A presheaf can be seen as a contravariant functor

Given a topological space $(X, mathscr{T})$ we have a category Open$(X)$ of open sets where the objects obj$($Open$(X))$ are the open sets of $X$ and the morphisms are inclusions mappings between subsets and none otherwise, i.e. if $U subseteq V$, then $i : U hookrightarrow V, x mapsto i(x) = x$. Verify that the data of a presheaf is precisely the data of a contravariant functor from the category of open sets of $X$ to the category of sets Sets.

A contravariant functor $mathscr{F}$ is a functor from the opposite category Open$^{op}(X)$ into Sets. In other words, the objects are open sets as in Open$(X)$ but the morphisms are arrow reversed, i.e. for $U subseteq V$ we have $i^{op} : V rightarrow U, x mapsto i^{op}(x) = x$. Hence, the morphisms become restrictions (am I right?).

So now we have for each $mathscr{F} : text{obj}(text{Open}(X)) rightarrow text{Sets}, U mapsto mathscr{F}(U)$. In other words, to each open set of $X$ there is a set $mathscr{F}(U)$ (the sections of $mathscr{F}$ over $U$).

We also have for any $U subseteq V$ such that $mathscr{F} : text{Hom}(V, U) rightarrow text{Hom}(mathscr{F}(V), mathscr{F}(U)), i^{op} rightarrow mathscr{F}(i^{op})$. In other words, for any restriction from $V$ to $U$, we have $mathscr{F} : mathscr{F}(V) rightarrow mathscr{F}(U)$. How does $mathscr{F} : mathscr{F}(V) rightarrow mathscr{F}(U)$ become the restriction map res$_{V, U} : mathscr{F}(V) rightarrow mathscr{F}(U)$ as required in a presheaf?

A functor is required to preserve identities, i.e. $mathscr{F}(text{id}_{U}) = text{id}_{mathscr{F}(U)}$. Since inclusion/restriction of $U$ to itself becomes $i(x) = x = text{id}(x)$.

And a functor also has to preserve compositions, that is, if $U subseteq V subseteq W$ such that we restrict $W$ to $V$ via $i^{op}$ and we restrict $V$ to $U$ via $j^{op}$, then by composition of mappings we can restrict $W$ directly to $U$ via $i^{op} circ j^{op}$ and by functoriality we must have $mathscr{F}(i^{op} circ j^{op}) = mathscr{F}(i^{op}) circ mathscr{F}(j^{op})$. So if res$_{V, U} = mathscr{F}(i^{op})$ then this fullfills the last criterion for a presheaf.