# Why is this map continuous? Group operation on étalé space of a sheaf

Let $$X$$ be some topological space and suppose we have a presheaf of abelian groups, $$F\colon \mathrm{\textbf{Op}}(X) \to \textbf{Ab}$$. Construct the corresponding étalé space as follows:

a) For each $$x \in X$$ set $$\mathscr{F}_x \colon= \mathrm{Lim}_{U \ni x} \, F(U)$$ and let $$\phi_{x}^{U}$$ be the homormorphism from $$F(U)$$ into the direct limit. Set $$\mathscr{F} = \sqcup \mathscr{F}_x$$.

b) For $$f \in F(U)$$ set $$[f,U] = \{ (\phi_{x}^{U}(f),x) | x \in U \}.$$ One can check that the sets $$[f,U]$$ for $$U$$ open in $$X$$ and $$f \in F(U)$$ form a basis for topology on $$\mathscr{F}$$, so give $$\mathscr{F}$$ the topology generated by them.

Let $$\pi$$ be the projection from $$\mathscr{F} \to X$$ that simply takes every element of $$\mathscr{F}_x$$ to $$x$$. One can check that $$\pi$$ is a local homeomorphism. One can also check that the map $$(f,x) \mapsto (-f,x)$$ is continuous from $$\mathscr{F}$$ to $$\mathscr{F}$$.

Now let $$\mathscr{F} + \mathscr{F}$$ represent the subspace of $$\mathscr{F} \times \mathscr{F}$$ consisting of pairs $$((f,x), (g,x))$$ for some $$x \in X$$. How can we confirm that the map $$((f,x),(g,x)) \mapsto (f+g,x)$$ is continuous from $$\mathscr{F} + \mathscr{F}$$ to $$\mathscr{F}$$?

For context, this construction is a functor from presheaves to category of étalé spaces and is one half of the composition of functors that yields sheaffification, and the equivalence of categories. My construction above is verbatim as in Serre's \textit{FAC}. I can't figure out how to show that group operation is continuous, however.

• Normally you show that each of the maps $g(x) = a \cdot x$ is continuous for any fixed $a \in$ the group. That is sufficient to prove that the full binary operation is continuous w.r.t. the product topology. – Shine On You Crazy Diamond Nov 12 '18 at 23:30
• Roll up and smoke adjoint, I think I just came up with a good argument. I am going to post below - please feel free to comment. – Prince M Nov 12 '18 at 23:38

## 1 Answer

How about this, we will shoot for pointwise continuity.

Let $$\psi$$ be the map from $$\mathscr{F} + \mathscr{F}$$ to $$\mathscr{F}$$ that takes $$((f,x),(g,x))$$ to $$(f+g,x)$$.

Let $$((f,x),(g,x)) \in \mathscr{F} + \mathscr{F}$$ and let $$[t,U]$$ be any basis element of $$\mathscr{F}$$ containing the image $$\psi((f,x),(g,x)) = (f+g,x)$$. Usually we would consider an arbitrary open set $$V$$ containing $$(f+g,x)$$, but it is sufficient to show it for a basis element.

Now since $$(f+g,x) \in [t,U]$$ this means that $$x \in U$$ and $$\phi_{x}^{U}(t) = f+g$$. By properties of the direct limit we know there exists open $$V$$ containing $$x$$ with some $$s_1 \in F(V)$$ such that $$\phi_{x}^{V}(s_1) = f$$, likewise we have an open $$W$$ containing $$x$$ with an $$s_2$$ in $$F(W)$$ such that $$\phi_x^{W}(s_2) = g$$.

This means that \begin{align*} f+g &= \phi^{V \cap W}_{x} \circ \phi^{V}_{V \cap W}(s_1) + \phi^{V \cap W}_{x} \circ \phi^{W}_{V \cap W}(s_2)\\ &= \phi^{V \cap W}_{x}( \phi^{V}_{V \cap W}(s_1) + \phi^{W}_{V \cap W}(s_2)). \end{align*}

Again by properties of the direct limit, this means that there exists some open $$O$$ containing $$x$$ such that $$O$$ is contained in $$U$$ and contained in $$V \cap W$$ such that

\begin{align*} \phi^{U}_{O}(t) &= \phi^{V \cap W}_{O}(\phi^{V}_{V \cap W}(s_1)+\phi^{W}_{V \cap W}(s_2))\\ &= \phi^{V}_{O}(s_1) + \phi^{W}_{O}(s_2). \end{align*}

This means that $$[\phi^{V}_{O}(s_1) + \phi^{W}_{O}(s_2),O] = [\phi^{U}_{O}(t),O] \subset [t,U]$$

So now take $$[\phi^{V}_{O}(s_1),O] \times [\phi^{W}_{O}(s_2),O] \cap \mathscr{F} + \mathscr{F}$$ which is open. As a set we have $$[\phi^{V}_{O}(s_1),O] \times [\phi^{W}_{O}(s_2),O] \cap \mathscr{F} + \mathscr{F} = \{(\phi^{O}_{x}(\phi^{V}_{O}(s_1)),x),(\phi^{O}_{x}(\phi^{W}_{O}(s_2)),x) : x \in O \}$$

The image of the set under $$\psi$$ is $$[\phi^{V}_{O}(s_1) + \phi^{W}_{O}(s_2),O]$$.