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I am trying to prove that the restriction functor $\mathbb{r}: {\rm Sh}(X) \to {\rm Sh}(\mathcal{B})$ is an equivalence of categories, and I came across this answer (here you can also find the full statement that I am trying to solve, also here is a question I asked also related to this problem).

However, my question is now a bit different: we wish to prove that $f'$ is a morphism of sheaves (you can see what this means more clearly in the answer I have linked). I did as they hinted and considered a $U' \subseteq U$, $x \in F(U)$, and again a open covering of $U$ by $\{V_i| i\in I\} \subseteq \mathcal{B}$. I have reached thus far: $$ (f'(U)(x){\mid_{U'}}){\mid_{V_i}}=(f'(U')(x{\mid_{U'}})){\mid_{V_i}}=f'(V_i)(x{\mid_{V_i}})=y_i $$

Now, by the injectivity condition part on $G$ (I think, given that we previously found that $y_i{\mid_{U_i \cap U_j}}=y_j{\mid_{U_i \cap U_j}}$), we have that $$ f'(U_i)(x){\mid_{U_i \cap U_j}}= f'(U_j)(x){\mid_{U_i \cap U_j}} $$

If I am wrong please correct me on this.

Is this enough to prove that $f'$ is a morphism of sheaves? And do you have any tips to prove that $\mathbb{r}(f')=f?$

Any help is really, really appreciated! Thanks for taking your time to read my question.

babu
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    By "injectivity part of the sheaf condition on $G$" I meant what other sources might call "separatedness of $G$": if you have a cover ${ V_i \mid i \in I }$ of $U'$, and two sections $y_1, y_2 \in G(U')$, then $(y_1) |{V_i} = (y_2) |{V_i}$ for each $i \in I$ implies $y_1 = y_2$. Now, try applying this for $y_1 = (f'(U)(x))|{U'}$ and $y_2 = f'(U')(x |{U'})$ with the canonical cover ${ V\in \mathcal{B} \mid V \subseteq U' }$. – Daniel Schepler Jan 30 '24 at 16:57
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    As for showing $r(f') = f$: by construction, whenever $U$ is open and $V \subseteq U$, $V \in \mathcal{B}$, then for any section $x \in F(U)$, you have $(f'(U)(x)) |_V = f(V)(x |_V)$. Now, if $U \in \mathcal{B}$ already, try the case $V = U$... – Daniel Schepler Jan 30 '24 at 19:35

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