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?


1 Answer 1


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 $f_{|U_i} = g_{|U_i}$.

  • $\begingroup$ (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 ! ) $\endgroup$
    – user171326
    Jun 17, 2017 at 20:58
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    $\begingroup$ 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 $f : \Bbb R \to \Bbb R$ ? $\endgroup$
    – user171326
    Jun 17, 2017 at 21:14
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    $\begingroup$ 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? $\endgroup$
    – lattice
    Jun 17, 2017 at 21:22
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    $\begingroup$ Exactly !! :) ${}$ $\endgroup$
    – user171326
    Jun 17, 2017 at 21:24
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    $\begingroup$ Yeah I don't think your teacher will abe to trick you again on this one :P $\endgroup$
    – user171326
    Jun 17, 2017 at 21:28

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