2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The functions may glue together to form an unbounded function. $\endgroup$ – Lord Shark the Unknown Oct 7 '18 at 19:34
  • 2
    $\begingroup$ The covering needs not be finite, so your $\max(\lvert f\rvert,\lvert g\rvert)$ is not really a bound. $\endgroup$ – Saucy O'Path Oct 7 '18 at 19:35
  • $\begingroup$ I don't see why? If you define $h$ (by gluing) as I did I don't see why it's unbounded? @LordSharktheUnknown $\endgroup$ – algebra1112 Oct 7 '18 at 19:36
  • 1
    $\begingroup$ Because $\max\{1,2,3,4,5,6,7,\cdots\}=\boxed{??}$. $\endgroup$ – Saucy O'Path Oct 7 '18 at 19:39
  • 2
    $\begingroup$ You may not be glueing two functions, you may be glueing infinitely many functions. $\endgroup$ – Lord Shark the Unknown Oct 7 '18 at 20:04
2
$\begingroup$

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 $$ s(N) = s_N(N) = N > \sup_{\mathbb{R}} |s|. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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