A cusp family defeats the missing source hypothesis

Setup and integrality of the source. Put $R=k[z]$, $S=R[x,y]\mathrm{/}(y^2-x^3)=k[x,y,z]\mathrm{/}(y^2-x^3)$, $X=\mathrm{Spec}S$, $Y=\mathrm{Spec}R$, and $f=\mathrm{Spec}\phi$ for $\phi (z)=z$; then $S\cong k[z]{\otimes }_{k}k[x,y]\mathrm{/}(y^2-x^3)$, so $X\cong C{\times }_k{\mathbb{A}}_k^1$ with $C=\mathrm{Spec}k[x,y]\mathrm{/}(y^2-x^3)$. The polynomial $y^2-x^3$ is irreducible in $k[x,y,z]$: viewed in $k[z][x][y]$ it is monic of $y$-degree $2$, so a factorization into nonunits must have $y$-degrees $(1,1)$ after comparing $y$-degrees and using that the units of $k[z][x]$ are the units of $k$ by [F17], giving $y+a$ and $y+b$ with $a,b\in k[z][x]$, whence $a+b=0$ and $ab=-x^3$, so $a^2=x^3$, impossible because the $x$-degree of $a^2$ is even whereas $3$ is odd; by [F16] this irreducible element of the unique factorisation domain $k[x,y,z]$ is prime, so $(f)$ is prime and $S$ is a domain by [F18], that is, $X$ is integral and in particular irreducible.

The section point and dominance. The evaluation $S\to k[z]$ with $x,y\mapsto 0$ and $z\mapsto z$ is surjective with kernel $\mathfrak{q}=(x,y)S$, so $S\mathrm{/}\mathfrak{q}\cong k[z]$ is a domain and $\mathfrak{q}$ is prime by [F18]; an element of $R$ lying in $\mathfrak{q}$ maps to $0$ in $S\mathrm{/}\mathfrak{q}\cong k[z]$, hence is $0$, so $\mathfrak{q}\cap R=(0)$ and $f(\mathfrak{q})=\eta$, the generic point $(0)$ of $Y$; by [F26] $\mathrm{Spec}k[z]$ is irreducible with generic point $\eta$, so the closure of the image of $f$ contains the closure of $\{\eta \}$, which is all of $Y$, and $f$ is dominant.

The target is smooth and both schemes are finite type over $k$, indeed finitely presented over $k$. The ring map $k\to k[z]$ has the standard smooth presentation $k[z]\cong (k[z]{)}_g$ with $n=1$, $c=0$ and $g=1$ in the sense of [F4] (the $c=0$ case is a localization of a polynomial ring), so at every point of $Y$ this chart exhibits a standard smooth presentation and $Y\to \mathrm{Spec}k$ is smooth by [F1]; the rings $k[z]\cong k[x]\mathrm{/}(0)$ and $S\cong k[x,y,z]\mathrm{/}(y^2-x^3)$ are quotients of polynomial rings by finitely generated (indeed zero or principal) ideals, so $Y$ and $X$ are finite type over $k$ by [F27] and finitely presented over $k$ by [F5].

Every fibre is the cusp. For every $\mathfrak{p}\in \mathrm{Spec}R$ the fibre ring is $S{\otimes }_R\kappa (\mathfrak{p})\cong \kappa (\mathfrak{p})[x,y]\mathrm{/}(y^2-x^3)$ by the first isomorphism of [F15] applied to $A=R$, the variables $x,y$, $I=(y^2-x^3)$ and $C=\kappa (\mathfrak{p})$; in particular the fibre over the generic point is the cusp over $\kappa (\eta )=R_{(0)}\mathrm{/}(0)\cong \mathrm{Frac}(R)=k(z)$ by [F23] and [F24], and the fibre over a closed point $\mathfrak{p}=(z-t)$ is the cusp over $k$.

The cusp ring over any field is a Noetherian domain, and the hypersurface $S$ is Noetherian. Let $\kappa$ be a field, $f=y^2-x^3$ and $B=\kappa [x,y]\mathrm{/}(f)$. The irreducibility argument of step 1.1 with the coefficient ring $\kappa [x]$ (whose units are ${\kappa }^{\times }$ by [F17]) shows that $f$ is irreducible in $\kappa [x,y]$; by [F16] $f$ is a prime element of the unique factorisation domain $\kappa [x,y]$, so $(f)$ is a prime ideal and $B$ is a domain by [F18]. The ring $\kappa [x,y]$ is Noetherian by [F20] and [F21], and for every ideal $J\subseteq B$ its preimage in $\kappa [x,y]$ is an ideal, hence generated by finitely many elements by [F19], whose images generate $J$ by [F22]; thus every ideal of $B$ is finitely generated and $B$ is Noetherian by [F19]. The specific hypersurface $S\cong k[x,y,z]\mathrm{/}(y^2-x^3)$ of step 1.1 is Noetherian as well: $k[x,y,z]$ is Noetherian by [F20] and [F21], and the quotient argument just given, using the correspondence of [F22] and the criterion of [F19], applies verbatim with $(y^2-x^3)$ in place of $(f)$.

Suppose $U\subseteq Y$ is nonempty open with $f^{-1}(U)\to U$ smooth. The generic point $\eta$ of $Y$ lies in $U$: if $\eta \mathrm{\notin }U$, the closed set $Y\setminus U$ contains $\eta$ and hence contains the closure of $\{\eta \}$, which is $Y$ by [F26], contradicting $U\mathrm{\ne }\mathrm{\varnothing }$. Therefore $\eta \in U$, and the point $\mathfrak{q}$ of step 1.2, which maps to $\eta$, lies in $f^{-1}(U)$.

Reduction to a principal affine chart. By [F25] there is $g_0\in R$ with $\eta \in D(g_0)\subseteq U$, so $g_0\mathrm{\ne }0$; smoothness is local on source and target by [F1], so the morphism $f^{-1}(D(g_0))\to D(g_0)$ is smooth, and its source is $D(\phi (g_0))=\mathrm{Spec}{S}_{\phi (g_0)}$ while its target is $\mathrm{Spec}{R}_{g_0}$, both finite type over $k$ by [F27]. Applying [F1] at the point $\mathfrak{q}$ and absorbing further principal localizations into the presentation by [F4] (the prime ${\mathfrak{q}}_B$ below is the image of $\mathfrak{q}$, and $\phi (g_0)\mathrm{\notin }\mathfrak{q}$ by step 1.2), we may take the affine chart $A=R_{g_0}\to B=S_{\phi (g_0)}$; the two isomorphisms of [F15] (the first with $C=R_{g_0}$, the second with $C=S$ and $M=\{g_0^n\}$) identify $B$ with $A[x,y]\mathrm{/}(y^2-x^3)$, a quotient of $A[x,y]$ by a principal, hence finitely generated, ideal, so [F5] makes $A\to B$ of finite presentation and [F1] and [F4] make it standard smooth at ${\mathfrak{q}}_B=\mathfrak{q}{S}_{\phi (g_0)}$.

The cusp local ring has dimension one. With $B$ as in step 2.1, $\dim \kappa [x,y]=2$ by [F11]; the element $f\mathrm{\ne }0$ is a nonzerodivisor because $\kappa [x,y]$ is a domain by [F16], so the prime $(f)$, minimal over itself, has height $1$ by [F12]; the height formula [F10] for the finite-type $\kappa$-domain $\kappa [x,y]$ gives $\mathrm{ht}((f))+\dim B=\dim \kappa [x,y]=2$, hence $\dim B=1$. The ideal $\mathfrak{m}=(x,y)B$ is maximal with $B\mathrm{/}\mathfrak{m}\cong \kappa$, so $\mathrm{ht}(\mathfrak{m})=\dim B=1$ by [F13] and $\dim B_{\mathfrak{m}}=\mathrm{ht}(\mathfrak{m})=1$ by [F14].

The source is not smooth over $k$. Working under the declared Axiom of Choice [F6], at the $k$-rational point $a=(x,y,z)$ of $X=\mathrm{Spec}S$ one has $\dim S=2$ by [F10], [F11] and [F12] applied to the finite-type $k$-domain $k[x,y,z]$, Noetherian by step 2.1, so $\dim {\mathcal{O}}_{X,a}=\mathrm{ht}((x,y,z))=\dim S=2$ by [F13] and [F14]; the Jacobian $(-3x^2,2y,0)$ of $S$ vanishes at $a$, so [F7] gives ${\dim }_k{T}_{a}X=3$, and [F8] with the local Noetherianness of $X$ from [F9] and step 2.1 excludes regularity at $a$. If $X\to \mathrm{Spec}k$ were smooth, then by [F1] it would be standard smooth at $a$, and since the structure map $k\to S$ is of finite presentation by step 1.3, clause 1 of [F2] would make the fibre $S{\otimes }_{k}k=S$ geometrically regular at $a$, hence by [F3] the local ring $S_a$ would be regular, contradicting $3\mathrm{\ne }2$; thus the source is itself singular at $a$.

The fibre of the chart is computed. By step 2.3, $B\cong A[x,y]\mathrm{/}(y^2-x^3)$ with $A\to B$ of finite presentation; the contraction ${\mathfrak{p}}_A={\mathfrak{q}}_B\cap A$ is $(0)$ because $\mathfrak{q}\cap R=(0)$ in step 1.2, and the fibre $B{\otimes }_A\kappa ({\mathfrak{p}}_A)$ is isomorphic to $\kappa ({\mathfrak{p}}_A)[x,y]\mathrm{/}(y^2-x^3)$ by the first isomorphism of [F15] applied over $A$, while $\kappa ({\mathfrak{p}}_A)=\mathrm{Frac}(A)=\mathrm{Frac}(R)=k(z)$ by [F23] and [F24].

The origin of the cusp is not regular. The point $\mathfrak{m}$ of $\mathrm{Spec}B$ is $\kappa$-rational, and the Jacobian of the single equation $f=y^2-x^3$ is the row $(-3x^2,2y)$, which vanishes at the origin, so [F7] with $n=2$ and $r=1$ gives $T_{\mathfrak{m}}\mathrm{Spec}B\cong \ker (0\text{\hspace{0.17em}⁣}:{\kappa }^2\to {\kappa }^1)$, a $\kappa$-vector space of dimension $2$. Since $B$ is Noetherian by step 2.1, $\mathrm{Spec}B$ is locally Noetherian by [F9], and [F8] says that $\mathfrak{m}$ is regular exactly when ${\dim }_{\kappa }{T}_{\mathfrak{m}}\mathrm{Spec}B=\dim {\mathcal{O}}_{\mathrm{Spec}B,\mathfrak{m}}=1$ by step 3.1; as $2\mathrm{\ne }1$ the origin is not regular.

The fibre of the chart is geometrically regular. Since $A\to B$ is of finite presentation and standard smooth at ${\mathfrak{q}}_B$ by step 2.3, clause 1 of [F2], applied under the declared Axiom of Choice [F6], gives that the fibre $B{\otimes }_A\kappa ({\mathfrak{p}}_A)$ is geometrically regular at ${\mathfrak{q}}_B$; with $\kappa ({\mathfrak{p}}_A)=k(z)$ by step 3.3, taking the trivial field extension $\kappa ({\mathfrak{p}}_A)\mathrm{/}\kappa ({\mathfrak{p}}_A)$ and the prime over the image of ${\mathfrak{q}}_B$ in [F3] gives that the local ring $(B{\otimes }_{A}k(z){)}_{{\mathfrak{q}}_B}$ is a regular local ring.

That local ring is the nonregular cusp local ring. By step 3.3 the fibre ring is $k(z)[x,y]\mathrm{/}(y^2-x^3)$ and ${\mathfrak{q}}_B$ corresponds to its maximal ideal $(x,y)$, so $(B{\otimes }_{A}k(z){)}_{{\mathfrak{q}}_B}\cong (k(z)[x,y]\mathrm{/}(y^2-x^3){)}_{(x,y)}$; by steps 2.1, 3.1 and 4.1 with $\kappa =k(z)$ this local ring is a domain of dimension $1$ whose tangent space has dimension $2$, hence is not regular, contradicting step 4.2.

Conclusion and scope. Steps 4.2 and 5.1 are contradictory, so no nonempty Zariski-open $U\subseteq Y$ has $f^{-1}(U)\to U$ smooth. The hypotheses of the refuted claim are met except its missing smooth-source hypothesis: $k$ has characteristic $0$, $X$ is integral and $Y$ is smooth over $k$ by steps 1.1 and 1.3, and $f$ is dominant by step 1.2, while every fibre is the cusp by step 1.4 and is singular at its origin by step 4.1; by step 3.2 the source $X$ is itself not smooth at $a$, which is exactly the omitted hypothesis. Hence target-open generic smoothness genuinely needs smoothness of the source, and the constant cusp family is a characteristic-zero counterexample.