# proving that a functor is an equivalence of categories

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!

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 $$j^*F(U\cap Y)=F(U)$$.