On sheafification and stalks

Question

I have a problem but it may be easy for you. So, please give me a lecture. Thank you.

Let $\mathcal{F}$ be a presheaf on a topological space $X$ and ${}^a\mathcal{F}$ a sheafification of $\mathcal{F}$:

${}^a\mathcal{F}(U):=\{s:U\to \bigoplus_{x\in X}\mathcal{F}_x\ |\ s\text{ is a section of }\pi:\mathcal{F}_x\ni a\to x\in X\}$,

where $U$ is an open set in $X$ and $\mathcal{F}_x$ is a stalk of $\mathcal{F}$ associated with $x\in U$.

I understood that ${}^a\mathcal{F}$ is a sheaf, but I did not understand that $({}^a\mathcal{F})_x \simeq \mathcal{F}_x$ for any $x\in X$. According to some texts, it is clear by definion. Why?

Answer 1

You have to use the topology of the etale space. It is the coarse topology such that for every $s\in {\cal F}(U)$, $s:U\rightarrow {}^a{\cal F}$ where $s(x)$ is the image $s_x$ of $s\in {\cal F}_x$ is continue. A base open subset of ${}^a{\cal F}$ is defined by $\{s_x\in {\cal F}(U)\}$ where $s$ is an element $s\in {\cal F}$.

Consider the map $f:{}^a{\cal F}_x\rightarrow {\cal F}_x$ such that for every $u\in {}^a{\cal F}_x$, take a section $s\in {}^a{\cal F}(V), V\subset U$ such that $u$ is the image of $s$ in ${}^a{\cal F}_x={\cal F}_x$. There exists $t\in {\cal F}(W)$ such that $u=t_x$. Write $f(u)=t_x$. $f$ is well-defined, if you consider $t'\in {\cal F}(W')$ such $t'_x=u=t_x$, there exists $A\subset W, A\subset W'$ such that the restriction of $t$ and $t'$ to $A$ are equal this implies that $f$ is well defined.

$f$ is surjective: for every $u\in {\cal F}_x$, consider $s\in {\cal F}(U)$ whose image in ${\cal F}_x$ is $u$, $t:U\rightarrow {}^a{\cal F}$ defined by $t(y)=s_y$ is an element of ${}^a{\cal F}(U)$, and the image of the element in ${}^a{\cal F}_x$ induced by $t$ by $f$ is $u$.

$f$ is injective. Suppose that $f(u)=f(v)=t_x$, without restricting the generality, we can suppose that there exists $s,s'\in{}^a{\cal F}(U)$ whose image in ${}^a{\cal F}_x$ is respectively $u$ and $v$. Remark that the map $p:U_t=\{t_x,x\in U\}\rightarrow U$ defined by $p(t_x)=x$ is injective. Write $W=s^{-1}(U_t)\cap{s'}^{-1}(U_t)$, it is an open subset which contains $x$, we deduce that the restriction of $s$ and $s'$ to $W$ are equal since $p$ is injective. This implies that $u=v$.

Answer 2

As said in the comments, your description of $\mathcal{F}^a$ is not completely correct. First, it should be the disjoint union instead of the direct sum of the $\mathcal{F}_x$. Then all sections of $\pi$ are not allowed, only those that satisfy this condition : $$ (1) \quad \forall x\in U, \exists V\ni x \text{ a neighborhood of $x$ in $U$ and }t\in\mathcal{F}(V) \text{ such that } \forall y\in V, s(y)=t_y $$ Otherwise you have too many sections, for example, if $\mathcal{F}=\mathcal{C}$ is the sheaf of continuous function on a space $X$, with your definition a section in $\mathcal{F}^a$ would consists of the choice of a germ of continuous function at every point, without conditions that these germs glue (and they might define a function which is not continuous).

Hence the good definition is the following : $$\mathcal{F}^a(U)=\{s:U\rightarrow\coprod_{x\in U}\mathcal{F}_x | s \text{ is a section of $\pi$ and satisfies condition $(1)$}\}$$

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Now it is easy to see that $\mathcal{F}^a_x=\mathcal{F}_x$. Indeed, if $s_x$ is a germ of a section in $\mathcal{F}^a_x$, then you can find a representative $(U,s)$ where $s\in\mathcal{F}^a(U)$. Now by condition $(1)$, there exists $t\in\mathcal{F}(V)$ such that $\forall y\in V, s(y)=t_y$. But this implies that $(U,s)$ and $(V,t)$ define the same germ. So $s\in\mathcal{F}_x$.

To be perfectly rigorous, check that what I just described is a well-defined map $\mathcal{F}^a_x\rightarrow\mathcal{F}_x$ which is the inverse of the obvious map $\mathcal{F}_x\rightarrow\mathcal{F}^a_x$.