Starting from a $B$-sheaf $\hat{\mathcal{F}}$, a sheaf defined on the basic open sets of a topological space $X$, I am asked to prove that the construction $$\mathcal{F}(U) = \varprojlim_{V \subset U, V \in B} \hat{\mathcal{F}}(V)$$ defines a sheaf.

I already shown that $\mathcal{F}$ defines a presheaf, where the restriction maps are given by the universal property of the inverse limit. However, I got stuck trying to show that it satisfies the sheaf axiom.

If a take an open cover $U = \bigcup_{i \in I}U_i$ by open sets $U_i \subset X$, and I take sections on $s_i = \mathcal{F}(U_i)$ such that $s_i = s_j$ on $U_i \cap U_j$ for all $i, j$, I am supposed to show that there is a unique section $s \in \mathcal{F}(U)$ such that restricted to $U_i$ is exactly $s_i$ for all $i$.

Now, a section $s_i \in \mathcal{F}(U_i)$ is actually a tuple of sections $(t_V)_V$ where $t_V \in \hat{\mathcal{F}}(V)$ for each $V \subset U, V \in B$, and such that the restriction maps satisfy $$\mathrm{res}_{U_i,U_j}(s_j) = s_i $$ for all $U_i \subset U_j$.

However, I don't know what to infer from this to prove the sheaf axiom. A hint rather than an answer would be very much appreciated. Thank you.


First, beware, you wrote "the basic open sets of a topological space $X$". I do not know if it is just a small mistake while writing, but there can be multiple basis for a given topology. Here we just chose one, that is $B$.

Now, to define $s \in \mathcal{F}(U)$ is the same as giving compatible sections $s_V$ for each $V$ a basic open set in $U$ (compatible in the following meaning : if $V_0$ is a subset of both $V_1$ and $V_2$, where all of these $V$ are basic open sets, then $s_{V_0}$ is the restriction of both $s_{V_1}$ and $s_{V_2}$). Now, given such a $V \subset U$, if we want $s$ to satisfy the restriction conditions on $U_i$, then $s_V$ should satisfy certain conditions. Can you see what they are and how to recover $s_V$ from them ? I will leave some hints in the following spoiler that you should not read before trying by yourself.

that is, $s_V$ restricted to $V \cap U_i$ should be equal to $s_i$ restricted to $V \cap U_i$. Also, notice that you can replace the cover $V = \bigcup (V \cap U_i)$ with another cover where any open set of the cover is a basic open set (I will leave this part to you). Now you have restricted your problem to the case where all open sets are basic open, which is exactly the $B$-sheaf axiom.

In the end, all you need to show is that the $s_V$ you got this way are compatible, so that you can recover $s \in \mathcal{F}(U)$. You will also have to show unicity, but the proof is in the same vein :

$s$ is uniquely determined by the $s_V$, and the $s_V$ are unique with the restriction conditions given by the ones on $s$.


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