Definition of a stalk why must we be careful about the equivalence realtion

Question

We define a stalk as follows. Let $\mathcal{F}$ be a sheaf of abelian groups on $X$ and let $p$ be a point in $X$.

The stalk $\mathcal{F}_p:\{(U,s)\mid p\in U \subset X \text{ an open set and } s\in \mathcal{F} \}/\sim$. Where $(U,s)\sim (U',s')$ if $s\mid_V=s'\mid_V$ for some $V\subset U\cap U'$ open.

Seemingly we cannot always replace $V$ with $U\cap U'$ as then this would not be an equivalence relation, but I cannot come up with an example on how this fails to be an equivalence relation.

Answer 1

Consider what happens if we have a $U''$ properly contained in $U\cap U'$. We could have pairs $(U,s)$, $(U', s')$, and $(U'', s'')$ such that $s$ agrees with $s''$ on $U\cap U''=U''$ and $s'$ agrees with $s''$ on $U'\cap U''=U''$, but $s$ and $s'$ don't agree on the larger set $U\cap U'$. So transitivity could fail in general if we don't "cut down" appropriately.

Answer 2

Let $X=\{a,b,p\}$ with the discrete topology, and let $\mathcal{F}$ be the constant sheaf with value $\Bbb Z$. Consider the following global sections, both of which assign $0$ to $p$:

• $\alpha$, which assigns $1$ to $a$ and $0$ to $b$,
• $\beta$, which assigns $0$ to $a$ and $1$ to $b$.

Using the relation of "agreeing where defined", these are both equivalent to the local section defined on $\{p\}$ which assigns $0$ to $p$, but they're not equivalent to each other. So the relation is not transitive.