Since $\mathcal{P}$ is \'etale 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. \medskip\noindent 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 $x' \in W(f')$. \medskip\noindent 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 \'etale morphism $Y' \to \mathbf{A}^n_Y$ over $Y$, see Morphisms, Lemma \ref{morphisms-lemma-smooth-etale-over-affine-space}. 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 \ref{lemma-etale-local-source-target} (and because \'etale coverings are smooth coverings) we see that $\mathcal{P}$ is \'etale local on the source-and-target. By Lemma \ref{lemma-etale-etale-local-source-target} 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. \medskip\noindent 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 \'etale localization on $Y$ to produce enough translations and it becomes a bit annoying. Instead we use the automorphisms of Lemma \ref{lemma-orbits} and \'etale 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 $n \geq 2$. \medskip\noindent 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 \ref{lemma-orbits}. 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 \ref{lemma-orbits} 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 \'etale at every point lying over $x$ and the morphism $T_p : \mathbf{A}^n_Y \to \mathbf{A}^n_Y$ is \'etale 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 \ref{lemma-etale-etale-local-source-target} (we've already seen $\mathcal{P}$ is \'etale local on the source-and-target). Since $T_p : \mathbf{A}^n_x \to \mathbf{A}^n_x$ is finite \'etale 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.