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I know that the presheaf of bounded functions is not a sheaf but I don't see why. I checked in wikipedia they say that this presheaf does not verify the axiom of "Glue". For me it verifies this axiom. Indeed, if U and V are open sets, and f and g are bounded functions on U and V respectively, and they agree on the intersection, then combining them in the obvious way - let h(x) = f(x) if x is in U, g(x) if x is in V, is a bounded function with the bound being max(|f|, |g|). Do you think I don't understand well what gluing means?

Thanks.

algebra1112
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  • The functions may glue together to form an unbounded function. – Angina Seng Oct 07 '18 at 19:34
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    The covering needs not be finite, so your $\max(\lvert f\rvert,\lvert g\rvert)$ is not really a bound. –  Oct 07 '18 at 19:35
  • I don't see why? If you define $h$ (by gluing) as I did I don't see why it's unbounded? @LordSharktheUnknown – algebra1112 Oct 07 '18 at 19:36
  • Maybe you're right, but I still don't understand we could take $max(f_{i})$ @SaucyO'Path – algebra1112 Oct 07 '18 at 19:39
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    Because $\max{1,2,3,4,5,6,7,\cdots}=\boxed{??}$. –  Oct 07 '18 at 19:39
  • I still don't agree, if we take for example $max{\frac{1}{2^{î}}}$ it's bounded. I see your point, but if we have everything is bounded I think we can define $max$ @SaucyO'Path – algebra1112 Oct 07 '18 at 19:43
  • Ok, feel free to ignore the point. –  Oct 07 '18 at 19:44
  • I am not ignoring anything I would like just to understand @SaucyO'Path – algebra1112 Oct 07 '18 at 19:46
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    You may not be glueing two functions, you may be glueing infinitely many functions. – Angina Seng Oct 07 '18 at 20:04
  • I see your point. But in the definition there is no restriction about the number of function. I agree with you but if you give me a mathematical proof of this I would be happy and it would be very useful @LordSharktheUnknown – algebra1112 Oct 07 '18 at 22:02
  • Maybe your confusion also arose from the fact that the values of an infinite amount of bounded functions can tend to infinity even though none of the values of the functions on the subsets of the cover is infinite anywhere. This amounts to the question why infinity is not included in the natural numbers even though the number of natural numbers is infinite. For more information on that see e.g. https://math.stackexchange.com/questions/36289/is-infinity-a-number – exchange Dec 05 '18 at 09:15

1 Answers1

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Let us take the $\mathcal{F}$ presheaf of bounded real functions on the real line $\mathbb{R}$. Then for each $U\subset \mathbb{R}$ open we have $$ U \mapsto \mathcal{F}(U) = \{ f\colon U \longrightarrow \mathbb{R} \mid \sup_U |f| < \infty \} $$ It is clearly a presheaf. Now let's see the sheaf requirements.

Fix $U\subset \mathbb{R}$ and an open covering $U=\cup_i U_i$.

  1. For $s,t \in \mathcal{F}(U)$, we need that $$ s|_{U_i} = t|_{U_i}, \forall i \Rightarrow s=t $$
  2. For a family $\{s_i \in \mathcal{F}(U_i)\}$ we need that $$ s_i|_{U_i\cap U_j} = s_j|_{U_i\cap U_j}, \forall i,j \Rightarrow \exists s \in \mathcal{F}(U) \mid s|_{U_i}= s_i $$

It is not hard to see that 1. holds. Now for 2. take $U_i = (i-2,i+2)$ an open interval for each $i\in \mathbb{Z}$. We have $\mathbb{R} = \cup U_i$. Define $$ s_i\colon U_i \longrightarrow \mathbb{R}, \, s_i(t) = t $$ Then $\sup_{U_i} |s_i| = \max\{|i-2|,|i+2|\}$ and $s_i \in \mathcal{F}(U_i)$. Suppose that there exists $s\in\mathcal{F}(\mathbb{R})$ such that $s|_{U_i}= s_i$. Let $N \in \mathbb{Z}$ be such that $N> \sup_{\mathbb{R}} |s|$. Then we have an absurd since $$ s(N) = s_N(N) = N > \sup_{\mathbb{R}} |s|. $$

Alan Muniz
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