**Proof.**
Since $\mathcal{P}$ is étale local on the source we see that $x \in W(f)$ if and only if the image of $x$ in $X \times _ Y Y'$ is in $W(X \times _ Y Y' \to Y')$. Hence we may assume the diagram in the lemma is cartesian.

Assume $x \in W(f)$. Since $\mathcal{P}$ is smooth local on the target we see that $(g')^{-1}W(f) = W(f) \times _ Y Y' \to Y'$ has $\mathcal{P}$. Hence $(g')^{-1}W(f) \subset W(f')$. We conclude $x' \in W(f')$.

Assume $x' \in W(f')$. For any open neighbourhood $V' \subset Y'$ of $y'$ we may replace $Y'$ by $V'$ and $X'$ by $U' = (f')^{-1}V'$ because $V' \to Y'$ is smooth and hence the base change $W(f') \cap U' \to V'$ of $W(f') \to Y'$ has property $\mathcal{P}$. Thus we may assume there exists an étale morphism $Y' \to \mathbf{A}^ n_ Y$ over $Y$, see Morphisms, Lemma 29.36.20. Picture

\[ \xymatrix{ X' \ar[r] \ar[d] & Y' \ar[d] \\ \mathbf{A}^ n_ X \ar[r]_{f_ n} \ar[d] & \mathbf{A}^ n_ Y \ar[d] \\ X \ar[r]^ f & Y } \]

By Lemma 35.32.6 (and because étale coverings are smooth coverings) we see that $\mathcal{P}$ is étale local on the source-and-target. By Lemma 35.32.9 we see that $W(f')$ is the inverse image of the open $W(f_ n) \subset \mathbf{A}^ n_ X$. In particular $W(f_ n)$ contains a point lying over $x$. After replacing $X$ by the image of $W(f_ n)$ (which is open) we may assume $W(f_ n) \to X$ is surjective. Claim: $W(f_ n) = \mathbf{A}^ n_ X$. The claim implies $f$ has $\mathcal{P}$ as $\mathcal{P}$ is local in the smooth topology and $\{ \mathbf{A}^ n_ Y \to Y\} $ is a smooth covering.

Essentially, the claim follows as $W(f_ n) \subset \mathbf{A}^ n_ X$ is a “translation invariant” open which meets every fibre of $\mathbf{A}^ n_ X \to X$. However, to produce an argument along these lines one has to do étale localization on $Y$ to produce enough translations and it becomes a bit annoying. Instead we use the automorphisms of Lemma 35.32.10 and étale morphisms of affine spaces. We may assume $n \geq 2$. Namely, if $n = 0$, then we are done. If $n = 1$, then we consider the diagram

\[ \xymatrix{ \mathbf{A}^2_ X \ar[r]_{f_2} \ar[d]_ p & \mathbf{A}^2_ Y \ar[d] \\ \mathbf{A}^1_ X \ar[r]^{f_1} & \mathbf{A}^1_ Y } \]

We have $p^{-1}(W(f_1)) \subset W(f_2)$ (see first paragraph of the proof). Thus $W(f_2) \to X$ is still surjective and we may work with $f_2$. Assume $n \geq 2$.

For any $1 \leq i, j \leq n$ with $i \not= j$ and $d \geq 0$ denote $T_{i, j, d}$ the automorphism of $\mathbf{A}^ n$ defined in Lemma 35.32.10. Then we get a commutative diagram

\[ \xymatrix{ \mathbf{A}^ n_ X \ar[r]_{f_ n} \ar[d]_{T_{i, j, d}} & \mathbf{A}^ n_ Y \ar[d]^{T_{i, j, d}} \\ \mathbf{A}^ n_ X \ar[r]^{f_ n} & \mathbf{A}^ n_ Y } \]

whose vertical arrows are isomorphisms. We conclude that $T_{i, j, d}(W(f_ n)) = W(f_ n)$. Applying Lemma 35.32.10 we conclude for any $x \in X$ the fibre $W(f_ n)_ x \subset \mathbf{A}^ n_ x$ is either $\mathbf{A}^ n_ x$ (this is what we want) or $\kappa (x)$ has characteristic $p > 0$ and $W(f_ n)_ x$ is the complement of a finite set $Z_ x \subset \mathbf{A}^ n_ x$ of closed points. The second possibility cannot occur. Namely, consider the morphism $T_ p : \mathbf{A}^ n \to \mathbf{A}^ n$ given by

\[ (x_1, \ldots , x_ n) \mapsto (x_1 - x_1^ p, \ldots , x_ n - x_ n^ p) \]

As above we get a commutative diagram

\[ \xymatrix{ \mathbf{A}^ n_ X \ar[r]_{f_ n} \ar[d]_{T_ p} & \mathbf{A}^ n_ Y \ar[d]^{T_ p} \\ \mathbf{A}^ n_ X \ar[r]^{f_ n} & \mathbf{A}^ n_ Y } \]

The morphism $T_ p : \mathbf{A}^ n_ X \to \mathbf{A}^ n_ X$ is étale at every point lying over $x$ and the morphism $T_ p : \mathbf{A}^ n_ Y \to \mathbf{A}^ n_ Y$ is étale at every point lying over the image of $x$ in $Y$. (Details omitted; hint: compute the derivatives.) We conclude that

\[ T_ p^{-1}(W) \cap \mathbf{A}^ n_ x = W \cap \mathbf{A}^ n_ x \]

by Lemma 35.32.9 (we've already seen $\mathcal{P}$ is étale local on the source-and-target). Since $T_ p : \mathbf{A}^ n_ x \to \mathbf{A}^ n_ x$ is finite étale of degree $p^ n > 1$ we see that if $Z_ x$ is not empty then it contains $T_ p^{-1}(Z_ x)$ which is bigger. This contradiction finishes the proof.
$\square$

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