How can I understand this definition of a presheaf?

Question

We have given the following definition of a presheaf:

A preasheaf $F$ on a topological space is a functor $$F:\text{Op}(X)^{\text{op}}\rightarrow \text{Sets}$$ Where $\text{Op}$ is the category of open sets.

Now I want to understand this a bit more. So I know that a functor maps objets to objects, but since the objects in $\text{Op}(X)^{\text{op}}$ are the same as the one in $\text{Op}(X)$ we have

• for all $U\subset X$ open $F(U)$ is a set

Now a functor maps morphisms to morphisms. But the morphisms in $\text{Op}(X)^{\text{op}}$ are the one from $\text{Op}(X)$ but with arrows in the other direction right? But the morphisms in $\text{Op}(X)$ are inclusions of opens sets. But now if I have $V\subset U$ in $\text{Op}(X)$ then this means that $U\subset V$ in $\text{Op}(X)^{\text{op}}$, but now what does happen if we apply the functor $F$?

Answer 1

It means that for each open set U inside X, $\mathcal{F}(U)$ is a set, and for two open sets $U \subset V$ we have a morphism of sets $\rho_{V,U} :\mathcal{F}(V) \to \mathcal{F}(U)$ such that $\mathcal{F}$ satisfies the following conditions:

1. $\rho_{U,U}=id_{\mathcal{F}(U)}$
2. For $W \subset V \subset U$, $\rho_{U,W}=\rho_{V,W} \circ \rho_{U,V}$

Answer 2

Generally the category Op$(X)$ has morphisms $V\to U$ iff $V\subset U$.

So the opposite category Op$(X)^{\rm op}$'s morphism are $U\to V$ iff $V\subset U$.

Now we take functor $F:$Op$(X)^{\rm op}\to $Sets, we get map $F(U)\to F(V)$ when $V\subset U$.

For example let $X=\mathbb{R}$ and $F(U):=\{f:U\to \mathbb{R}|f$:continuous } for open subset $U\subset \mathbb{R}$, and by taking restriction of function there is $F(U)\to F(V)$ for $V\subset U$.