# On sheafification and stalks

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?

• Where did you get that definition from? First time I see it. When I was introduced to it we defined ${}^a\mathcal{F}(U)$ as a subset of $\prod_{x\in U}\mathcal{F}_x$. It's maybe the same thing, I'm just curios. Commented Aug 14, 2017 at 16:30
• Also I'm not sure about your condition in the set. A section of a morphism is usually something like this: you have $\pi:A \to B$ and $s:B\to A$ with $\pi\circ s=id_B$. But in your condition $s$ and $\pi$ don't map to the same spaces. Can you clarify? Commented Aug 14, 2017 at 16:43
• As remarked by @Maik, your definition of ${}^a\mathcal{F}(U)$ is completely false. Commented Aug 14, 2017 at 17:28
• Possible duplicate of Sheafification of a presheaf through the etale space Commented Aug 14, 2017 at 17:49

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

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 $\mathcal{F}_x\rightarrow\mathcal{F}^a_x$.

• How does $(U,s)$ and $(V,t)$ define the same germ? They live in different spaces. Commented Sep 17, 2020 at 14:44
• @AlJebr The explanation was given in the line just below. Is it not clear ? Commented Sep 17, 2020 at 17:21
• I'm sorry, where is the explanation? I see that $s$ lives in $\mathcal F^a(U)$ but $t$ lives in $\mathcal F(V)$. How can $(U,s)$ and $(V,t)$ define the same germ if the direct limit is taken over different spaces? I guess I should ask: What do you mean by they define the same germ? Commented Sep 17, 2020 at 21:07
• @AlJebr The explanation was "To be perfectly rigorous, check that what I just described is a well-defined map $\mathcal{F}^a_x\to\mathcal{F}_x$ which is the inverse of the obvious map $\mathcal{F}_x\to\mathcal{F}^a_x$." What I meant is the class of $(V,t)$ is a germ in $\mathcal{F}_x$, and the canonical map $\phi:\mathcal{F}\to\mathcal{F}^a$ maps this germ to a germ in $\mathcal{F}^a_x$ which is still represented by $(V,t)$ (or rather $(V,\phi(t))$). Then $(U,s)=(V,\phi(t))$ in $\mathcal{F}^a_x$ and the map $(U,s)\to (V,t)$ well defined and the inverse of $\phi$. Commented Sep 17, 2020 at 21:45
• Ok. Thanks. I think I get it now. This is all new to me so I think I was getting thrown off by the abuse of notation. I'm not used to it yet. I think I see it now though. Thanks. Commented Sep 17, 2020 at 22:10

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 $u=v$.

I use the notation in here.

We want to show that for any $$x\in X$$, $$\phi_x$$ is bijection.

injectivity: Let $$\phi_x(s_x)= \phi(t_x)$$ (where $$s\in \mathcal{F}(U), t\in \mathcal{F}(V)$$).

Since $$\phi_x(s_x)=(\tilde{s})_x, \phi_x(t_x)=(\tilde{t})_x$$, $$\tilde{s}|_W=\tilde{t}|_W$$ for some open neighbourhood $$x\in W\subset U\cap V$$ by definition of germ. So for any $$p\in W$$, $$(p, s_p)=\tilde{s}(p)=\tilde{t}(p)=(p, t_p)$$, therefore $$s_p=t_p$$. Especially, you get $$s_x=t_x$$, considering a case $$p=x$$.

surjectivity: Take arbitrary $$s_x\in \mathcal{F}^{+}_x$$ (where $$s\in \mathcal{F}^{+}(U)$$). We write $$s(x)=(x, t_x)$$(where $$t\in \mathcal{F}(V)$$), then by continuity of $$s$$ on $$U$$, for some open neighbourhood $$x\in W\subset U$$ such that $$s(W)\subset [V, t]$$ since $$[V, t]$$ is open neighbourhood of $$s(x)$$. That is, for every $$y\in W$$, $$s|_W(y)=s(y)=(y, t_y)$$. therefore we get $$\phi(V)(t)|_W=s|_W$$ and, \begin{align} s_x =( \phi(V)(t))_x = \phi_x(t_x) \in \operatorname{Im} \phi_x. \end{align}