Let \(A\) be an abelian group and let \(X\) be a topological space. Define the constant presheaf associated to \(A\) on \(X\) to be the presheaf \(U \mapsto A\) for all \(U \neq \varnothing\), with all restriction maps the identity. Show that the constant sheaf \(\mca\) defined in the text is the sheaf associated to this presheaf.
Recall that the constant sheaf \(\mca\) of the text is defined by giving \(A\) the discrete topology and setting \(\mca(U)\) to be the group of continuous maps \(U \to A\).
Solution
Write \(\tilde\mca\) for the constant presheaf of the question and \(\mca\) for the sheaf of the text.
Define a morphism of presheaves \(F: \tilde\mca \to \mca\) by giving, for each open \(U\), the map \[ F_U: \tilde\mca(U) = A \to \mca(U) = C^0(U, A) \] which sends \(a \in A\) to the constant function with value \(a\). Since \(A\) carries the discrete topology, every constant function is continuous, so \(F_U\) lands in \(\mca(U)\). The maps \(F_U\) are compatible with restriction, because restricting a constant function gives the constant function with the same value, so \(F\) is a morphism of presheaves.
Now compare stalks. For every \(p \in X\) we have \(\tilde\mca_p = A\), since every restriction map of \(\tilde\mca\) is the identity. On the other side, a continuous map from a small enough neighbourhood of \(p\) into the discrete group \(A\) is constant near \(p\), so \(\mca_p = A\) as well, and \(F_p: A \to A\) is the identity. Thus \(F_p\) is an isomorphism for every \(p\).
A morphism of presheaves which is an isomorphism on all stalks induces an isomorphism on the associated sheaves. By the universal property of sheafification, \(F\) factors through \((\tilde\mca)^+\), and the induced map \((\tilde\mca)^+ \to \mca\) is an isomorphism on stalks between two sheaves, hence an isomorphism. So \(\mca\) is the sheafification of the constant presheaf.