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
Best 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.