$\theta: \mathcal {F}\rightarrow \mathcal {G}$ is an isomorphism iff $\theta_x:\mathcal {F}_x\to\mathcal {G}_x $ is an isomorphism for any $x $.

Question

Let $\mathcal {F}$, $\mathcal{G}$ be sheaves.

Show that $\theta: \mathcal {F}\to \mathcal {G}$ is an isomorphism iff $\theta_x:\mathcal {F}_x\to \mathcal {G}_x $ is an isomorphism for any $x$.

I have to use the next Lemma:

$\phi:Hom (\mathcal {F},\mathcal {G})\to \prod_{x}Hom (\mathcal{F}_x, \mathcal {G}_x) $,

$\phi (\theta)=(\theta_x)_{x\in X} $.

Then $\phi$ is injective and the image is

{$(\theta_x)_{x\in X}|\forall U\subseteq X$ open, $\forall s\in \mathcal {F }(U), \exists ! t\in \mathcal {G}(U) $ s.t. $t_x=\theta_x (s_x), \forall x\in U $}.

I have trouble with "$\Leftarrow $".

Answer 1

You can follow your nose and prove this directly. I've written a proof below which probably uses some form of the lemma you mentioned.

The stalk $F_x$ can be thought of as the group of equivalence classes $(t,V)$, where $V$ is an (open) neighborhood of $x$ and $t \in F(V)$. We have $(t_1, V_1) = (t_2,V_2)$ if and only if there exists an open neighborhood $W$ of $x$, contained in $V_1 \cap V_2$, such that $t_1|_W = t_2|_W$. If $(t,V) \in F_x$, then $\theta_x(t,V)$ is defined to be $(\theta(V)t,V)$. This is well defined.

Assume $F_x \rightarrow G_x$ is an isomorphism for all $x$. It's not difficult to show that each map $F(W) \rightarrow G(W)$ is injective. Let's show that that each map $\theta(U):F(U) \rightarrow G(U)$ is surjective. Let $s \in G(U)$. For each $x \in U$, let $s(x)$ be the image of $s$ in $G_x$. Now $\theta_x$ is surjective, so there exists a neighborhood $V_x$ of $x$ and an element $(t_x,V_x) \in F_x$ (for $t_x \in F(V_x)$) such that $\theta_x(t_x,V_x) = s(x)$. Using the definition of $\theta_x$, we can choose $V_x$ so that it is contained in $U$ and so that $\theta(V_x)t_x = s|_{V_x}$.

I claim that $t_x$ and $t_y$ agree on $W := V_x \cap V_y$ for all $x, y \in U$. Indeed, $\theta(W)$ is injective, and maps $t_x|_W - t_y|_W$ to $s|_W - s|_W = 0$. Since $V_x : x \in U$ forms an open cover of $U$, the sheaf axiom tells us that there exists a $t \in F(U)$ such that $t|V_x = t_x$ for all $x \in U$.

We now have $\theta(U)t = s$, because they agree on the open cover $V_x$ of $U$.