The Stacks project

Proof. Given an étale morphism $V \to Y$ where $V$ is a scheme choose a scheme $U$ and a surjective étale morphism $U \to X \times _ Y V$. Consider the commutative diagram

Let $u \in U$ with images $x \in |X|$, $y \in |Y|$, and $v \in V$. Then $f$ is Gorenstein at $x$ if and only if $U \to V$ is Gorenstein at $u$ (by definition). Moreover the morphism $U_ v \to X_ v = (X_ V)_ v$ is surjective étale. Hence the scheme $U_ v$ is Gorenstein if and only if the algebraic space $X_ v$ is Gorenstein. Thus the equivalence of (1), (2), and (3) follows from the corresponding equivalence for morphisms of schemes, see Duality for Schemes, Lemma 48.24.4 by a formal argument.

Proof of equivalence of (a) and (b). The corresponding equivalence for flatness is Morphisms of Spaces, Lemma 67.30.8. Thus we may assume $f$ is flat at $x$ when proving the equivalence. Consider a diagram and $x, y, u, v$ as above. Then $\mathcal{O}_{Y, \overline{y}} \to \mathcal{O}_{X, \overline{x}}$ is equal to the map $\mathcal{O}_{V, v}^{sh} \to \mathcal{O}_{U, u}^{sh}$ on strict henselizations of local rings, see Properties of Spaces, Lemma 66.22.1. Thus we have

by Algebra, Lemma 10.156.4. Thus we have to show that the Noetherian local ring $\mathcal{O}_{U, u}/\mathfrak m_ v \mathcal{O}_{U, u}$ is Gorenstein if and only if its strict henselization is. This follows immediately from Dualizing Complexes, Lemma 47.22.3 and the definition of a Gorenstein local ring as a Noetherian local ring which is a dualizing complex over itself. $\square$