When is a sheaf determined by its stalks

Question

Let $\mathcal{F}$ be a sheaf on a topological space. Of course, it is not true in general that $\mathcal{F}$ is entirely determined by its stalks, that is, there can be another sheaf $\mathcal{G}$ with $\mathcal{F}_x \simeq \mathcal{G}_x$ for any $x$, but $\mathcal{G} \not \simeq \mathcal{F}$. Usually, the above isomorphisms must come from a morphism between the sheaves.

However, I am wondering if there could be any condition on $\mathcal{F}$ or on the topological space $X$ for it to hold.

For instance, let $\mathcal{F}$ be a sheaf on a contractible space $X$, and suppose that $\mathcal{F}_x = \mathbb{C}$ for any $x \in X$. Is $\mathcal{F}$ the constant sheaf ?

Thanks a lot !

Answer 1

As you already noticed this is not true in general. However, the correct condition is the following :

Call a sheaf $F$ locally constant if there is a covering $U_i$ with $F_i := F_{|U_i} \cong \underline{\Bbb C}_{|U_i}^r$ for some $r \in \mathbb N$.

For locally constant sheaves, we have the following theorem :

Theorem : There is an equivalence of categories between locally constant sheaves with stalks $\Bbb C^r$ and representations $\rho : \pi_1(X) \to GL_r(\Bbb C)$.

In particular, on any simply connected space the property you asked is true.

To prove the theorem, you first notice that when $F$ is locally constant, $F$ is constant if and only if there is a non-zero global section $s \in F(X)$ (i.e $s_x \neq 0$ for all $x \in X$). Then, the difficult part is to construct a non-zero global section. You do it by induction on the cover, and if the section can't be extended, you can show that there is a loop $\gamma$ such that $F_{|\gamma}$ is not constant. Shrinking $\gamma$ gives a contradiction.

This is only the idea of the proof and one needs more care, see here for a complete proof.