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?


  • $\begingroup$ The functions may glue together to form an unbounded function. $\endgroup$ Oct 7, 2018 at 19:34
  • 3
    $\begingroup$ The covering needs not be finite, so your $\max(\lvert f\rvert,\lvert g\rvert)$ is not really a bound. $\endgroup$
    – user562983
    Oct 7, 2018 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 $\endgroup$ Oct 7, 2018 at 19:36
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    $\begingroup$ Because $\max\{1,2,3,4,5,6,7,\cdots\}=\boxed{??}$. $\endgroup$
    – user562983
    Oct 7, 2018 at 19:39
  • 3
    $\begingroup$ You may not be glueing two functions, you may be glueing infinitely many functions. $\endgroup$ Oct 7, 2018 at 20:04

1 Answer 1


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


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