Showing that sheafification induces an isomorphism on stalks (without using adjoints)

Question

Let $\mathscr{F}$ be a presheaf and let $\mathscr{F}^+$ be its sheafification, defined as follows: $$\mathscr{F}^+(U) = \left\{(s_p) \in \prod_{p \in U}\mathscr{F}_p : (*)\right\}$$ where the condition $(*)$ says that for all $p \in U$ there is an open neighbourhood $V \subseteq U$ of $p$ and $t \in \mathscr{F}(V)$ such that $t_q = s_q$ for all $q \in V$ ($t_q$ is the germ of $t$).

I'm having a hard time showing that sheafification $\mathscr{F} \to \mathscr{F}^+$ induces an isomorphism on stalks. Suppose $s_x$ is a germ of $\mathscr{F}$ at $x,$ so there is a representative $(s,U)$ with $x \in U$ and $s \in \mathscr{F}(U).$ Then sheafification takes this to the representative $(\prod_{p\in U}s_p,U).$

I have two questions. How do we show that this map is an isomorphism and what would be the inverse map? Also, is there a simpler/nicer representative for $(\prod_{p\in U}s_p,U)?$ For example, I'd like to think that we can project $\prod_{p\in U}s_p$ to $s_x,$ but I'm not sure how we can define $s_x$ as a section of $\mathscr{F}^+.$

Answer 1

I am not sure I understand your second question. Note in my answer of the first question, I will not write $\prod_{p\in U} s_p$, but $(s_p)_{p\in U}$ : in general, if $(A_i)_{i\in I}$ is a family of sets, then the elements of $\prod_{i\in I} A_i$ are families $(a_i)_{i\in I}$ where $a_i\in A_i$. In particular, you can look at its "coordinates" : the $x$-coordinate of $(s_p)_{p\in U}$ is $s_x$.

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The map $\Phi:\mathcal{F}_x\to\mathcal{F}^+_x$ is by definition the following :

• for an element $s_x\in\mathcal{F}_x$, first take a representative $(s,U)$ where $s\in\mathcal{F}(U)$.
• send this representative to the family $(s_p)_{p\in U}\in\mathcal{F}^+(U)$
• take the class of $(s_p)_{p\in U}$ in $\mathcal{F}^+_x$.

Now let us show that this map is a bijection.

• The map $\Phi$ is into, indeed, if $s_x,t_x$ are sent to the same element, this means that the class of $(s_p)_{p\in U}$ is equal to the class of $(t_p)_{p\in V}$. But by definition of germs, this means that there exists $W\subset U\cap V$, an open neighborhood of $x$, such that $(s_p)|_W=(t_p)|_W$. But $(s_p)|_W$ is just $(s_p)_{p\in W}$ and similarly $(t_p)_{p\in V}|_W=(t_p)_{p\in W}$. This in turns, implies that for all $p\in W$, $s_p=t_p$. So in particular, with $p=x$, $s_x=t_x$. This proves the injectivity.
• The map $\Phi$ is onto, indeed, let $s_x\in\mathcal{F}^+_x$, then fix a representative $(s,U)$ where $s\in\mathcal{F}^+(U)$. So by definition $s$ is a family $s=(s_p)_{p\in U}$ which satisfies condition $(*)$. This implies that there exists a neighborhood $V\subset U$ of $x$ such that $(s_p)_{p\in V}$ is of the form $(t_p)_{p\in V}$ for a section $t\in\mathcal{F}(V)$. I claim that $s_x=\Phi(t_x)$. Indeed, let us compute $\Phi(t_x)$. Choose the representative $(t,V)$, then consider the class of $(t_p)_{p\in V}$. Since $(s_p)_{p\in U}$ and $(t_p)_{p\in V}$ coincide on a neighborhood of $x$(=V), they define the same class which is $s_x$. This proves that $s_x=\Phi(t_x)$, so this proves that $\Phi$ is onto.