Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a topological space. $U\subset X$ open.
$\mathfrak{B}(U) = \{f:U\to \mathbb{R}| f \textrm{ continuous and bounded}\}$ is a presheaf.

I would like to see the sheafification of this presheaf. I have a solution from my tutorial but I don't understand all the steps:

$\mathfrak{B}^g(U):= \{(f_x)\in\prod\limits_{x\in U}\mathfrak{B}_x| \forall x\in U, \exists W\subset U, x\in W, g\in \mathfrak{B}(W) \textrm{ s.t. }\forall w\in W: g_W = f_W\}$.
$\mathfrak{B}_x = \{f_x:U_x\to\mathbb{R}|U_x\textrm{ ''small enough'' s.t. } f_x\textrm{ continuous and bounded}\}$.
But if $U_x$ is for example the preimage of $B_\epsilon(f(x))$ of $f:U\to \mathbb{R}$ then $f_x$ is bounded and hence $\mathfrak{B}_x=\{f_x:U_x\to\mathbb{R}| f_x \textrm{ continuous}\}$.

(I don't understand what he wanted to tell with this part, and also I am not sure if I wrote $B$ correctly or if it should be $\mathfrak{B}$. We defined $B_\epsilon(0)=\{x\in\mathbb{R}: |x|<\epsilon\}$ in the proof for showing that $\mathfrak{B}$ is not a sheaf.)

Now choose $W=U_x$ for a $x\in U$. So $g=f_x$ and hence $g_w=f_w$ $\forall w\in W\Rightarrow \mathfrak{B}^g(U)=\{f:U\to\mathbb{R}| f \textrm{ continuous}\}$.

I would be really happy if someone could help me understand this example. Maybe you also have a better explanaition of the sheafification?

Thanks and best, Luca

share|improve this question
A better explanation? The difficulty would be to find a worse one... –  Georges Elencwajg Mar 16 '14 at 23:00
I am struggeling at the point where he sais for $U_x$ we have that $f_x$ is bounded and so $B_x=\{f_x:U_x\to\mathbb{R}\}$. So he sais that $f_x$ is bounded on $U_x$ for all $f_x:U_x\to\mathbb{R}$? Do you know what I mean? And yes, I am really happy to have this tutor :) –  Luca Mar 16 '14 at 23:02
Here is the correct order for the quantifiers: given a continuous function $f:U\to \mathbb R$, then for every $x\in U$ there exists an open neighbourhood $U_x\subset U$ of $x$ such that $f$ is bounded on $U_x$. This is what is meant with "continuous functions are locally bounded" in my answer. By the way, I am criticizing a mathematical explanation, not your tutor. –  Georges Elencwajg Mar 16 '14 at 23:24
ok! Thank you, It's now clearer. :) is it OK if I accept Hurkyl's answer? Just because it's like a more detailed version of yours.. –  Luca Mar 16 '14 at 23:29
I confess I feel a little silly that I was so busy sketching how the argument goes that I forgot to mention why we would have even thought it should be true in the first place! –  Hurkyl Mar 16 '14 at 23:36

2 Answers 2

up vote 4 down vote accepted

I think you're using $\mathfrak{B}^g$ to mean the sheafification of $\mathfrak{B}$? I'll assume so.

There is an evident morphism $\mathfrak{B} \to \mathcal{C}$ from the presheaf of bounded continuous functions to the sheaf of continuous functions. This factors uniquely into $\mathfrak{B} \to \mathfrak{B}^g \to \mathcal{C}$, where the first map is the canonical map for the sheafification.

The goal of this proof is to understand these morphisms by understanding how it looks on the stalks of the sheaves.

The key point is that sheafification induces isomorphisms on all of the stalks. And furthermore, a morphism of sheaves is an isomorphism if and only if it induces an isomorphism on all of the stalks.

Thus, by showing that $\mathfrak{B} \to \mathcal{C}$ induces isomorphisms on the stalks of these presheaves, we can conclude that $\mathcal{C}$ is a sheafification of $\mathfrak{B}$.

The key idea to manage this proof is the defining property of continuity: if $f$ is continuous at $x$, you can ensure that $f(y)$ is as close to $f(x)$ as you like by restricting $y$ to an appropriate neighborhood of $x$.

share|improve this answer

The sheafification of your presheaf $\mathfrak{B}$ is the sheaf $\mathcal C$ of continuous functions on $X$.
This is immediate once you realize that every continuous $U\to \mathbb R$ is locally bounded.

Edit: a few words about sheafification in general
Given a presheaf $\mathcal F$ on a topological space $X$, one associates to it a sheaf $\mathcal F^+$, its sheafification.
The construction is a bit opaque for a beginner, involving families of germs of sections on open subsets $U\subset X$.
If, however, by a stroke of luck $\mathcal F\subset \mathcal G$ somehow happens to be a subpresheaf of a sheaf $\mathcal G$, sheafification of $\mathcal F$ becomes very easy: one just defines $$\mathcal F^+(U)=\{g\in \mathcal G(U)|g \; \text {is locally in} \; \mathcal F\}$$ the last condition meaning that there exists a covering $(U_i)$ of $U$ (depending on $g$ !) such that $g|U_i\in \mathcal F(U_i)\subset \mathcal G(U_i)$.
So my (too) crisp answer meant that you should apply this to $\mathcal F=\mathfrak{B}$ and $\mathcal G=\mathcal C$.

And now for the good news: this construction actually works for all presheaves $\mathcal F$.
The trick is to construct a huge sheaf $\mathcal G$ in which to embed an arbitrary presheaf $\mathcal F$.
Well, just take $\mathcal G(U)=\prod_{x\in U} \mathcal F_x$ with the obvious restrictions.
The embedding is of course given by $$ \mathcal F(U)\hookrightarrow \mathcal G(U)=\prod_{x\in U} \mathcal F_x: s\mapsto (s_x)_{x\in U}$$
And this is the explanation for the recipe, apparently drawn out of the blue, everybody gives for sheafifying a presheaf.

share|improve this answer
What do you mean with 'depending on g !' ? –  Ryan Mar 18 '14 at 13:06
Dear @Ryan, I mean exactly what I write: the covering $(U_i)$ of $U$ depends on the element $g$. For example if you take a $g$ already in the original presheaf, $g\in \mathcal F(U)$, you may take the covering with just one piece, namely $U$ itself. But in Luca's question for example, if you take $X=U=\mathbb R$ and $g(x)=x$, the covering might be $U_i=(i-2,i+2)$ or $U_i=(-i,+i)$ but you MUST have infinitely many different $U_i$'s in the covering . –  Georges Elencwajg Mar 18 '14 at 15:07
Thank you for answering! Shouldn't it be for every cover of U? You only want g in F(U) if you can use the glueing axiom for every cover. –  Ryan Mar 18 '14 at 17:56
Dear @Ryan: no, checking that one suitable coverof $U$ exists for each $g$ suffices. That $\mathcal F^+$ then satisfies the sheaf axioms for all covers is automatic. –  Georges Elencwajg Mar 18 '14 at 19:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.