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I would like to prove that a functor is an equivalence of categories. I tried to use that equivalence of categories is the same as fully faithful plus "surjective". I found a difficulty verfying this properties. Here what I am trying to prove. Given $Y$ subset of a topological space $X$. We suppose that $Y$ induces a bijection between open subset of $X$ and open subset of $Y$ by the map $U \rightarrow U \cap Y$. Let $j: Y \rightarrow X$ an inclusion. We denote by $Sh(X)$ and $Sh(Y)$ the categories of sheaves on X and Y. Show that the functor $j^{-1}: Sh(X) \rightarrow Sh(Y)$ is an equivalence of categories. I would appreciate your answers. Thanks!

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You have to show that $j^*:Sh(X)\rightarrow Sh(Y$ is fully faithful and essentially surjective.

Let $F,G$ sheaves defined on $X$, suppose that $j^*(F)=j^*(G)$. Let $U$ be an open subset of $X$, $j^*F(U\cap Y)=j^*G(U\cap Y)=F(j^{-1}(U\cap Y))=G(j^{-1}(U\cap Y))$. We have $j^{-1}(U\cap Y)\cap Y=U\cap Y$. Since $U\rightarrow U\cap Y$ is bijective $j^{-1}(U\cap Y)=U$, we deduce that $F(U)=G(U)$.

Let $H$ be an element of $Sh(Y)$, define $H'(U)=H(U\cap Y)$. It is a sheaf defined on $X$, we deduce that $j^*$ is faithfull and surjective.

$j^*$ is also fully faithfull since $Hom(j^*(F),j^*(G))(U\cap Y)=Hom(F(U),G(U))$ since $j^*F(U\cap Y)=F(U)$.

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  • $\begingroup$ It is the definition of the pullback of a sheaf. $\endgroup$ – Tsemo Aristide Oct 8 '18 at 21:51

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