# Hartshorne Exercise III.2.7(a): sheaf cohomology of constant sheaf $\Bbb Z$ on $S^1$ in the usual topology

I am trying to solve this exercise:

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

(a) Show that $$H^1(S^1,\mathbb Z)\simeq \mathbb Z$$, using our definition of cohomology.

I have tried to construct an injective resolution: like Proposition 2.2, Let $$I^0=\prod_{x\in S^1}j_*(\mathbb Q)$$. But then I don't know how to calculate its stalk. So I have difficulties in build the $$I^1$$... If I just use its discontinuous sections to build a flasque resolution, I also can't calculate the stalk...

Could you provide some help or give a complete answer? Using Cech cohomology is also accepted. Thanks!

There's no need to compute a full resolution of $$\Bbb Z$$ - a short exact sequence is enough to solve the problem if you're clever about it.

Let $$\def\cF{\mathcal{F}} \def\ZZ{\mathbb{Z}} \def\RR{\mathbb{R}} \def\cQ{\mathcal{Q}} \def\G{\Gamma} \def\coker{\operatorname{coker}} \cF$$ be the flasque sheaf which assigns to each $$U\subset S^1$$ the set of all functions $$U\to\RR$$. Embed $$\ZZ\to\cF$$ in the obvious way and let $$\cQ$$ be the quotient. Taking the long exact sequence in cohomology arising from $$0\to \ZZ\to \cF\to \cQ\to 0,$$ we note that $$H^1(S^1,\cF)=0$$ implying $$H^1(S^1,\ZZ)=\coker(\G(S^1,\cF)\to\G(S^1,\cQ))$$. By exercise II.1.3, any section $$s\in\G(S^1,\cQ)$$ is the image of a family $$\{(s_i,U_i)\}_{i\in I}$$ with $$s_i\in\cF(U_i)$$ where $$U_i$$ form an open cover of $$S^1$$ and $$(s_i-s_j)|_{U_i\cap U_j}$$ is a section of $$\ZZ_{U_i\cap U_j}$$. Since $$S^1$$ is compact, we may assume $$I$$ is finite; after subdividing, throwing away redundant elements, and reordering we may assume that our cover consists of connected open subsets so that $$U_i$$ only intersects $$U_{i-1}$$ and $$U_{i+1}$$ with indices interpreted modulo $$|I|$$.

Now I claim that it suffices to consider $$|I|=3$$. Let $$n_{i+1}$$ be the value of $$s_{i+1}-s_i$$ on $$U_i\cap U_{i+1}$$. Replacing $$s_{i+1}$$ with $$s_{i+1}-n_{i+1}$$, which does not change the image of $$s_{i+1}$$ in $$\cQ(U_{i+1})$$, we see that $$s_i=s_{i+1}$$ on $$U_i\cap U_{i+1}$$. Therefore we can glue $$s_i$$ and $$s_{i+1}$$ to form a section of $$\cF$$ over $$U_i\cup U_{i+1}$$ without changing its image in $$\cQ$$. Repeating this process for $$i=3,\cdots,|I|-1$$, we see that we can glue the sections $$s_i$$ on $$U_3\cup\cdots\cup U_{|I|}$$ so that we're only looking at $$\{(s_1,U_1),(s_2,U_2),(s_3,U_3\cup\cdots\cup U_{|I|})\}$$.

If we have a section $$\{(s_1,U_1),(s_2,U_2),(s_3,U_3)\}$$, by the same logic we may assume that $$s_1=s_2$$ on $$U_1\cap U_2$$ and $$s_2=s_3$$ on $$U_2\cap U_3$$. Therefore up to adding a global section of $$\cF$$, the global sections of $$\cQ$$ are exactly those of the form $$\{(0,U_1),(0,U_2),(n,U_3)\}$$ for $$n\in\ZZ$$ and opens $$U_i$$ satisfying our ordering and intersection assumptions. Since any two such sections are equivalent up to an element of $$\G(S^1,\cF)$$ iff their $$n$$s match, we see that the cokernel is exactly $$\ZZ$$.

### Using Cech cohomology

Take the open cover $$U_{up} \cup U_{down}$$ of $$S^1$$ where $$U_{up}$$ covers a bit more than top half of circle, and $$U_{down}$$ covers a bit more than bottom half of circle, both $$U_{up}, U_{down}$$ are connected. Note that $$U_{up} \cap U_{down} = U_{east} \cup U_{west}$$ is then a disjoint union of two connected open intervals, around "East" and "West" pole of the circle.

Cech cohomology means you want to compute the cohomology of this complex: $$0 \to C^0 = \mathbb{Z}(U_{up}) \oplus \mathbb{Z}(U_{down}) \to C^1 = \mathbb{Z}(U_{up} \cap U_{down}) \to 0$$ with restriction map being $$(a,b) = b-a$$.

Now note that

• for any open $$U$$, $$\mathbb{Z}(U)$$ is locally constant $$\mathbb{Z}$$-valued functions on $$U$$. When $$U$$ is connected, this forces $$\mathbb{Z}(U)$$ to be constant functions on $$U$$.
• This means $$\mathbb{Z}(U_{up})$$ and $$\mathbb{Z}(U_{down})$$ are isomorphic to $$\mathbb{Z}$$. This also means $$\mathbb{Z}(U_{up} \cap U_{down}) \cong \mathbb{Z} \oplus \mathbb{Z}$$, corresponding to the constant value on the "east" piece and the "west" piece respectively.
• The Cech cohomology sequence is then $$0 \to \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z} \to 0$$ where the middle map is $$(a,b) \to (b-a, b-a)$$. It's then clear that $$H^1(S^1, \mathbb{Z}) \cong \mathbb{Z}$$. (since kernel is everything, image is those where two coordinates are equal)