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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).

So, we have the functor $\mathbb{r}$ such that $$ \mathbb{r}:{\rm Sh}(X)(F,G) \to {\rm Sh}(\mathcal{B})(r(F),r(G)),\quad \text{where} \quad f \mapsto r(f) $$

where $f$ is a natural transformation between sheaves. Now, we wish to prove the injectivity of this functor, which means, that we want to verify that for all $f,g \in {\rm Sh}(F,G)$, $\mathbb{r}(f)= \mathbb{r}(g)$, we have that $f=g$.

My first question here comes from my lack understanding on how they consider as their hypothesis for injectivity that $f(B)=g(B)$ whenever $B \in \mathcal{B}$. Is $\mathbb{r}(f)= \mathbb{r}(g) \Leftrightarrow f(B)=g(B)$? If so why?

Then, they say the following,

Now, by the hypothesis, $$f(x) {\mid_{V_i}} = f(V_i)(x {\mid_{V_i}}) = g(V_i)(x {\mid_{V_i}}) = g(x) {\mid_{V_i}}$$ for each $i$;

Where does that first equality come from? I can't seem to figure it out.

If anyone can help me in these two question I would be the most grateful!

Any help is really appreciated.

babu
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1 Answers1

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$f$ and $g$ are maps of sheaves from $F$ to $G$. As such, they consist of a collection of maps $f(U)\colon F(U)\to G(U)$ for every open $U$ of $X$ that satisfy naturality with respect to restriction maps, and therefore, if $f(U)=g(U)$ for all open $U$ in $X$, then $f=g$. But $\mathrm{r}(f)$ just restricts the collection $\{f(U)\colon FU\to GU\mid U\in\mathrm{Ouv}(X)\}$ to the collection $\{f(B)\colon FB\to GB\mid B\in\mathcal{B}\}$, so $\mathrm{r}(f)=\mathrm{r}(g)$ just says that $f(B)=g(B)$ for all $B\in\mathcal{B}$.

Now, notationally the equality $f(x)|_{V_i}=f(V_i)(x|_{V_i})$ can be a bit confusing. For $x\in FU$, the left-hand side stands for $\left(f(U)(x)\right)|_{V_i}$. By naturality of $f$ with respect to restriction maps, this equals $f(V_i)(x|_{V_i})$.

Daniël Apol
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