My question is mainly along the lines of another post:
Surjectivity on stalks implies surjectivity on sheaves
I’m trying to prove that given a map of sheaves $\psi : \mathcal{F} \rightarrow \mathcal{G}$ the followings true:
$(\mathcal{Im}\psi)_p = \mathcal{G}_p$ for every p $\Rightarrow \mathcal{Im}(U) = \mathcal{G}(U)$ for any open subset U of X
I found a solution
https://www.math.arizona.edu/~cais/CourseNotes/AlgGeom04/Hartshorne_Solutions.pdf
Which states that $\psi$ is surjective iff $\mathcal{Im}\psi = \mathcal{G}$ iff $(\mathcal{Im}\psi)_p = \mathcal{G}_p$ For every p iff $\mathcal{Im}\psi_p = \mathcal{G}_p$ For every p iff $\psi_p$ is surjective
Most of this is fine, but why does step 2 hold? I.e why is it true that
$\mathcal{Im}\psi = \mathcal{G}$ Iff $(\mathcal{Im}\psi)_p = \mathcal{G}_p$ For every p
I can see why the forward implication holds but why the backward?
The answer and comments in the post above didn’t seem to involve the fact any specific properties of the image sheaf (as was pointed out in the comments) and so the answer seemed weird (given that the statement shouldn’t hold for arbitrary sheafs F and G)
My questions are then
1) what is the proof for the reverse implication of the above mentioned step 2?
2) where (if at all) is there a mistake in the answer/comments in the above mentioned post? And if there are no mistakes, then where in both inclusions in the proof for the reverse implication are the hypotheses (that we are working in the image sheaf) used?
3) is there a more elegant approach to proving the initial surjectivity statement?