isomorphism of presheaves vs isomorphism of sheaves

Question

Is it true that an isomorphism of presheaves between two presheaves induces an isomorphism of sheaves between their sheafifications?

Answer 1

Yes. Let $\mathcal{F}_1,\mathcal{F}_2$ be the pre-sheaves, and $\tilde{\mathcal{F}_i}$, $i=1,2$ be the sheafifications. Consider $$\tilde{\mathcal{F}}_1\gets \mathcal{F}_1\to \mathcal{F}_2\to \tilde{\mathcal{F}_2} $$ This induces by definition of $\tilde{\mathcal{F}}_1$ a unique map $\tilde{\mathcal{F}_1}\to \tilde{\mathcal{F}_2}$. Taking the inverse of the map $\mathcal{F}_1\to \mathcal{F_2}$ we get a map $\tilde{\mathcal{F}_2}\to \tilde{\mathcal{F}_1}$. Using the uniques it follows by a standard argument that these two maps are inverses, and hence $\tilde{\mathcal{F}_2}\cong \tilde{\mathcal{F}_1}$

Answer 2

Yes; this follows purely from the fact that sheafification is a functor. Functors always send isomorphisms to isomorphisms because they preserve composition and identities. More generally, functors preserve left and right inverses.

Theorem: Let $F: \mathsf{C} \to \mathsf{D} $ be a functor. If $f: X \to Y$ is an arrow in $\mathsf{C}$ with a left (resp. right) inverse $g: Y \to X$, then $F(g)$ is a left (resp. right) inverse for $F(f)$.

Proof: If $g \circ f = \mathrm{id}$ then $$\mathrm{id}_{F(X)} = F(\mathrm{id}_{X}) = F(g \circ f) = F(g) \circ F(f).$$