**Proof.**
One direction of the lemma follows from Lemma 48.21.7. To prove the converse, we may assume $f^!\mathcal{O}_ Y$ has a unique nonzero cohomology sheaf. Let $y = f(x)$. Let $\xi _1, \ldots , \xi _ n \in X_ y$ be the generic points of the fibre $X_ y$ specializing to $x$. Let $d_1, \ldots , d_ n$ be the dimensions of the corresponding irreducible components of $X_ y$. The morphism $f : X \to Y$ is Cohen-Macaulay at $\eta _ i$ by More on Morphisms, Lemma 37.22.7. Hence by Lemma 48.21.7 we see that $d_1 = \ldots = d_ n$. If $d$ denotes the common value, then $d = \dim _ x(X_ y)$. After shrinking $X$ we may assume all fibres have dimension at most $d$ (Morphisms, Lemma 29.28.4). Then the only nonzero cohomology sheaf $\omega = H^{-d}(f^!\mathcal{O}_ Y)$ is flat over $Y$ by Lemma 48.21.4. Hence, if $h : X_ y \to X$ denotes the canonical morphism, then $Lh^*(f^!\mathcal{O}_ Y) = Lh^*(\omega [d]) = (h^*\omega )[d]$ by Derived Categories of Schemes, Lemma 36.22.8. Thus $h^*\omega [d]$ is the dualizing complex of $X_ y$ by Lemma 48.18.4. Hence $X_ y$ is Cohen-Macaulay by Lemma 48.23.1. This proves $f$ is Cohen-Macaulay at $x$ as desired.
$\square$

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