Hartshorne's Exercise II. 2.15 (fully faithful functor)

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?

Step 1: Suppose $f: V \rightarrow W$ is a morphism of affine varieties, and let $f^{\#}: A(W) \rightarrow A(V)$ be the induced homomorphism of affine coordinate rings. Let $Y$ be a closed irreducible subset of $V$ corresponding to the prime ideal $P$ of $A(V)$. Then the the zero set of $\left(f^{\#}\right)^{-1} (P)$ in $W$ is precisely the closure of $f(V)$ in $W$ - prove this by showing directly that any closed set of $W$ that contains $f(V)$ must contain the zero set of $\left(f^{\#}\right)^{-1} (P)$.

Step 2: Let $V,W$ be affine varieties and let $(\phi,\phi^{\#}):(t(V),\alpha_* \mathcal{O}_V) \rightarrow (t(W),\alpha_* \mathcal{O}_W)$ be a morphism of schemes ($\alpha_*$ is as in the proof of Proposition II.2.6). Since we have an isomorphism of schemes $(\psi_V,\psi^{\#}_V):(t(V),\alpha_*\mathcal{O}_V) \cong (Spec A(V),A(V))$, and similarly $(\psi_W,\psi^{\#}_W):(t(W),\alpha_*\mathcal{O}_W) \cong (Spec A(W),A(W))$, we have an induced morphism of schemes $(\gamma,\gamma^{\#}):(Spec A(V),A(V)) \rightarrow (Spec A(W),A(W))$, and a commutative diagram:

$$ \require{AMScd} \begin{CD} (t(V),\alpha_*\mathcal{O}_V) @>(\phi,\phi^{\#})>> (t(W),\alpha_*\mathcal{O}_W)\\ @VV({\psi_V},{\psi^{\#}_V})V @VV(\psi_W,\psi^{\#}_W)V \\ (Spec A(V),A(V)) @>(\gamma,\gamma^{\#})>> (Spec A(W),A(W)) \end{CD} $$

By restricting $\gamma^{\#}$ on global sections we get a ring homomorphism $f^{\#}: A(W) \rightarrow A(V)$, which in turn induces a morphism of varieties $f: V \rightarrow W$. This $f$ is precisely the restriction of $\phi$ to the closed points of $t(V)$. The proof of Proposition II.2.3(c), implies that $\gamma$ is given by contracting under $f^{\#}$ prime ideals of $A(V)$ to prime ideals of $A(W)$. So if $Y$ is a closed and irreducible subset of $V$ corresponding to prime ideal $P \in Spec A(V)$, we will have that $\gamma(P) = (f^{\#})^{-1}(P)$, or by the commutativity of the diagram $\phi(Y) = \psi_W^{-1} \circ \gamma \circ \psi_V(Y) = \mathcal{Z}\left((f^{\#})^{-1}(P)\right)$, where $\mathcal{Z}(\cdot)$ is the zero-set operator. Step 1 now guarantees that $\mathcal{Z}\left((f^{\#})^{-1}(P)\right)$ is the closure of $f(Y)$ in $W$.

Step 3: Now let V, W by any varieties and $(\phi,\phi^{\#}):(t(V),\alpha_* \mathcal{O}_V) \rightarrow (t(W),\alpha_* \mathcal{O}_W)$ a morphism of schemes. If we cover $W$ by affine opens $W=\bigcup_j U_j$, the scheme $(t(V),\alpha_* \mathcal{O}_V)$ is covered by the affine schemes $(t(U_j),\alpha_* \mathcal{O}_{U_j})$. Now $\phi^{-1}(\alpha(U_j))$ is an open set in $t(V)$, say $\phi^{-1}(\alpha(U_j)) = t(V) - t(Y_j)$, with $Y_j$ closed and irreducible in $V$. Covering $V-Y_j$ by affine opens $V-Y_j = \bigcup_{i} V_{ij}$ we have that the scheme $\left(\phi^{-1}(\alpha(U_j)), \alpha_* \mathcal{O}_V|_{\phi^{-1}(\alpha(U_j))}\right)$ is covered by the affine schemes $(t(V_{ij}), \alpha_* \mathcal{O}_{V_{ij}})$ and so $(\phi,\phi^{\#})$ is described locally by morphisms of affine schemes $(\phi_{ij},\phi_{ij}^{\#}) :(t(V_{ij}), \alpha_* \mathcal{O}_{V_{ij}}) \rightarrow (t(U_j),\alpha_* \mathcal{O}_{U_j})$. Finally, using Step 2 each $(\phi_{ij},\phi_{ij}^{\#})$ is induced by some $f_{ij}: V_{ij} \rightarrow U_j$, which glued together induce a morphism of varieties $f: V \rightarrow W$. It can be seen by reversing the above local arguments that $f$ induces $(\phi,\phi^{\#})$.

I have some partial progress in the case where $W$ is affine.

First, assume $V, W$ are both affine varieties. Show the natural map given by $t$, $Hom(V, W) \to Hom(tV, tW)$ is equal to the composition

\begin{align*} Hom_{Var}(V, W) \to Hom_{fg \text{ } k-alg}(AW, AV) \to Hom_{Sch/k}(Spec AV, Spec AW) \to Hom_{Sch/k}(tV, tW), \end{align*} where the first two maps are the canonical maps are given by pullback, and the third map is the natural identification of prime ideals and irreducible closed subsets, given by $I(\bullet)$ and $V(\bullet)$. All three of these maps are bijective. To show the map on $Hom$ induced by $t$ is equal to the above composite, you may want to use the lemma:

Lemma: Let $f: V \to W$ is a morphism of varieties, and let $g: AW \to AV$ be the induced map by pullback. If $P$ is a prime in $AV$ corresponding to closed irreducible set $C$ in $V$, then the prime $g^{-1}P$ corresponds to $\overline{f(C)}$.

Proof: Nullstellensatz.

Second, we now just assume the second variety is affine, and the first is any kind of variety. I'll call the first variety $X$ and the second $Y$. The goal is to reduce to the first situation where both varieties are affine. Suppose for simplicity that $X$ has an open cover of two affine varieties $U$, $V$. Note $X$ has a basis of open affines, and it is not hard to extend this part of the proof to the situation where we might not have a two element open cover. By pasting lemma for varieties, $Hom(X, Y)$ is the pullback of $Hom(U, Y)$ and $Hom(V, Y)$ over $Hom(U \cap V, Y)$. In plainer language, $Hom(X, Y)$ is in bijection with pairs of morphisms in $Hom(U, Y) \times Hom(V, Y)$ that agree over $Hom(U \cap V, Y)$. Since we also have pasting lemma for schemes, $Hom(tX, tY)$ is also the pullback of $Hom(tU, tY)$ and $Hom(tV, tY)$ over $Hom(t(U \cap V), tY)$. Consider this diagrams:

$$ \require{AMScd} \begin{CD} Hom(X, Y) @>>> Hom(U, Y)\\ @VVV @VVV \\ Hom(V, Y) @>>> Hom(U \cap V, Y) \end{CD} $$

$$ \require{AMScd} \begin{CD} Hom(tX, tY) @>>> Hom(tU, tY)\\ @VVV @VVV \\ Hom(tV, tY) @>>> Hom(t(U \cap V), tY) \end{CD} $$

I meant for this diagram to be a cube, where the front face is the top diagram of maps of varieties, the back face is the bottom diagram of maps of schemes, and the maps from the front face to the back face are maps induced by $t$.

Verify that all 6 sides of the diagram commutes. Note that from the first part $Hom(U, Y) \to Hom(tU, tY)$ is bijective (and similarly for $V$). Then chase the diagram to see that a compatible choice of morphisms from $U$ and $V$ into $Y$ maps into a compatible choice of morphisms from $tU$ and $tV$ into $tY$. Conversely, take a compatible choice of morphisms from $tU$ and $tV$ into $tY$. These two morphisms of schemes come from two morphism of varieties; check they agree on $U \cap V$, by using the fact that $t$ is faithful.

Third, somehow do this for when both varieties are any kind of variety. I'm not sure how to continue from here. Maybe someone has a suggestion?

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 $t(\phi) = f$.