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.

locally finitecover $U = \cup_{p_i \mid i \in I} U(p_i)$ (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