Lemma 89.4.2. Let $X, x_ i, U_ i \to X, u_ i$ be as in (89.3.0.1) and assume $f : Y \to X$ corresponds to $g_ i : Y_ i \to U_ i$ under $F$. Then there exists a factorization
\[ Y = Z_ m \to Z_{m - 1} \to \ldots \to Z_1 \to Z_0 = X \]
of $f$ where $Z_{j + 1} \to Z_ j$ is the blowing up of $Z_ j$ at a closed point $z_ j$ lying over $\{ x_1, \ldots , x_ n\} $ if and only if for each $i$ there exists a factorization
\[ Y_ i = Z_{i, m_ i} \to Z_{i, m_ i - 1} \to \ldots \to Z_{i, 1} \to Z_{i, 0} = U_ i \]
of $g_ i$ where $Z_{i, j + 1} \to Z_{i, j}$ is the blowing up of $Z_{i, j}$ at a closed point $z_{i, j}$ lying over $u_ i$.
Proof.
A blowing up is a representable morphism. Hence in either case we inductively see that $Z_ j \to X$ or $Z_{i, j} \to U_ i$ is representable. Whence each $Z_ j$ or $Z_{i, j}$ is a decent algebraic space by Decent Spaces, Lemma 68.6.5. This shows that the assertions make sense (since blowing up is only defined for decent spaces). To prove the equivalence, let's start with a sequence of blowups $Z_ m \to Z_{m - 1} \to \ldots \to Z_1 \to Z_0 = X$. The first morphism $Z_1 \to X$ is given by blowing up one of the $x_ i$, say $x_1$. Applying $F$ to $Z_1 \to X$ we find a blowup $Z_{1, 1} \to U_1$ at $u_1$ is the blowing up at $u_1$ and otherwise $Z_{i, 0} = U_ i$ for $i > 1$. In the next step, we either blow up one of the $x_ i$, $i \geq 2$ on $Z_1$ or we pick a closed point $z_1$ of the fibre of $Z_1 \to X$ over $x_1$. In the first case it is clear what to do and in the second case we use that $(Z_1)_{x_1} \cong (Z_{1, 1})_{u_1}$ (Lemma 89.3.3) to get a closed point $z_{1, 1} \in Z_{1, 1}$ corresponding to $z_1$. Then we set $Z_{1, 2} \to Z_{1, 1}$ equal to the blowing up in $z_{1, 1}$. Continuing in this manner we construct the factorizations of each $g_ i$.
Conversely, given sequences of blowups $Z_{i, m_ i} \to Z_{i, m_ i - 1} \to \ldots \to Z_{i, 1} \to Z_{i, 0} = U_ i$ we construct the sequence of blowing ups of $X$ in exactly the same manner.
$\square$
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