I have a general question about the motivation behind to definition the smooth morphisms
as we know it from algebraic geometry. The most common
definition of a smooth morphism $: X \to Y$ between two smooth Noetherian schemes
$X,Y$ is:


$f$ is smooth if and only if


(i) $f$ is flat and locally of finite presentation


(ii) for every $y \in Y$ the fiber $X \times\_Y k(y)$ is smooth variety over $k(y)$


I read recently that the motivation of smooth morphisms in algebraic geometry arise as a kind of imitation attempt
of a maps called "submersions" in differential geometry. Namely if $X,Y$ are smooth manifolds
then a submerison $s: X \to Y$ is a surjective, proper $C^{\infty}$ map
for which for every $x \in X$ the induced differention
$D\_xf: T\_x X \to T\_y Y$ is surjective. The Ehresmann's lemma says that such submersion is moreover a locally trivial fibration. That seems to coincide with the intuition that flaness is something like a continuous family of neighbored fibers.


My Question is if there exist a definition of a smooth morphism $: X \to Y$
in algebraic geometry world, which emphesize more immediately that the motivation for
smoothness in algeom arise from submersions in differential geometry?


Let look again at definition above. (ii) seems reasonable, since this tells
that every fiber of $f$ is smooth, ie morally a manifold. But that reason that the point
(i) arise immediately from the differential geometry isn't immediately clear if one not believes that flatness makes families "continuous". Morally "continuity of fibers" (= flatness) should be a consequence (like by of Ehresman lemma in differential geo), not an immediate "part" of definition.


Can the flatness requirement be replaced in algeobraic definition by requirement
that the induced differention
$D\_xf: T\_x X \to T\_y Y$ is surjective? And is this equivalent of flatness in algebraic setting?


The reason is that I conjecture that this could be true is that we can surely
define the tangent space of $X$ at every $x$ *pure algebraically* as the dual
of the stalk $\Omega\_{X,x}$ or equivalently as $\{ \phi \in \operatorname{Hom}
(\operatorname{Spec}k[\epsilon], X) \ \vert \text{ Im } \phi = \{x\} \}$.


The Question is if in algebraic setting the surjectivity of algebraic
$D\_xf: T\_x X \to T\_y Y$ at every $x$ is equivalent to flatness of $f$?



Smooth morphism (algebraic geometry) vs. Submersion (differential geo) & Ehresman's Lemma
One of the many equivalent definitions of smoothness of a morphism $f\colon X\to Y$ of varieties over a field $k$ is that $f$ is smooth if and only if it is *formally smooth*. The latter means the following: given any square-zero extension of $k$-algebras $S\to R$ and a commuting square
$\require{AMScd}$
\begin{CD}
\mathrm{Spec}(R) @>x>> X\\
@V V V @VV f V\\
\mathrm{Spec}(S) @>y>> Y
\end{CD}
of $k$-schemes, there exists at least one diagonal map $\mathrm{Spec}(S)\to X$ which splits the square into two commuting triangles.


Let's unpack what this means in the particular case $R=k$ and $S=k[\varepsilon]$ the ring of dual numbers over $k$ (i.e. with $\varepsilon^2=0$). As the OP already noted, a morphism $\mathrm{Spec}(k[\varepsilon])\to Y$ is the same as a $k$-rational tangent vector in $Y$. The base point of such a tangent vector is the composite $\mathrm{Spec}(k)\to\mathrm{Spec}(k[\varepsilon])\to Y$.


So what formal smoothness tells us in this case is that for any $k$-rational point $x$ of $X$ and any $k$-rational tangent vector $\overrightarrow y$ of $Y$ based at $f(x)$, there is at least one $k$-rational tangent vector $\overrightarrow x$ of $X$, based at $x$, such that $f(\overrightarrow x)=\overrightarrow y$. In other words, the map $D\_xf\colon T\_xX\to T\_{f(x)}Y$ is surjective.


Thus, we see that a smooth map of varieties induces surjective maps on tangent spaces. However, in full generality (permitting singular varieties), the two conditions are *not* equivalent. For example, consider the case when $X$ is the singular affine curve with equation $xy=0$, and $Y=\mathrm{Spec}(k)$ is the point. Since the tangent space of $Y$ is trivial, the structure morphism $X\to Y$ must induce a surjection on tangent spaces. On the other hand, $X\to Y$ is certainly not a smooth map since $X$ is not smooth! The failure of formal smoothness can be seen directly: the map $k[\delta]/(\delta^3)\to k[\varepsilon]$ is a square-zero extension, but the $k[\varepsilon]$-valued point $(x,y)=(\varepsilon,\varepsilon)$ of $X$ does not lift to a point valued over $k[\delta]/(\delta^3)$.


So for general $k$-varieties, smoothness implies surjectivity on tangent spaces, but not conversely. On the other hand, I think that formal smoothness does what you asked for in the first question: it's a definition which is very similar to that of a submersion which captures the ``correct'' definition of smoothness.




---


On the other hand, if you assume that all your varieties are smooth, then the story becomes much simpler: a morphism $f\colon X\to Y$ of smooth $k$-varieties is smooth if and only if it induces a surjection on tangent spaces at each point of $X$. We've seen the left-to-right implication already. For the converse implication, suppose that $f\colon X\to Y$ induces a surjection on tangent spaces. To show that $f$ is smooth, it suffices to prove that $f$ is flat and that the relative cotangent sheaf $\Omega^1\_{X/Y}$ is locally free.


The first of these -- flatness of $f$ -- is addressed by R. van Dobben de Bruyn's comment (using miracle flatness).


For the second of these, we have an exact sequence
$$
f^\*\Omega^1\_{Y/k} \to \Omega^1\_{X/k} \to \Omega^1\_{X/Y} \to 0 \,.
$$
Surjectivity of $f$ on tangent spaces is equivalent to saying that $f^\*\Omega^1\_{Y/k} \to \Omega^1\_{X/k}$ is an injection on fibres. This makes $\Omega^1\_{X/Y}$ into a coherent sheaf, all of whose fibres have the same dimension, which then implies that it is locally free (this is e.g. Lemma 10.77.3 in the Stacks project).


