Definition 6.30.1 (009I)—The Stacks project

Definition 6.30.1. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology on $X$.

A presheaf $\mathcal{F}$ of sets on $\mathcal{B}$ is a rule which assigns to each $U \in \mathcal{B}$ a set $\mathcal{F}(U)$ and to each inclusion $V \subset U$ of elements of $\mathcal{B}$ a map $\rho ^ U_ V : \mathcal{F}(U) \to \mathcal{F}(V)$ such that $\rho ^ U_ U = \text{id}_{\mathcal{F}(U)}$ for all $U \in \mathcal{B}$ whenever $W \subset V \subset U$ in $\mathcal{B}$ we have $\rho ^ U_ W = \rho ^ V_ W \circ \rho ^ U_ V$.
A morphism $\varphi : \mathcal{F} \to \mathcal{G}$ of presheaves of sets on $\mathcal{B}$ is a rule which assigns to each element $U \in \mathcal{B}$ a map of sets $\varphi : \mathcal{F}(U) \to \mathcal{G}(U)$ compatible with restriction maps.

Comments (2)

Comment #3218 by Kevin Buzzard on March 07, 2018 at 17:09

Your definition of presheaf seems non-standard to me. Do you not want to also demand that the restriction from $U$ to $U$ is the identity map? [this is not implied by what you have at the time of writing; for example if $A$ is your favourite set with two elements and $a$ is your favourite element of this set, then we can define $\mathcal{F}(U)=A$ for all $U$ and we can let $\rho^U_V$ be the constant map sending everything to $a$ for all $U$ and $V$ ]

Comment #3320 by Johan on May 17, 2018 at 17:49

Dear Kevin, OK I finally got to your comment! Of course, this is an omission and I didn't really intend to have a nonstandard definition of a presheaf! Fixed here.

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