This is not a research level problem for sure. But, similar question by some one else $2$ years back on Stack exchange has not received any attention. So, I thought it does not suit there.

Suppose $\mathcal{F}$ is a presheaf on a paracompact space $X$. Suppose further that $\mathcal{F}$ satisfies the gluing axiom.

Then, $\mathcal{F}(U)\rightarrow\widetilde{F}(U)$ is surjective for each open $U$ in $X$. This is mentioned in Ramanan's Global calculus but they use Escape etale definition of associated sheaf $\widetilde{F}$ where as I use compatible sections definition. They use something called shrinking and all that which did not look interesting for me.

So, a random element in $\widetilde{F}(U)$ is of the form $(s_p)\in \bigsqcup_{p\in U}\mathcal{F}_p$ such that, for each $p\in U$ there exists an open set $U(p)$ containing $p$ and a section $s(p)\in \mathcal{F}(U(p))$ and $s(q)\in \mathcal{F}(U(q))$ such that $s(p)_y=s_y$ for all $y\in U(p)$ and that $s(q)_y=s_y$ for all $y\in U(q)$.

Suppose $y\in U(p)\cap U(q)$ then, we have $s(p)_y=s(q)_y=s_y$.

Thus, we have $s(p)_y=s(q)_y$ for all $y\in U(p)\cap U(q)$ which means there exists an open set $Z(y)$ containing $y$ such that $s(p)|_{Z(y)}=s(q)|_{Z(y)}$ for each $y\in U(p)\cap U(q)$.

This means there is an open cover $\{Z(y)\}$ for $U(p)\cap U(q)$ such that $s(p)|_{Z(y)}=s(q)|_{Z(y)}$ for all $y$.

Suppose I can prove that this imply $s(p)|_{U(p)\cap U(q)}=s(q)|_{U(p)\cap U(q)}$.

Then, by gluing axiom, there exists $s\in \mathcal{F}(U)$ such that $s|_{U_p}=s(p)$ for all $p$ which then imply $s\mapsto(s_p)_{p\in U}$ under the map $\mathcal{F}(U)\rightarrow \widetilde{\mathcal{F}}(U)$ which then prove surjectivity.

So, it all boils down to proving $s(p)|_{U(p)\cap U(q)}=s(q)|_{U(p)\cap U(q)}$ given that $s(p)|_{Z(y)}=s(q)|_{Z(y)}$ for open cover $\{Z(y)\}$ of $U(p)\cap U(q)$. It is like proving uniquenes axiom.

Any comments are welcome which gives an idea of where I can use paracompactness.

  • Paracompactness implies that there is a locally finite cover $U = \cup_{p_i \mid i \in I} U(p_i)$ (my $U(p_i)$ may be smaller than your $U(p_i)$). This means that each $U(p_i)$ only intersects finitely many $U(p_j)$. I think it is not likely that $s(p)|_{U(p) \cap U(q)} = s(q)|_{U(p) \cap U(q)}$, but you should be able to use the local finiteness to select a single small enough cover where the intersections actually match. (it's not immediately obvious to me how to make this argument). – dorebell Sep 13 at 1:52
  • @dorebell I did not completely understand your comment... As it is paracompact, every open cover has locally finite refinement... are you saying there is a smaller open cover $\{V_\alpha\}$ of $U$ where $s(p)|_{V_\alpha\cap V_\beta}=s(q)|_{V_\alpha\cap V_\beta}$... – Praphulla Koushik Sep 13 at 3:26
  • Your goal is to build such a ${V_\alpha}$. The open cover $U = \cup_p U(p)$ has a locally finite refinement $U = \cup_{i \in I} V_i$. This means that for each $i \in I$, $V_i \subseteq U(p_i)$ for some $p_i$. We might as well assume (by renaming $p_i$) that $p_i \in V_i$, and we can now rename the $V_i$ to $U(p_i)$. So without loss of generality, your cover $\{U(p)\}$ has a locally finite subcover $\cup_i U(p_i)$. Now your goal is to use this local finiteness property to find the $V_\alpha$. This seems to be the importance of the paracompactness condition, but I'll let you finish the proof. – dorebell Sep 13 at 3:42
  • @dorebell I am completely lost... I do not see how to use paracompactness here... only thing i know about paracompactness is that every cover has a locally finite refinement.... – Praphulla Koushik Sep 13 at 4:26
  • @Ben I do not yet see how this is natural... Can you kindly give a naive reason why this could be true... Not the result that you said but the one I asked.. – Praphulla Koushik Sep 13 at 20:58

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