A dominant map has a surjective differential on a dense source open

The variety $Y$ is nonempty and irreducible, so it has a nonempty affine chart $Y_0=\mathrm{Spec}R$ [F4], and by dominance [F3] the open subset $f^{-1}(Y_0)\subseteq X$ is nonempty; it therefore contains a nonempty affine chart $X_0=\mathrm{Spec}S\subseteq X$ [F4]. The rings $R=k[Y_0]$ and $S=k[X_0]$ are finitely generated $k$-algebras given by finitely many coordinate classes [F5], and because $Y_0$ and $X_0$ are nonempty open subsets of irreducible spaces they are themselves irreducible [F8], so $R$ and $S$ are nonzero integral domains [F6]. The restriction $f{\mathrm{\mid }}_{X_0}\text{\hspace{0.17em}⁣}:X_0\to Y_0$ is a morphism of affine varieties, so its pullback is a $k$-algebra homomorphism $\phi \text{\hspace{0.17em}⁣}:R\to S$ [F7]. The fraction fields of the charts identify canonically with $k(Y)$ and $k(X)$, and dominance makes $f^{\ast }\text{\hspace{0.17em}⁣}:k(Y)\hookrightarrow k(X)$ injective [F9]; carrying the affine pullback through these identifications exhibits the map $\mathrm{Frac}R\to \mathrm{Frac}S$ induced by $\phi$ as $f^{\ast }$, so $\phi$ is injective. Write $K:=\mathrm{Frac}R=k(Y)$ and $L:=\mathrm{Frac}S=k(X)$.

Write the coordinate generators of $S$ as $s_1,\dots ,s_m$, so that $S=k[s_1,\dots ,s_m]$ and $L=k(s_1,\dots ,s_m)$ is finitely generated over $k$ [F5, F10]; in the same way $K=k(r_1,\dots ,r_n)$ for the coordinate generators of $R$ [F5, F10]. Since $\phi$ is injective, $K\subseteq L$, so $L\mathrm{/}K$ is finitely generated [F10]. By [F11], $\dim Y={\mathrm{trdeg}}_{k}K$ and $\dim X={\mathrm{trdeg}}_{k}L$ are finite, so the tower $k\subseteq K\subseteq L$ has, by [F12], $r:={\mathrm{trdeg}}_{K}L={\mathrm{trdeg}}_{k}L-{\mathrm{trdeg}}_{k}K=\dim X-\dim Y\ge 0.$ Moreover, since $k\subseteq R$ and the $s_i$ generate $S$ as a $k$-algebra, they generate $S$ as an $R$-algebra: the $R$-subalgebra they generate is a $k$-subalgebra containing every $s_i$, hence equals $S$ [F5].

The field $K$ contains $k$ and so has characteristic $0$, hence is perfect [F13]. By [F14] the finitely generated extension $L\mathrm{/}K$ has a separating transcendence basis $t_1,\dots ,t_r$; a separating transcendence basis is in particular a transcendence basis, so its length is ${\mathrm{trdeg}}_{K}L=r$ as computed in step 2.1. By [F15] the differentials $\mathrm{d}{t}_1,\dots ,\mathrm{d}{t}_r$ form an $L$-basis of ${\mathrm{\Omega }}_{L\mathrm{/}K}$. In particular ${\dim }_L{\mathrm{\Omega }}_{L\mathrm{/}K}=r$; for $r=0$ the empty list is the basis and ${\mathrm{\Omega }}_{L\mathrm{/}K}=0$.

By step 2.1 the elements $s_1,\dots ,s_m$ generate $S$ as an $R$-algebra, so by the universal property [F21] there is a surjective $R$-algebra map $R[x_1,\dots ,x_m]\to S$ with $x_i\mapsto s_i$, whose kernel we call $I$. By [F20], ${\mathrm{\Omega }}_{R[x_1,\dots ,x_m]\mathrm{/}R}$ is free with basis $\mathrm{d}{x}_1,\dots ,\mathrm{d}{x}_m$, and the sequence $I\mathrm{/}{I}^2\to S{\otimes }_{R[x_1,\dots ,x_m]}{\mathrm{\Omega }}_{R[x_1,\dots ,x_m]\mathrm{/}R}\to {\mathrm{\Omega }}_{S\mathrm{/}R}\to 0$ is exact. Therefore ${\mathrm{\Omega }}_{S\mathrm{/}R}$ is a quotient of the free $S$-module with basis $\mathrm{d}{x}_1,\dots ,\mathrm{d}{x}_m$, and in particular it is generated as an $S$-module by the $m$ elements $g_1,\dots ,g_m$, where $g_i$ is the image of $\mathrm{d}{x}_i$.

Apply the localization clause of [F16] to the injective ring map $\phi \text{\hspace{0.17em}⁣}:R\to S$ with $U=S\setminus \{0\}$ and $V=R\setminus \{0\}$: the image of $V$ in $S$ is contained in $U$ because $\phi$ is injective, and $V^{-1}R=K$, $U^{-1}S=L$ are the fraction fields [F19]. This gives an isomorphism $U^{-1}{\mathrm{\Omega }}_{S\mathrm{/}R}\cong {\mathrm{\Omega }}_{L\mathrm{/}K}$, and [F17] identifies $U^{-1}{\mathrm{\Omega }}_{S\mathrm{/}R}$ with $L{\otimes }_S{\mathrm{\Omega }}_{S\mathrm{/}R}$. Hence ${\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_{S}L\cong {\mathrm{\Omega }}_{L\mathrm{/}K}$ and, by step 3.1, the $L$-vector space ${\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_{S}L$ has dimension $r$.

For each $i$, the element $\mathrm{d}{t}_i$ of ${\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_{S}L=U^{-1}{\mathrm{\Omega }}_{S\mathrm{/}R}$ from step 4.1 has the form ${\omega }_i\mathrm{/}{h}_i$ with ${\omega }_i\in {\mathrm{\Omega }}_{S\mathrm{/}R}$ and $0\mathrm{\ne }{h}_i\in S$ [F18]. If $\lambda \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{S\mathrm{/}R}\to {\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_{S}L$ is the localization map, then $\lambda ({\omega }_i)=h_i\mathrm{d}{t}_i$, which is nonzero because $h_i\mathrm{\ne }0$ in the field $L$ and $\mathrm{d}{t}_i$ is part of an $L$-basis (step 3.1). The two $L$-spans agree, ${\mathrm{span}}_L(\lambda ({\omega }_1),\dots ,\lambda ({\omega }_r))={\mathrm{span}}_L(\mathrm{d}{t}_1,\dots ,\mathrm{d}{t}_r)={\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_{S}L,$ since each $\lambda ({\omega }_i)=h_i\mathrm{d}{t}_i$ lies in the span of the $\mathrm{d}{t}_i$ and each $\mathrm{d}{t}_i=h_i^{-1}\lambda ({\omega }_i)$ lies in the span of the $\lambda ({\omega }_i)$.

Each generator $g_j$ of step 3.2 satisfies $\lambda (g_j)\in {\mathrm{span}}_L(\lambda ({\omega }_1),\dots ,\lambda ({\omega }_r))$ by step 5.1, say $\lambda (g_j)={\sum }_{i=1}^r{c}_{ji}\lambda ({\omega }_i)$ with $c_{ji}\in L$. Since $L=\mathrm{Frac}S=(S\setminus \{0\}{)}^{-1}S$ [F19], the finitely many coefficients have a common denominator: $c_{ji}=a_{ji}\mathrm{/}h$ with $a_{ji}\in S$ and $0\mathrm{\ne }h\in S$. Then $\lambda (h{g}_j-{\sum }_i{a}_{ji}{\omega }_i)=0$, so by the kernel criterion for localizations [F18] there is $0\mathrm{\ne }{u}_j\in S$ with $u_j(h{g}_j-{\sum }_i{a}_{ji}{\omega }_i)=0$ in ${\mathrm{\Omega }}_{S\mathrm{/}R}$. Put $H:=h{\prod }_{j=1}^m{u}_j\mathrm{\ne }0$. In the localization ${\mathrm{\Omega }}_{S\mathrm{/}R}[1\mathrm{/}H]$ the element $u_{j}h$ is invertible, and the relations rewrite as $g_j={\sum }_i(u_j{a}_{ji}\mathrm{/}(u_{j}h)){\omega }_i$ with coefficients in $S_H$. Hence ${\mathrm{\Omega }}_{S\mathrm{/}R}[1\mathrm{/}H]$ is generated over $S_H$ by ${\omega }_1,\dots ,{\omega }_r$. (For $r=0$, step 5.1 gives $\lambda (g_j)\in {\mathrm{span}}_L\mathrm{\varnothing }=0$, so $u_j{g}_j=0$ for suitable $0\mathrm{\ne }{u}_j\in S$ and ${\mathrm{\Omega }}_{S\mathrm{/}R}[1\mathrm{/}H]=0$ with $H:={\prod }_j{u}_j$, generated by the empty family.)

Let $\mathfrak{p}\in \mathrm{Spec}S$ satisfy $H\mathrm{\notin }\mathfrak{p}$. Applying [F25] with $R\to S$ the localization $S\to S_H$, $N=\kappa (\mathfrak{p})$ and $M={\mathrm{\Omega }}_{S\mathrm{/}R}$, and identifying $S_H{\otimes }_S{\mathrm{\Omega }}_{S\mathrm{/}R}$ with ${\mathrm{\Omega }}_{S\mathrm{/}R}[1\mathrm{/}H]$ via [F17], shows ${\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (\mathfrak{p})\cong \kappa (\mathfrak{p}){\otimes }_{S_H}{\mathrm{\Omega }}_{S\mathrm{/}R}[1\mathrm{/}H]\mathrm{.}$ The right-hand side is generated as a $\kappa (\mathfrak{p})$-vector space by the images of ${\omega }_1,\dots ,{\omega }_r$, because ${\mathrm{\Omega }}_{S\mathrm{/}R}[1\mathrm{/}H]$ is generated over $S_H$ by these $r$ elements (step 6.1) and $\kappa (\mathfrak{p}){\otimes }_{S_H}-$ is right exact [F22]. A vector space spanned by $r$ elements contains a basis inside that spanning set [F23] and therefore has dimension at most $r$ [F24]. Thus ${\dim }_{\kappa (\mathfrak{p})}({\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (\mathfrak{p}))\le r\mathrm{.}$

Let $X_{\mathrm{r}\mathrm{e}\mathrm{g}}$ and $Y_{\mathrm{r}\mathrm{e}\mathrm{g}}$ be the regular loci of the schemes $X$ and $Y$ [F35]. The field $k$ is perfect [F13], and $X$, $Y$ are reduced finite-type $k$-schemes via [F2]; hence by [F36] the two regular loci are open, and each meets every irreducible component of its scheme in a dense open subset. Since $X$ and $Y$ are irreducible, $X_{\mathrm{r}\mathrm{e}\mathrm{g}}$ and $Y_{\mathrm{r}\mathrm{e}\mathrm{g}}$ are nonempty dense open subsets of $X$ and $Y$. The preimage $f^{-1}(Y_{\mathrm{r}\mathrm{e}\mathrm{g}})\subseteq X$ is nonempty, because the dense image $f(X)$ meets the nonempty open set $Y_{\mathrm{r}\mathrm{e}\mathrm{g}}$ [F3, F8], and it is open. The principal open $D(H)=\{\mathfrak{p}\in \mathrm{Spec}S:H\mathrm{\notin }\mathfrak{p}\}$ is an open subset of the chart $X_0$ [F38] and it is nonempty because $H\mathrm{\ne }0$ (step 6.1), so $D(H)$ is a nonempty open subset of $X$. Define $U:=D(H)\cap X_{\mathrm{r}\mathrm{e}\mathrm{g}}\cap f^{-1}(Y_{\mathrm{r}\mathrm{e}\mathrm{g}})\subseteq X,V:=Y_{\mathrm{r}\mathrm{e}\mathrm{g}}\subseteq Y\mathrm{.}$ The three sets displayed are nonempty open subsets of the irreducible space $X$ [F8], so $U$ is a nonempty open subset of $X$ with $U\subseteq X_{\mathrm{r}\mathrm{e}\mathrm{g}}$ and $f(U)\subseteq V$.

Fix a closed point $x\in U$ and put $y=f(x)\in V=Y_{\mathrm{r}\mathrm{e}\mathrm{g}}$; both are $k$-rational points of the respective schemes [F2]. Let ${\mathfrak{m}}_x\subseteq S$ and ${\mathfrak{m}}_y\subseteq R$ be the corresponding maximal ideals. The local rings are ${\mathcal{O}}_{Y,y}=R_{{\mathfrak{m}}_y}$ and ${\mathcal{O}}_{X,x}=S_{{\mathfrak{m}}_x}$ [F31], and the induced map of local rings is the localization of $\phi$ at these primes, which sends $a$ to $\phi (a)$ [F29] and is local [F30]; hence the cotangent map ${\bar{f}}_x^{\mathrm{♯}}\text{\hspace{0.17em}⁣}:C_{y}Y\to C_{x}X$ of [F27] sends the class of $a\in {\mathfrak{m}}_y$ to the class of $\phi (a)$. The cotangent isomorphism [F26] at the $k$-rational points, applied on the charts $\mathrm{Spec}R$ and $\mathrm{Spec}S$ and combined with [F32], gives identifications ${\theta }_y\text{\hspace{0.17em}⁣}:C_{y}Y\cong {\mathrm{\Omega }}_{R\mathrm{/}k}{\otimes }_R\kappa (y),{\theta }_x\text{\hspace{0.17em}⁣}:C_{x}X\cong {\mathrm{\Omega }}_{S\mathrm{/}k}{\otimes }_S\kappa (x),$ both sending $[a]\mapsto \mathrm{d}a\otimes 1$. Let $\alpha \text{\hspace{0.17em}⁣}:S{\otimes }_R{\mathrm{\Omega }}_{R\mathrm{/}k}\to {\mathrm{\Omega }}_{S\mathrm{/}k}$ be the functoriality map of F16, $b\otimes \mathrm{d}a\mapsto b\mathrm{d}\phi (a)$; it is the first map of the exact sequence $S{\otimes }_R{\mathrm{\Omega }}_{R\mathrm{/}k}\stackrel{\alpha }{\to }{\mathrm{\Omega }}_{S\mathrm{/}k}\to {\mathrm{\Omega }}_{S\mathrm{/}R}\to 0$ of [F28]. Base changing this sequence along $S\to \kappa (x)$ and using the identification $(S{\otimes }_R{\mathrm{\Omega }}_{R\mathrm{/}k}){\otimes }_S\kappa (x)\cong {\mathrm{\Omega }}_{R\mathrm{/}k}{\otimes }_R\kappa (x)$ of [F25], together with the fact that $R\to \kappa (x)$ is evaluation at $y$, yields the exact sequence ${\mathrm{\Omega }}_{R\mathrm{/}k}{\otimes }_R\kappa (y)\stackrel{{\alpha }_x}{\to }{\mathrm{\Omega }}_{S\mathrm{/}k}{\otimes }_S\kappa (x)\to {\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (x)\to 0$ [F22]. For $a\in {\mathfrak{m}}_y$ one computes ${\alpha }_x({\theta }_y([a]))={\alpha }_x(\mathrm{d}a\otimes 1)=\mathrm{d}\phi (a)\otimes 1={\theta }_x([\phi (a)])={\theta }_x({\bar{f}}_x^{\mathrm{♯}}([a]))$, so under the identifications ${\theta }_y$, ${\theta }_x$ the map ${\alpha }_x$ is precisely the cotangent map ${\bar{f}}_x^{\mathrm{♯}}$; in particular $\mathrm{rank}{\alpha }_x=\mathrm{rank}{\bar{f}}_x^{\mathrm{♯}}$.

By [F27], $d_{x}f$ is the transpose of ${\bar{f}}_x^{\mathrm{♯}}$, so by [F33] its rank equals $\mathrm{rank}{\bar{f}}_x^{\mathrm{♯}}$; all tangent and cotangent spaces here are finite-dimensional [F39, F26]. Hence $\mathrm{rank}{d}_{x}f=\mathrm{rank}{\alpha }_x$. Since ${\mathrm{\Omega }}_{R\mathrm{/}k}{\otimes }_R\kappa (y)\to {\mathrm{\Omega }}_{S\mathrm{/}k}{\otimes }_S\kappa (x)\to {\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (x)\to 0$ is exact (step 8.1), the rank-nullity theorem [F34] applied to ${\alpha }_x$ gives $\mathrm{rank}{d}_{x}f={\dim }_k({\mathrm{\Omega }}_{S\mathrm{/}k}{\otimes }_S\kappa (x))-{\dim }_{\kappa (x)}({\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (x))={\dim }_k{T}_{x}X-{\dim }_{\kappa (x)}({\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (x)),$ the last equality because ${\theta }_x$ identifies ${\mathrm{\Omega }}_{S\mathrm{/}k}{\otimes }_S\kappa (x)$ with $C_{x}X$ [F26] and $C_{x}X$ is the dual of the finite-dimensional space $T_{x}X$ [F39].

Since $x\in X_{\mathrm{r}\mathrm{e}\mathrm{g}}$, the classical dimension test [F35] gives ${\dim }_k{T}_{x}X={\dim }_{x}X$, and since $X$ is irreducible its only irreducible component is $X$ itself, so ${\dim }_{x}X=\dim X$ by [F37]. Likewise $y\in Y_{\mathrm{r}\mathrm{e}\mathrm{g}}$ gives ${\dim }_k{T}_{y}Y={\dim }_{y}Y=\dim Y$ by [F35, F37]. The point $x$ lies in $D(H)$, so $H\mathrm{\notin }{\mathfrak{m}}_x$ and step 7.1 bounds ${\dim }_{\kappa (x)}({\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (x))\le r$. Therefore step 9.1 and $r=\dim X-\dim Y$ (step 2.1) give $\mathrm{rank}(d_{x}f)=\dim X-{\dim }_{\kappa (x)}({\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (x))\ge \dim X-r=\dim Y={\dim }_k{T}_{y}Y\mathrm{.}$ A linear map has rank at most the dimension of its target, so $\mathrm{rank}(d_{x}f)={\dim }_k{T}_{y}Y$ and $d_{x}f\text{\hspace{0.17em}⁣}:T_{x}X\to T_{y}Y$ is surjective. This holds at every closed point $x$ of the nonempty open set $U$, and $f(U)\subseteq V\subseteq Y_{\mathrm{r}\mathrm{e}\mathrm{g}}$, which is the assertion.

Boundary and scope dispositions. Empty: $X$ and $Y$ are nonempty because irreducible means nonempty [F8], so the charts and the open $U$ of step 7.2 are nonempty, and no empty-case convention is needed; the sets $X_{\mathrm{r}\mathrm{e}\mathrm{g}}$, $Y_{\mathrm{r}\mathrm{e}\mathrm{g}}$ are nonempty by [F36], and if $Y$ is a point then $r=\dim X$, and the separating basis in step 3.1 is empty exactly when $\dim X=0$. Zero: the case $r=\dim X-\dim Y=0$ is covered by the empty-list convention of steps 3.1 and 6.1: then ${\mathrm{\Omega }}_{L\mathrm{/}K}=0$, ${\mathrm{\Omega }}_{S\mathrm{/}R}[1\mathrm{/}H]=0$, so step 7.1 gives ${\dim }_{\kappa (x)}({\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (x))=0$, and step 10.1 concludes $\mathrm{rank}(d_{x}f)=\dim X=\dim Y$ with no modification; likewise $\dim Y=0$ forces $Y$ to be a single point in the present irreducible setting, $T_{y}Y=0$, and surjectivity is the equality $\mathrm{rank}=0$ already obtained. One: nothing in the argument divides by a natural number or assumes a generator count $\ge 1$; the lists $s_1,\dots ,s_m$, $r_1,\dots ,r_n$, $g_1,\dots ,g_m$ and $t_1,\dots ,t_r$ may have length one or zero, and the length-one case $r=1$ has $\dim X-\dim Y=1$; generically finite maps instead have $r=0$, and both cases are covered by the same argument. Degenerate: the proof does not require $\phi$ to be surjective or the charts to be smooth, and it does not require $f$ to be finite or flat; the degenerate dominant case $X\to Y=\mathrm{Spec}k$ (so $K=k$, $r=\dim X$, $H$ and the nonempty open $U$ are obtained as in steps 6.1 and 7.2) is covered by steps 6.1-10.1, and characteristic $0$ is essential, the Frobenius example $k[u]\to k[t]$, $u\mapsto t^p$ showing failure in characteristic $p$ and lying outside the hypothesis of characteristic $0$. Endpoints: the two inequalities used in step 10.1 are the endpoint bounds ${\dim }_{\kappa (x)}({\mathrm{\Omega }}_{S\mathrm{/}R}{\otimes }_S\kappa (x))\le r$ of step 7.1 and $\mathrm{rank}(d_{x}f)\le {\dim }_k{T}_{y}Y$; at $r=0$ the first is tight and at $r=\dim X$ the second is tight, in both cases producing the stated equality rather than a strict inequality. Nonempty-choice: AC is declared in [F1] and is used in this proof only through the AC-assuming suppliers [F36] (regular loci), [F33] (transpose rank), [F23] (bases inside spanning sets), [F2] (the classical-scheme dictionary), [F7] (affine antiequivalence), [F6] (domain criterion), [F9] (function fields of charts) and [F11] (dimension equals transcendence degree), each cited at the step that uses it; the field-theoretic steps 3.1, 3.2-6.1 and the linear algebra of steps 8.1-10.1 make no further choice. Biconditional directions: no biconditional is asserted by this lemma; the only implications are the chain of equalities and the single inequality of step 10.1, whose forward reading gives surjectivity, and no converse is claimed.