Affine Morphisms: affine-local on target

A morphism $\pi: X \to Y$ is affine if for every affine open $U \subset Y$, $\pi^{-1}(U)$ is affine.

Clearly, affine morphisms are quasi-compact and quasi-separated (since affine schemes are).

Proposition. $\pi$ is affine iff there is a cover of $Y$ by affine opens $U_{\alpha}$ such that $\pi^{-1}(U_{\alpha})$ is affine.

The proposition has some nonobvious consequence. Recall that if $Z \subset \text{Spec} A$ is globally cut out by an equation (i.e. $Z = \mathbb{V}(V)$) then its complement is affine. It turns out that is is also true if $Z$ is locally cut out by an equation.

Corollary. Let $Z$ be a closed subset of $X = \text{Spec} A$. Suppose $X$ can be covered by affine open sets on each of which $Z$ is cut out by one equation. Then the complement of $Z$ is affine.

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Proof of Proposition. By the Affine Communication Lemma, it suffices to check that the condition $\pi^{-1}(U)$ (variable $U$) is affine-local. This is equivalent to checking the following to criteria.

1. Suppose $\pi^{-1}(\text{Spec} B) = \text{Spec} A$ is affine. Then for all $g \in B$, $\pi^{-1}(\text{Spec} B_g)$ is affine.
   Proof. We have $\pi$ restricts to a map $\text{Spec} A \to \text{Spec} B$ which must be induced from some $\phi: B \to A$. Localization gives $\phi_g: B_g \to A_\phi(g)$  which must induce the restriction of $\pi: \text{Spec} A_\phi(g) \to \text{Spec} B_g$. So $\pi^{-1}(\text{Spec} B_g) = \text{Spec} A_{\phi(g)}.$
2. Suppose $(g_1, \ldots, g_n) = B$ and $\pi^{-1}(\text{Spec} B_{g_i}) = \text{Spec} A_i$ is affine. Then $\pi^{-1}(\text{Spec} B)$ is affine.
   Proof. Let $Z = \pi^{-1}(\text{Spec} B)$. Let $A = O_X(Z)$. We want to show that $Z$ is affine, i.e. $(Z, O_Z)$ is isomorphic  to $(\text{Spec} A, O_{\text{Spec} A})$ via the canonical map $\alpha: Z \to  \text{Spec} A$ which is induced by $A \to O_Z(Z).$ (Think of the case $Z$ is affine. For general case, use gluing).
   
   The factorization $B \to A \to O_Z(Z)$ gives a factorization $$\pi: Z \xrightarrow{\alpha} \text{Spec} A \xrightarrow{\beta} \text{Spec} B.$$ As $\alpha$ and $\beta$ are both surjective, and since $D(g_i)$'s cover $\text{Spec} B$, we must have $\beta^{-1}(D(g_i))$'s cover $\text{Spec} A$.
   
   Thus it suffices to show that $\alpha|_{\pi^{-1}(D(g_i)} : \pi^{-1}(D(g_i)) \to \beta^{-1}(D(g_i))$ is an isomorphism for each $i$. Then by gluing, $\alpha$ is an isomorphism $Z \to \text{Spec} A$ (or by stalk).
   
   Abusing notation, we denote both $\beta^{\sharp}g_i$ (an element of $A$) and $\pi^{\sharp}g_i$ (an element of $O_Z(Z)$) by $f_i$. Then $\beta^{-1}D(g_i) = D(f_i) = \text{Spec} A_{f_i} \subset \text{Spec} A$.
   
   On the other hand, $\pi^{-1}(D(g_i)) = Z_{f_i} = \text{Spec} A_i$. It suffices to show that $\alpha^{\sharp}$ induces an isomorphism $A_{f_i} \to A_i$.
   
   Indeed, notice that $Z$ is quasi-compact and quasi-separated since these notion are affine-local, and we have assume that $\pi$ is affine (hence quasi-compact and quasi-separated) on an open cover of $\text{Spec} B$. Then by the QCQS Lemma, we have the canonical map $A_{f_i} = (O_Z(Z))_{f_i} \to O_Z(Z_{f_i}) = A_i$ is an isomorphism.

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Proof of Corollary.

Suppose $(U_{\alpha})_{\alpha}$ is an affine cover of $X= \text{Spec} A$ such that on each $U_{\alpha}$, $Z$ is cut out by an equation $f_{\alpha}$. Let $V = X - Z$ and $\pi: V \to X$ be the inclusion map. Then $\pi^{-1}(U_{\alpha}) = U_{\alpha} \backslash Z$ is affine so $\pi$ is an affine map.  Thus $\pi^{-1} (X)$ must also be affine.  QED.