Let \((X, \OO_X)\) be a locally ringed space, and let \(f: \mcl \to \mcm\) be a surjective map of invertible sheaves on \(X\). Show that \(f\) is an isomorphism.
Hint
Reduce to a question of modules over a local ring by looking at the stalks.
Solution
1. For every \(x\in X\), the induced homomorphism \(f_x:\mcl_x\to\mcm_x\) is an isomorphism of \(\OO_{X,x}\)-modules.
Put \(R=\OO_{X,x}\). The stalks of both invertible sheaves are free \(R\)-modules of rank one. Choose generators \(e\) and \(h\), and write \(f_x(e)=ah\) with \(a\in R\). Surjectivity of the sheaf map implies surjectivity on stalks, so there is \(b\in R\) with \(f_x(be)=h\). Consequently \(ba=1\). The ring is commutative, so \(a\) is a unit with inverse \(b\), and \(ch\mapsto bce\) is the inverse module map.
2. The stalkwise isomorphisms make \(f\) an isomorphism of sheaves.
For an open set \(U\), any section of \(\mcl(U)\) whose image under \(f\) is zero has zero germ at every point by step 1, and is therefore zero. Thus \(f\) is injective on sections over every open set.
Let \(s\in\mcm(U)\). Surjectivity at each stalk gives local lifts: a lift of the germ at \(x\) is represented by a section near \(x\), and equality with the germ of \(s\) becomes equality of sections after shrinking that neighborhood. On overlaps, these local lifts agree by injectivity on sections. They therefore glue to a unique section \(t\in\mcl(U)\) with \(f(t)=s\). Uniqueness makes this inverse assignment compatible with restrictions and with the \(\OO_X\)-module operations. It defines the inverse sheaf morphism, as in P-AGH215ISOINJSURJ.
3. Q.E.D.
Step 1 proves invertibility on every stalk, and step 2 constructs the inverse morphism of sheaves.