Subsheaves and stalks This question has already been asked in some other posts but has not been clearly addressed. Suppose $F,G\subseteq H$ are two subsheaves of $H$ where $H$ is a sheaf on a topological space $X$. Then how does $F_x=G_x$ for any $x\in X$ implies $F=G$, or equivalently $F(U)=G(U)$ for any open $U\subseteq X$?
 A: It is a general fact that if a morphism of sheaves induces an isomorphism on all the stalks, then it is an isomorphism. Working with strict equalities with sheaves is a bit awkward since there is so much data. I believe this is how it would be done using equality at the stalks.
If $s \in F(U)$ is a section, then $[s,U] \in F_p$ for all $p \in U$. Then, since $F_p = G_p$, we know that there is some $U_p$ so that $[s|_{U_p}, U_p] \sim [s, U]$ is in $G_p$ (using the equality, some representative of $[s, U]$ is in $G_p$) so that $s|_{U_p} \in G(U_p)$. Then $\{U_p\}_{p \in U}$ is an open cover of $U$ with $s|_{U_p}$ all agreeing on intersections so by the sheaf condition, $s \in G(U)$. This shows that $F(U) \subseteq G(U)$ and the reverse direction is symmetric.
A: Question: "Then how does Fx=Gx for any x∈X implies F=G, or equivalently F(U)=G(U) for any open U⊆X?"
Answer: Note: If $E,F$ are two sheaves of abelian groups on a topological space $X$ and if $E_x \cong F_x$ for all $x\in X$, this does not imply that $E \cong F$ as sheaves of abelian groups. If there is a map
$$\phi:E \rightarrow F$$
such that the induced map
$$\phi_x:E_x \rightarrow F_x$$
is an isomorphism for all $x$ this implies that $E \cong F$. The isomorphism $E_x \cong F_x$ must be induced by a map $\phi$ of sheaves for the result to hold.
