My question is in context of
Sheaves with the same stalks are not necessarily isomorphic
another question posted earlier.
Why is it true that given two subsheaves F and F’ of a sheaf G on a topological space X, $F_p = F’_p$ for any p in X implies $F = F’$. I want to understand this rigorously using the definitions as I’m still trying to get used to the language of stalks and sheaves.
I tried proving this as follows:
Assume $F_p = F’_p$.
We will show F(U) $\subseteq$ F’(U) for any U $\subseteq$ X open.
Let s $\in$ F(U) Then given any p in X, consider the germ of s in $F_p$. Now since $F_p = F’_p$, we have that $\in F’_p$ and so by definition of $F’_p$ we get that s is in F’(U)
(where F’$_p = \{<U,s_u> \textrm{s.t } p \in U \subseteq X \textrm{open and } s \in F’(U)\}$ modulo the standard equivalence relation)
Thus F(U) $\subseteq$ F’(U) and the other inclusion follows similarly.
This seems incorrect to me because as far as I understand, I haven’t used the condition that F and F’ are subsheaves of the same sheaf. The statement clearly doesn’t hold for any arbitrary sheaves F and F’. So my first question is about what is wrong with this argument and what would be a correct proof for this?
Additionally, my second question/concern is that I don’t fully understand in what sense the stalks are ‘equal’, since viewing in F$_p$ versus in F’$_p$ should technically be different objects since the restriction maps in F don’t need to be compatible with the restriction maps in F’.