This is the analogue of Duality for Schemes, Section 48.25.
Lemma 76.27.1. The property of morphisms of germs of schemes is étale local on the source-and-target (Descent, Definition 35.33.1).
\begin{align*} & \mathcal{P}((X, x) \to (S, s)) = \\ & \text{the local ring } \mathcal{O}_{X_ s, x} \text{ of the fibre is Noetherian and Gorenstein} \end{align*}
Proof. Given a diagram as in Descent, Definition 35.33.1 we obtain an étale morphism of fibres $U'_{v'} \to U_ v$ mapping $u'$ to $u$, see Descent, Lemma 35.33.5. Thus $\mathcal{O}_{U_ v, u} \to \mathcal{O}_{U'_{v'}, u'}$ is the localization of an étale ring map. Hence the first is Noetherian if and only if the second is Noetherian, see More on Algebra, Lemma 15.44.1. Then, since $\mathcal{O}_{U'_{v'}, u'}/\mathfrak m_ u \mathcal{O}_{U'_{v'}, u'} = \kappa (u')$ (Algebra, Lemma 10.143.5) is a Gorenstein ring, we see that $\mathcal{O}_{U_ v, u}$ is Gorenstein if and only if $\mathcal{O}_{U'_{v'}, u'}$ is Gorenstein by Dualizing Complexes, Lemma 47.21.8. $\square$
Definition 76.27.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume the fibres of $f$ are locally Noetherian (Divisors on Spaces, Definition 71.4.2).
Let $x \in |X|$, and $y = f(x)$. We say that $f$ is Gorenstein at $x$ if $f$ is flat at $x$ and the equivalent conditions of Morphisms of Spaces, Lemma 67.22.5 hold for the property $\mathcal{P}$ described in Lemma 76.27.1.
We say $f$ is a Gorenstein morphism if $f$ is Gorenstein at every point of $X$.
Here is a translation.
Lemma 76.27.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume the fibres of $f$ are locally Noetherian. The following are equivalent
$f$ is Gorenstein,
$f$ is flat and for some surjective étale morphism $V \to Y$ where $V$ is a scheme, the fibres of $X_ V \to V$ are Gorenstein algebraic spaces, and
$f$ is flat and for any étale morphism $V \to Y$ where $V$ is a scheme, the fibres of $X_ V \to V$ are Gorenstein algebraic spaces.
Given $x \in |X|$ with image $y \in |Y|$ the following are equivalent
$f$ is Gorenstein at $x$, and
$\mathcal{O}_{Y, \overline{y}} \to \mathcal{O}_{X, \overline{x}}$ is flat and $\mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{y}}\mathcal{O}_{X, \overline{x}}$ is Gorenstein.
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. $\square$
Lemma 76.27.4. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$. Assume that the fibres of $f$, $g$, and $g \circ f$ are locally Noetherian. Let $x \in |X|$ with images $y \in |Y|$ and $z \in |Z|$.
If $f$ is Gorenstein at $x$ and $g$ is Gorenstein at $f(x)$, then $g \circ f$ is Gorenstein at $x$.
If $f$ and $g$ are Gorenstein, then $g \circ f$ is Gorenstein.
If $g \circ f$ is Gorenstein at $x$ and $f$ is flat at $x$, then $f$ is Gorenstein at $x$ and $g$ is Gorenstein at $f(x)$.
If $f \circ g$ is Gorenstein and $f$ is flat, then $f$ is Gorenstein and $g$ is Gorenstein at every point in the image of $f$.
Proof. Working étale locally this follows from the corresponding result for schemes, see Duality for Schemes, Lemma 48.25.6. Alternatively, we can use the equivalence of (a) and (b) in Lemma 76.27.3. Thus we consider the local homomorphism of Noetherian local rings
whose fibre is
and we use Dualizing Complexes, Lemma 47.21.8. $\square$
Lemma 76.27.5. Let $S$ be a scheme. Let $f : X \to Y$ be a flat morphism of locally Noetherian algebraic spaces over $S$. If $X$ is Gorenstein, then $f$ is Gorenstein and $\mathcal{O}_{Y, f(\overline{x})}$ is Gorenstein for all $x \in |X|$.
Proof. After translating into algebra using Lemma 76.27.3 (compare with the proof of Lemma 76.27.4) this follows from Dualizing Complexes, Lemma 47.21.8. $\square$
Lemma 76.27.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume the fibres of $f$ are locally Noetherian. Let $Y' \to Y$ be locally of finite type. Let $f' : X' \to Y'$ be the base change of $f$. Let $x' \in |X'|$ be a point with image $x \in |X|$.
If $f$ is Gorenstein at $x$, then $f' : X' \to Y'$ is Gorenstein at $x'$.
If $f$ is flat at $x$ and $f'$ is Gorenstein at $x'$, then $f$ is Gorenstein at $x$.
If $Y' \to Y$ is flat at $f'(x')$ and $f'$ is Gorenstein at $x'$, then $f$ is Gorenstein at $x$.
Proof. Denote $y \in |Y|$ and $y' \in |Y'|$ the image of $x'$. Choose a surjective étale morphism $V \to Y$ where $V$ is a scheme. Choose a surjective étale morphism $U \to X \times _ Y V$ where $U$ is a scheme. Choose a surjectiev étale morphism $V' \to Y' \times _ Y V$ where $V'$ is a scheme. Then $U' = U \times _ V V'$ is a scheme which comes equipped with a surjective étale morphism $U' \to X'$. Choose $u' \in U'$ mapping to $x'$. Denote $u \in U$ the image of $u'$. Then the lemma follows from the lemma for $U \to V$ and its base change $U' \to V'$ and the points $u'$ and $u$ (this follows from the definitions). Thus the lemma follows from the case of schemes, see Duality for Schemes, Lemma 48.25.8. $\square$
Lemma 76.27.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is flat and locally of finite presentation. Let Then $W$ is open in $|X|$ and the formation of $W$ commutes with arbitrary base change of $f$: For any morphism $g : Y' \to Y$, consider the base change $f' : X' \to Y'$ of $f$ and the projection $g' : X' \to X$. Then the corresponding set $W'$ for the morphism $f'$ is equal to $W' = (g')^{-1}(W)$.
\[ W = \{ x \in |X| : f\text{ is Gorenstein at }x\} \]
Proof. Choose a commutative diagram
Let $u \in U$ with image $x \in |X|$. Then $f$ is Gorenstein at $x$ if and only if $U \to V$ is Gorenstein at $u$ (by definition). Thus we reduce to the case of the morphism $U \to V$ of schemes. Openness is proven in Duality for Schemes, Lemma 48.25.11 and compatibility with base change in Duality for Schemes, Lemma 48.25.9. $\square$
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