Lemma 38.5.6. With assumption and notation as in Definition 38.5.5 we have:

1. If $S' \subset S$ is a standard open neighbourhood of $s$, then setting $X' = X_{S'}$, $Z'_ k = Z_{S'}$ and $Y'_ k = Y_{S'}$ we obtain a standard shrinking.

2. Let $W \subset Y_ n$ be a standard open neighbourhood of $y$. Then there exists a standard shrinking with $Y'_ n = W \times _ S S'$.

3. Let $U \subset X$ be an open neighbourhood of $x$. Then there exists a standard shrinking with $X' \subset U$.

Proof. Part (1) is immediate from Lemmas 38.5.4 and 38.4.7.

Proof of (2). For convenience denote $X = Y_0$. We apply Lemma 38.4.7 (2) to find a standard shrinking $S', Y'_{n - 1}, Z'_ n, Y'_ n$ of the one step dévissage of $\mathop{\mathrm{Coker}}(\alpha _{n - 1})/Y_{n - 1}/S$ at $y_{n - 1}$ with $Y'_ n = W \times _ S S'$. We may repeat this procedure and find a standard shrinking $S'', Y''_{n - 2}, Z''_{n - 1}, Y''_{n - 1}$ of the one step dévissage of $\mathop{\mathrm{Coker}}(\alpha _{n - 2})/Y_{n - 2}/S$ at $y_{n - 2}$ with $Y''_{n - 1} = Y'_{n - 1} \times _ S S''$. We may continue in this manner until we obtain $S^{(n)}, Y^{(n)}_0, Z^{(n)}_1, Y^{(n)}_1$. At this point it is clear that we obtain our desired standard shrinking by taking $S^{(n)}$, $X^{(n)}$, $Z_ k^{(n - k)} \times _ S S^{(n)}$, and $Y_ k^{(n - k)} \times _ S S^{(n)}$ with the desired property.

Proof of (3). We use induction on the length of the complete dévissage. First we apply Lemma 38.4.7 (3) to find a standard shrinking $S', X', Z'_1, Y'_1$ of the one step dévissage of $\mathcal{F}/X/S$ at $x$ with $X' \subset U$. If $n = 1$, then we are done. If $n > 1$, then by induction we can find a standard shrinking $S''$, $Y''_1$, $Z''_ k$, and $Y''_ k$ of the complete dévissage $(Z_ k, Y_ k, i_ k, \pi _ k, \mathcal{G}_ k, \alpha _ k, z_ k, y_ k)_{k = 2, \ldots , n}$ of $\mathop{\mathrm{Coker}}(\alpha _1)/Y_1/S$ at $x$ such that $Y''_1 \subset Y'_1$. Using Lemma 38.4.7 (2) we can find $S''' \subset S'$, $X''' \subset X'$, $Z'''_1$ and $Y'''_1 = Y''_1 \times _ S S'''$ which is a standard shrinking. The solution to our problem is to take

$S''', X''', Z'''_1, Y'''_1, Z''_2 \times _ S S''', Y''_2 \times _ S S''', \ldots , Z''_ n \times _ S S''', Y''_ n \times _ S S'''$

This ends the proof of the lemma. $\square$

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