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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’.

KReiser
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User095746
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  • Equal means equal as elements of the stalk $G_p$. – Qiaochu Yuan Sep 19 '19 at 03:15
  • So when we say that “for two arbitrary sheafs, equality of stalks does not imply equality of the sheaves” we’re using ‘equal’ in a different sense then? As some sort of isomorphism of abelian groups / rings / whatever the algebraic structure of the stalk is? – User095746 Sep 19 '19 at 03:26
  • Yes, that's an imprecise statement. The precise statement is that if $F, G$ are two abstract sheaves, a collection of isomorphisms $\phi_p : F_p \cong G_p$ of stalks does not necessarily lift to an isomorphism $F \cong G$. – Qiaochu Yuan Sep 19 '19 at 05:24
  • I see. Thank you for the clarification! Though how does the lift work when the sheaves are subsheaves of another fixed sheaf? – User095746 Sep 19 '19 at 05:35
  • With the given hypotheses what you'll want to show is that if $F_p = F_p'$ for all $p$ then $F(U) = F'(U)$ for all $U$ (as subsets of $G(U)$) and then the desired isomorphism is just literally the identity map. – Qiaochu Yuan Sep 19 '19 at 05:37
  • Let us continue this discussion in chat. – User095746 Sep 19 '19 at 05:48
  • I tried showing that via the argument in the original post, and I don’t understand why that argument doesn’t work (because it seems like it shouldn’t for the reason mentioned). And so I’m unsure how to fix it. – User095746 Sep 19 '19 at 11:42
  • Hint: two sections $s,s'\in G(U)$ are equal if and only if $s_p=s'_p$ for all $p\in U$. What does this tell you if $F(U)\not=F'(U)$ (as subsets of $G(U)$)? – Ben Sep 20 '19 at 13:03

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