On the sheaf of discontinuous sections

Question

Let $X$ be a topological space and $\mathcal F$ be a sheaf. Godement defined the sheaf of discontinuous sections $\mathcal G(\mathcal F)$ by putting, for an open set $U\subseteq X$, $\mathcal G(\mathcal F)=\prod_{x\in U}\mathcal F_x$, with the natural restriction maps. If $X_{\text{disc}}$ is the set $X$ with the discrete topology, this Wikipedia's article says that we have $\mathcal G(\mathcal F)\simeq p_*p^{-1}\mathcal F$, where $p:X_{\text{disc}}\to X$ is the identity, $p_*$ is the direct image of $p$ and $p^{-1}$ is the inverse image of $p$. I have tried to prove this directly computing the stalks but failed.

How can one prove the isomorphism $\mathcal G(\mathcal F)\simeq p_*p^{-1}\mathcal F$?

Answer 1

1. For any sheaf on any topological space $X$, if $U=\coprod U_i$ is a disjoint union of open sets then $\mathcal{F}(U)=\prod\mathcal{F}(U_i)$ as it can easily be seen from the gluing condition of a sheaf. Thus on a discrete space $X$, we have $U=\coprod_{x\in U}\{x\}$ and so $\mathcal{F}(U)=\prod_{x\in U}\mathcal{F}(\{x\})$. Moreover $\mathcal{F}(\{x\})=\mathcal{F}_x$ since the collection of neighborhood of $x$ has a smallest element, namely $\{x\}$. Hence over a discrete space $X$, for any sheaf $\mathcal{F}$, we have $\mathcal{F}(U)=\prod_{x\in U}\mathcal{F}_x$.

2. Now let us compute the stalks of $p^{-1}\mathcal{F}$. If $x\in X$, denote $i_x:\{x\}\to X$ the inclusion, then $$(p^{-1}\mathcal{F})_x=i_x^{-1}p^{-1}\mathcal{F}=(pi_x)^{-1}\mathcal{F}=i_x^{-1}\mathcal{F}=\mathcal{F}_x$$ In other words, the stalks of $p^{-1}\mathcal{F}$ are the same as the stalks of $\mathcal{F}$.

3. Combining 1. and 2. together, we find that $$(p^{-1}\mathcal{F})(U)=\prod_{x\in U}(p^{-1}\mathcal{F})_x=\prod_{x\in U}\mathcal{F}_x$$

4. Finally, take the pushforward of $p^{-1}\mathcal{F}$ and remember that $p$ is just the identity over the underlying sets : $$(p_*p^{-1}\mathcal{F})(U)=(p^{-1}\mathcal{F})(p^{-1}(U))=(p^{-1}\mathcal{F})(U)=\prod_{x\in U}\mathcal{F}_x=\mathcal{G(F)}(U)$$ Therefore, we have an isomorphism $\mathcal{G(F)}\simeq p_*p^{-1}\mathcal{F}$.