The Stacks project

Proof. The equivalence of (1) and (2) is Lemma 53.16.1.

Assume (3). We will argue by descending induction on $i$ that all singularities of $U_ i$ are multicross. This is true for $U_ n$ as $U_ n$ has no singular points. If $U_ i$ is gotten from $U_{i + 1}$ by glueing $a, b \in U_{i + 1}$ to a point $c \in U_ i$, then we see that

is the set of elements having the same residue classes in $k$. Thus the number of branches at $c$ is the sum of the number of branches at $a$ and $b$, and the $\delta$-invariant at $c$ is the sum of the $\delta$-invariants at $a$ and $b$ plus $1$ (because the displayed inclusion has codimension $1$). This proves that (2) holds as desired.

Assume the equivalent conditions (1) and (2). We may choose an open $U \subset X$ such that $x$ is the only singular point of $U$. Then we apply Lemma 53.15.4 to the normalization morphism

All we have to do is show that in none of the steps we are squishing a tangent vector. Suppose $U_{i + 1} \to U_ i$ is the smallest $i$ such that this is the squishing of a tangent vector $\theta$ at $u' \in U_{i + 1}$ lying over $u \in U_ i$. Arguing as above, we see that $u_ i$ is a multicross singularity (because the maps $U_ i \to \ldots \to U_0$ are glueing of pairs of points). But now the number of branches at $u'$ and $u$ is the same and the $\delta$-invariant of $U_ i$ at $u$ is $1$ bigger than the $\delta$-invariant of $U_{i + 1}$ at $u'$. By Lemma 53.16.1 this implies that $u$ cannot be a multicross singularity which is a contradiction. $\square$