Given two subsheaves $\mathcal{F}$ and $\mathcal{G}$ of a sheaf $\mathcal{O}$ on a topological space X, prove that $\mathcal{F}_p = \mathcal{G}_p$ for every p $\in$ X $\Rightarrow$ $\mathcal{F}(U) = \mathcal{G}(U)$ for every open $U \subseteq$ X
I've been trying to prove this in context of showing that if $\mathcal{F}_p \xrightarrow{\phi_p} \mathcal{F'}_p \xrightarrow{\psi_p} \mathcal{F''_p}$ is exact (as a sequence of stalks) at $\mathcal{F'_p}$ then $\mathcal{F} \xrightarrow{\phi} \mathcal{F'} \xrightarrow{\psi} \mathcal{F''}$ is exact (as a sequence of sheaves) at $\mathcal{F'}$.
I got till the point where we have that $(Im(\phi))_p = (Ker(\psi))_p$ for every p in X and I want to use the above proposition to get that $Im(\phi) = Ker(\psi)$ as subsheaves of $\mathcal{F'}$ but I have no idea how to prove the proposition rigorously.
Any help would be appreciated!