is $\prod_{p\in S^1}i_p(\mathbb{Z})$).

For (b), you cannot mimick example 4.0.4 to compute the Cech cohomology. In fact, the higher Cech cohomology of this sheaf vanishes. Let $U,V$ your cover (which is good), and consider the Cech complex $$ R(U)\times R(V)\rightarrow R(U\cap V)$$ given by $(f,g)\mapsto f_{|U\cap V}-g_{|U\cap V}$.

The kernel of this map are couple of functions $(f,g)$ that agree on the intersection, and hence patch together to form a unique function on $S^1$. So $\overset{\vee}{H^0}(\{U,V\},R)=R(S^1)$ as expected.

But I claim that the map is onto, so that $\overset{\vee}{H^1}(\{U,V\},R)=0$. Indeed, let $f\in R(U\cap V)$, also let $u,v$ be two functions on $S^1$ such that $u$ has (compact) support in $U$, $v$ has (compact) support in $V$ and $u+v=1$. In particular the function $u$ is zero in a neighborhood of $S^1\setminus V$ and $v$ is zero in a neighborhood of $S^1\setminus U$. The function $uf$ can then be extended to $U$ because it is zero in a neighborhood of $U\setminus U\cap V$. Similarily, the function $vf$ can be extended to $V$. And $(uf,-vf)\mapsto uf+vf=f$ so that the map is onto.

Here $u,v$ are called a partition of unity. A sheaf like $R$ with partitions of unity is called a fine sheaf. This argument (of a similar one with more than two open sets in the cover) shows that the higher Cech cohomology groups of a fine sheaf over a paracompact space vanish. On a paracompact space, Cech and derived functor cohomology agree so the cohomology of a fine sheaf is trivial.

The difference with the example 4.0.4 is that there is no partition of unity in a constant sheaf.

  • $\begingroup$ For (a), you are certainly right, but I am confused because it is a theorem (Hartshorne exercise II.1.2) that a sequence of sheaves is exact if and only if it is exact on stalks. And certainly the stalks seem exact here. What is the problem with this reasoning? $\endgroup$
    – Potato
    Nov 18, 2015 at 14:33
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    $\begingroup$ The stalk of $\prod_{p\in X}i_p(A)$ at $p$ is not $A$. The stalk functor may well be exact, it does not commute with infinite products. The sheaf $\prod_{p\in X}i_p(A)$ is the sheaf of all functions $X\rightarrow A$ (as if $X$ where discrete). There are germs of functions $s,s':X\rightarrow A$ defined in a neighborhood of $p$ which does not agree in any smaller neighborhood, even if $s(p)=s'(p)$. So the natural map $\mathbb{Z}\rightarrow\prod_{p\in P}i_p\mathbb{Z}$ is not an iso at $p$. $\endgroup$
    – Roland
    Nov 18, 2015 at 14:49
  • $\begingroup$ Fantastic, thank you. $\endgroup$
    – Potato
    Nov 18, 2015 at 14:56
  • $\begingroup$ So this comes a little late but how can we take an injective resolution for (a) after all? The exercise says "compute it with the definition" so it should be possible but I have tried everything and had no luck. I don't want to make a separate question because it would be a duplicate.. $\endgroup$ May 16, 2020 at 6:55
  • $\begingroup$ @IoannisZolas I am sure you can ask a separate question. You can link to this one and say that this thread does not give the injective resolution so does not answer your question, and therefore this is not a duplicate. I don't have much time right now to think about it, but maybe later today I will. So if you prefer to wait, I may try to answer it. $\endgroup$
    – Roland
    May 16, 2020 at 8:24

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