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!
Best Answer
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 $\ZZ$.