Question about tag 03L7 in the stacks project (fpqc covers)

Question

The stacks project omits most of the proof of tag 03L7. I think $(1) \Rightarrow (2) \Leftrightarrow (3)$ is pretty clear, but how do you prove $(2), (3) \Rightarrow (1)$?

Say $(2)$ holds and $U \subseteq T$ is affine open. We can choose finitely many $i$ and quasi-compact opens $U_i \subseteq T_i$ such that $\bigcup_if_i(U_i) = U$. What we want is to show that the $U_i$ can actually be chosen to be affine. Since each $U_i$ is quasi-compact we can take a finite cover $\{W_{ij}\}_j$ by open affines and get $\bigcup_{ij}f_i(W_{ij}) = U$. There's finitely many $W_{ij}$ as needed, but we also need there to be only one affine open per $i$, which we don't currently have.

If the proof is omitted then it must be fairly easy, I must be missing something obvious here but I can't see what.

Answer 1

I'm going to rewrite the conditions here because I hate flipping back and forth. Let $\{f_i:T_i\to T\}_{i\in I}$ be a family of morphisms of schemes. The claim is that the following are equivalent:

1. $f_i$ is flat for all $i\in I$ and for every affine open $U\subset T$, there exists $n\geq 0$, a map $a:\{1,\cdots,n\}\to I$, and affine opens $V_j\subset T_{a(j)}$, $j=1,\cdots,n$, with $\bigcup_{j=1}^n f_{a(j)}(V_j)=U$.
2. Each $f_i$ is flat and for every affine open $U\subset T$ there exist quasi-compact opens $U_i\subset T_i$, which are almost all empty, such that $U=\bigcup f_i(U_i)$.
3. Each $f_i$ is flat and there exists an affine open covering $T=\bigcup_{\alpha\in A} U_\alpha$ and for each $\alpha \in A$ there exist $i_{\alpha,1},\cdots,i_{\alpha,n(\alpha)}$ and quasi-compact opens $U_{\alpha,j}\subset T_{i_\alpha,j}$ such that $U_\alpha=\bigcup_{j=1,\cdots,n(\alpha)} f_{i_{\alpha,j}}(U_{\alpha,j})$.

As each condition requires $f_i$ flat for all $i$, we can focus on the second portion of each statement.

1 implies 2: let $U_i=\bigcup_{j\mid a(j)=i} V_j$, which works as finite unions of affine opens are quasi-compact and at most $n$ of the $U_i$ are not empty.

2 implies 3: given an affine open $U_\alpha$, set $U=U_\alpha$ and then apply 2 to get finitely many nonempty quasi-compact opens $U_i\subset T_i$ with $U=\bigcup f_i(U_i)$. Picking indices, this exactly gives 3.

3 implies 1: Let $U_\beta = D(z)\subset U_\alpha$ for some fixed $\alpha$ and $z$. Then the condition of 3 also holds for $U_\beta$: write each $U_{\alpha,j}$ as a finite union of affine opens, then recognize that the preimage of $U_\beta$ in each $U_{\alpha,j}$ is just the union of a principal affine open of each of these smaller affine opens in $U_{\alpha,j}$, so the preimage of $U_\beta$ in each $U_{\alpha,j}$ is again quasi-compact and so the condition still holds.

The upshot of this last paragraph is that for any affine open $U\subset T$ we can get a cover of $U$ by sets of the form $U_\alpha$: since the principal affine opens generate the topology on $U_\alpha$, we can write the open set $U\cap U_\alpha\subset U_\alpha$ as a union of some principal affine opens and toss those in to our cover as per the above. Once we have that, we may take a finite subcover of the $U_\alpha$s since $U$ is quasi-compact, then cover each $U_{\alpha,j}$ by finitely many affine opens to get a finite list of affine opens of which the union of the images is $U$. We're done.

Note that there's no requirement that the map $a$ in 1 be injective - that's how you can get multiple affine opens in a single $T_i$ involved.