Lemma 20.20.2. Let $X$ be an irreducible topological space. Then $H^ p(X, \underline{A}) = 0$ for all $p > 0$ and any abelian group $A$.
Proof. Recall that $\underline{A}$ is the constant sheaf as defined in Sheaves, Definition 6.7.4. It is clear that for any nonempty open $U \subset X$ we have $\underline{A}(U) = A$ as $X$ is irreducible (and hence $U$ is connected). We will show that the higher Čech cohomology groups $\check{H}^ p(\mathcal{U}, \underline{A})$ are zero for any open covering $\mathcal{U} : U = \bigcup _{i\in I} U_ i$ of an open $U \subset X$. Then the lemma will follow from Lemma 20.11.8.
Recall that the value of an abelian sheaf on the empty open set is $0$. Hence we may clearly assume $U_ i \not= \emptyset $ for all $i \in I$. In this case we see that $U_ i \cap U_{i'} \not= \emptyset $ for all $i, i' \in I$. Hence we see that the Čech complex is simply the complex
\[ \prod _{i_0 \in I} A \to \prod _{(i_0, i_1) \in I^2} A \to \prod _{(i_0, i_1, i_2) \in I^3} A \to \ldots \]
We have to see this has trivial higher cohomology groups. We can see this for example because this is the Čech complex for the covering of a $1$-point space and Čech cohomology agrees with cohomology on such a space. (You can also directly verify it by writing an explicit homotopy.) $\square$
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