Lemma 6.4.2. Let $X$ be a topological space. The category of presheaves of sets on $X$ has products (see Categories, Definition 4.14.6). Moreover, the set of sections of the product $\mathcal{F} \times \mathcal{G}$ over an open $U$ is the product of the sets of sections of $\mathcal{F}$ and $\mathcal{G}$ over $U$.

Proof. Namely, suppose $\mathcal{F}$ and $\mathcal{G}$ are presheaves of sets on the topological space $X$. Consider the rule $U \mapsto \mathcal{F}(U) \times \mathcal{G}(U)$, denoted $\mathcal{F} \times \mathcal{G}$. If $V \subset U \subset X$ are open then define the restriction mapping

$(\mathcal{F} \times \mathcal{G})(U) \longrightarrow (\mathcal{F} \times \mathcal{G})(V)$

by mapping $(s, t) \mapsto (s|_ V, t|_ V)$. Then it is immediately clear that $\mathcal{F} \times \mathcal{G}$ is a presheaf. Also, there are projection maps $p : \mathcal{F} \times \mathcal{G} \to \mathcal{F}$ and $q : \mathcal{F} \times \mathcal{G} \to \mathcal{G}$. We leave it to the reader to show that for any third presheaf $\mathcal{H}$ we have $\mathop{\mathrm{Mor}}\nolimits (\mathcal{H}, \mathcal{F} \times \mathcal{G}) = \mathop{\mathrm{Mor}}\nolimits (\mathcal{H}, \mathcal{F}) \times \mathop{\mathrm{Mor}}\nolimits (\mathcal{H}, \mathcal{G})$. $\square$

There are also:

• 2 comment(s) on Section 6.4: Abelian presheaves

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).