# Sheafification of the presheaf of continuous functions between two topological spaces

The following question (as well as the notations that I use) is from an exercise of my algebraic geometry class:

Let $X$ and $Y$ be topological spaces. Determine the sheafification $\tilde{\mathcal{F}}$ of the presheaf $\mathcal{F}$ of continuous functions from $X$ to $Y$.

We have proven in our lecture that the sheafification of a presheaf can be generally constructed in terms of the stalks,

$\tilde{\mathcal{F}}(U) = \{(s_{x})\in\prod_{x\in U}\mathcal{F}_{x}|\ \forall x\in U\ \exists\ W\subset U, t\in\mathcal{F}(W):x\in W, s_{w}=t_{w}\ \forall w\in W\},$

where (for $s\in\mathcal{F}(U)$) $s_x$ denotes the equivalence class of $(U,s)$ in $\mathcal{F}_{x}$. So in this notation, $t$ would be a continuous function from $W$ to $Y$, with the induced topology on $W$, and the condition means that $(s_x)$ is locally continuous, meaning that for every $x\in U$ there exists an open neighbourhood of $x$ where $(s_x)$ is continuous.

Hence, $\tilde{\mathcal{F}}$ is the sheaf of "locally continuous functions" from $X$ to $Y$.

Is this correct? If so, what actually are "locally continuous functions"? I have trouble distinguishing between them and (globally) continuous functions, especially since we are talking about topological (so not necessarily metric) spaces, where everything seems a bit more abstract to me.

Also, I feel like there is some issue with the formal correctness: in my understanding $s_x$ is (for any $x\in U$) an equivalence class of continuous functions, which are pairwise the same on some neighbourhood of $x$, in particular their value in $x$ is the same. So I would rather define $\tilde{\mathcal{F}}(U)$ to be

$\tilde{\mathcal{F}}(U) = \{f_s:U\to Y|s\in\mathcal{F}(U),\text{ for all }x\in U\text{ it holds }f_s(x)=g(x)\text{ for some }g\in s_{x}\}.$

Is this just a lack of interest in the formal correctness (in favor of shortage of notation), or did I get it wrong? I feel like the notation in our class is often rather sloppy, but perhaps this is just usual in algebraic geometry because otherwise it would become too confusing?

This presheaf is already a sheaf : indeed, continuity is a local condition.

More precisely, a function is continuous if and only if there is a covering $U_i$ of $X$ such that each $f_{|U_i}$ is continuous. Moreover, it is clear that $f=g$ if and only if there is a covering $U_i$ such that $f_{|U_i} = g_{|U_i}$.

• (And I must say I'm pretty surprised you got such exercise, as usually continuous functions/smooth functions are the typical motivating example for introducing sheaf ! )
– user171326
Jun 17, 2017 at 20:58
• Yes it's a very trick question ! Also, it's good to think of sheafification in term of "making sections locals". A more interesting example is the following : what is the sheafification of the presheaf of bounded functions $f : \Bbb R \to \Bbb R$ ?
– user171326
Jun 17, 2017 at 21:14
• Probably locally bounded functions?^^ Actually my next exercise is to determine the sheafification of the presheaf of continuous and bounded functions $f:X\to\mathbb{R}$, so since continuity already is a local property, and since a continuous function is always locally bounded, this sheafification would be the sheaf of continuous functions, right? Similarly for my previous exercise, the sheafification of the presheaf of constant functions $f:X\to E$ (where $E$ is any set) is the sheaf of locally constant functions, right? Jun 17, 2017 at 21:22
• Exactly !! :) ${}$
– user171326
Jun 17, 2017 at 21:24
• Yeah I don't think your teacher will abe to trick you again on this one :P
– user171326
Jun 17, 2017 at 21:28