Past exam of the introduction to schemes course of the University of Oxford / 2021 / 2 / c

In an irreducible topological space, any two nonempty open subsets intersect. Let $W\subseteq X$ be a nonempty open subset. Two nonempty subsets open in $W$ are also open in $X$, and hence intersect; thus $W$ is itself an irreducible topological space. If $W$ were disconnected, a separation into two disjoint nonempty open subsets would contradict this intersection property. Hence $W$ is a connected space. The empty open subset is connected as well under the usual convention.

Discard empty members of $\mathcal{U}$ as in (b). Every finite intersection $U_{i_0}\cap \cdots \cap U_{i_q}$ is nonempty by repeated use of the intersection property and is an open subset, hence a connected space. A locally constant function on this intersection is constant. It follows that the constant sheaf hasand every restriction map between these nonempty intersections is the identity on $A$. The whole Čech cochain complex is therefore the same as for the indexed open cover of a point having the same index set. The cochain homotopy constructed in (b) applies component by component and provesThe argument concerns the constant sheaf of locally constant functions; arbitrary coefficient sheaves on an irreducible topological space need not have the same Čech cohomology.