$p_*(F) \simeq \prod_{s \in S} F_s$ where $F$ is a sheaf, $S$ is discrete space.

Question

Let $S$ be a discrete topological space. Let $p:X \times S \rightarrow X$ be a projection. $F$ is a sheaf on $X\times S$ and let $F_s=i_s^*(F),$ where $i_s(y)=(y,s) $. Show that $p_*(F) \simeq \prod_{s \in S} F_s.$

In first sight this statement looks easy, but when I tried to make this exercise, I don't have idea how associate $F(p^{-1}(U))$ with product and $colim$ at right side (I think, here $colim$ commute with product because we work under filtered category, I'm right?). Could anyone give me any hint how to do this (I think easy) exercise? I would be grateful.

Answer 1

Here is another proof using two facts : $i_s^*{i_s}_*$ is naturally isomorphic to the identity functor (because $i_s$ is an immersion) and $i_s^*$ commute with arbitrary product (this is obvious, but it holds only because $i_s$ is an open immersion, and in fact in this case, $i_s^*$ has a left adjoint).

Now we can prove something stronger : the natural map $\varphi:F\rightarrow \prod_{s\in S}{i_s}_*i_s^*F=\prod{i_s}_*F_s$ is an isomorphism. To prove this, it is enough to prove it locally, and we will use the open covering $(X\times\{s\})_s$. Now apply $i_s^*$ to $\varphi$. Using the two fact above, you see that $\varphi$ is locally an isomorphism.

To prove your claim, simply apply $p_*$ to $\varphi$. Since $p_*$ commutes with arbitrary product and $p_*{i_s}_*=(p\circ i_s)_*=\operatorname{id}_*$, you have : $$p_*F\overset{\varphi}\longrightarrow p_*\prod {i_s}_*F_s\simeq \prod p_*{i_s}_*F_s=\prod F_s$$

Answer 2

The first observation is that for a discrete space $S$, $X \times S \cong \coprod_{s \in S} X \times \{ s \}$. Since $F$ is in particular a contravariant functor, we know that $F$ of a coproduct will pull out as a product; remembering this, let $U \subset X$ open: \begin{equation} \begin{split} p_* F(U) &= F( p^{-1} (U)) \\ &= F \big( \coprod_{s \in S} U \times \{ s \} \big) \\ &= \prod_{s \in S} F \big( U \times \{ s \} \big) \\ &= \prod_{s \in S} \varinjlim_{V \ni s} \big( F (U \times V ) \big) \\ &= \prod_{s \in S} i^*_s F (U) \\ &= \prod_{s \in S} F_s (U) \end{split} \end{equation} And we are done.