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This is exercise 4 of chapter 2, Mac lane Moerdijk which I stuck in....

For a basis $B$ of the topology on a space $X$, the restriction functor $r: Sh(X)\rightarrow Sh(B)$ is an equivalence of categories. Iam looking a proof using this hint in the book: define a quasi-inverse $s:Sh(B)\rightarrow Sh(X)$ for $r$ as follows. Given a sheaf $F$ on $B$ and an open set $U\subset X$ consider the cover $\{ B_i : i\in I\}$ of $U$ by all basic open sets $B_i\in B$ which are contained in $U$. Define $s(F)(U)$ by the equilizer

$s(F)(U)\rightarrow \Pi F(B_i) \rightrightarrows \Pi F(B_i \cap B_j)$

I think a sheas on a space defines uniquely a sheaf on the basis also if I have a sheaf on basis, I can cover each open set by the elements of basis(in the definition of a basis we have the are closed under finite intersections so I may use the intersections to work on a finer space?) and this gives a unique sheaf on $X$ .....

Would you please help me to solve this using the info given on the quistion? I know this problem has answers here but I dont see $s(F)(U)$ in any of them and this confuses me....

user468730
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  • Does this answer your question : https://math.stackexchange.com/questions/3560073/for-basis-mathcalb-of-the-topology-on-x-the-restriction-functor-mathbf?rq=1 – Roland Apr 11 '21 at 18:25
  • I think not....there the problem is solved by essential surjectivity and fully faithfulness.... Not by defining an inverse – user468730 Apr 11 '21 at 20:19

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