Show that a morphism of sheaves is an isomorphism if and only if it is both injective and surjective.
Solution
Let \(\phi: \mcf \to \mcg\) be a morphism of sheaves. Being injective and surjective says exactly that the sequence \[ 0 \to \mcf \mapsvia{\phi} \mcg \to 0 \] is exact as a sequence of sheaves. By the stalkwise criterion for exactness, this holds if and only if \[ 0 \to \mcf_P \mapsvia{\phi_P} \mcg_P \to 0 \] is exact for every \(P\), that is, if and only if every \(\phi_P\) is an isomorphism of abelian groups.
It remains to see that a morphism of sheaves which is an isomorphism on all stalks is an isomorphism. This is standard: injectivity on stalks makes \(\phi\) injective on sections, and surjectivity on stalks lets one lift any section of \(\mcg\) locally and then glue the lifts, which are unique by the injectivity just established. The resulting inverse maps on sections are compatible with restriction and define an inverse morphism \(\inverseof{\phi}\).