TITLE: Hartshorne's Exercise II. 2.15 (fully faithful functor)
QUESTION [10 upvotes]: I'm struggling with the last part of the exercise. Namely, let $V,W$ be any two varieties over a field $k$. We build the functor $t$, which induces a natural map 
$$
\mbox{Hom}_{Var}(V,W)\xrightarrow{\beta} \mbox{Hom}_{Sch}(tV,tW)
$$
we need to prove the map is bijective (i.e. we want to show that $t$ is fully faithful). 
Here's what I've done so far: first, let $f:V\to W$. Recall that the map $\beta(f)$ between topological spaces sends an irreducible closed subset of $V$, say $C$, to $\overline{f(C)}\in tW$. We have a bijection between open sets $U$ in $V$ and open sets $tU$ in $tV$, this allows us to define $\beta(f)^{\sharp}$ between the sheaves.
Injectivity is easy: using the two preceeding parts of the exercise we see that a morphism of schemes sends point with residue field $k$ to points with residue field $k$, and these points correspond exactly to the points of the varieties. Hence if $\beta(f)=\beta(g)$, in particular for all points $p\in V$, we have $f(p)=\overline{f(p)}=\beta(f)(p)=\beta(g)(p)=\overline{g(p)}=g(p)$. Furthermore, the sheaves are not problematic (using the definition of push forward, and the fact that $f,g$ coincide on the variety). Hence $f=g$ as morphisms of varieties.
Now comes surjectivity and my problem: take a morphism of schemes $F:tV\to tW$. This sends closed points to closed points. Its restriction to closed points induces $f:V\to W$. This map is continuous, because $F$ is continuous and the topologies are in a bijection. Also, the pull-back of regular functions is induced nicely by $F$. 
The last thing to prove would be the following: that $\beta(f)=F$. In particular, I want to show that for an irreducible closed subset $C$ of $V$, $F(C)=\overline{f(C)}$. Can someone provide some insight on how to do this?

REPLY [2 votes]: Haven't read thoroughly all answers, but this claim has a pretty simple proof. You would like to prove the surjectivity of $t(V)\longrightarrow t(W)$.
As a first step assume that $W$ is an affine variety, making $t(W)$ an affine scheme. Then by (Ex. 2.4) on the structure of morphisms into affine schemes, we know that the entire morphism's structure is determined by the map on global sections, that is by $\Gamma(t(W),\mathcal{O}_{t(W)})\longrightarrow \Gamma(t(V),\mathcal{O}_{t(V)})$. By the definition of
$\Gamma(t(W),\mathcal{O}_{t(W)})$, this is exactly $A(W)$, the coordinate ring of $W$, and $\Gamma(t(V),\mathcal{O}_{t(V)}) = \mathcal{O}_V(V)$ is the ring of global regular functions of $V$. Therefore, our map of schemes induces a ring map $A(W)\longrightarrow \mathcal{O}_V(V)$, which induces a morphism of varieties from $V$ to $W$. Denote this morphism by $\phi$. Now the functorial construction of $t$ shows that $t(\phi):t(V)\longrightarrow t(W)$ is a morphism, which, at the level of global sections, is identical to $\phi^{\#}_W$, that is, to the "map $\phi$ on global sections". Since this map induces the same map as our original map $\Gamma(t(W),\mathcal{O}_{t(W)})\longrightarrow \Gamma(t(V),\mathcal{O}_{t(V)})$ at the level of global sections, and as this map (because of Ex. 2.4) determines the entire morphism's structure, we conclude that our original morphism of schemes must coincide with $t(\phi)$.
As a final step, we show how the claim for affine $t(W)$'s implies the claim for arbitrary $t(W)$'s. The variety $W$ can be covered by open varieties $W_i$, each of which giving rise to an affine scheme $t(W_i)$ and we may use the $t(W_i)$ as an affine open cover of $W$. We restrict the morphism $f:t(V)\longrightarrow t(W)$ to a bunch of morphisms $f_i:t(V_i)\longrightarrow t(W_i)$, where the $W_i$ are affine opens that cover $W$ and the $V_i$ are opens that cover $V$. Because the $t(W_i)$'s are affine schemes, we have just shown that there exists a morphism of varieties $\phi_i: V_i\longrightarrow W_i$ such that $t(\phi_i) = f_i$, and we would like to show that these morphisms glue together in order to show that there exists a morphism $\phi: V\longrightarrow W$ such that $t(\phi) = f$. Let $i,j$ be a pair of indices, we need to show that $\phi_i|_{V_i\cap V_j} = \phi_j|_{V_j\cap V_i}$, but we know that $t(\phi_i)|_{t(V_i\cap V_j)} = t(\phi_j)|_{t(V_j\cap V_i)}$, and that the mapping $\phi\mapsto t(\phi)$ is injective, as Hartshorne mentions this is the easy part of the bijection's proof. Hence, by injectivity it follows that $\phi_i|_{V_i\cap V_j} = \phi_j|_{V_j\cap V_i}$ and therefore we conclude that the $\phi_i$ glue together to give a morphism $\phi: V\longrightarrow W$, such that for all $i$, $t(\phi)|_{t(V_i)} = t(\phi|_{V_i}) = t(\phi_i) = f_i = f|_{t(V_i)}$, which implies that $t(\phi) = f$.