# Does formal smoothness work compatibly across morphisms?

This question is about the formal smoothness property for schemes. A morphism $X\to S$ is formally smooth if for every affine $S$-scheme $Y$ and every subscheme $Y_0\subset Y$ cut out by a nilpotent ideal, morphisms $Y_0\to X$ always extend to $Y\to X$. (Here all morphisms are $S$-morphisms.)

Let's suppose $f\colon X'\to X$ is an $S$-morphism between schemes which are each formally smooth over $S$. Say we have a commutative diagram of $S$-schemes $$\begin{array}{ccc} Y_0' &\to & Y_0 \\ \downarrow & & \downarrow\\ Y' &\to & Y \end{array}$$ with $Y_0$ and $Y_0'$ cut out by nilpotent ideals, and another commutative diagram $$\begin{array}{ccc} Y_0' &\to & Y_0 \\ \downarrow & & \downarrow\\ X' &\to & X \end{array}.$$ Since $X\to S$ is formally smooth, the morphism $Y_0\to X$ extends to a morphism $Y\to X$. My question is, does the morphism $Y_0'\to X'$ extend to a morphism $Y'\to X'$ in such a way that $$\begin{array}{ccc} Y' &\to & Y \\ \downarrow & & \downarrow\\ X' &\to & X \end{array}$$ commutes?

I'm currently thinking not, because (in the absence of further assumptions) you don't have any control over the way these morphisms lift. Am I overlooking something simple?

• It seems that you'd want X' \to X to be smooth here (at least if you wanted an easy yes.) As stated, it's easy to give a counterexample by taking X' a closed subscheme, and choosing the map from Y so that the induced map from Y' doesn't factor through X'. Dec 17, 2013 at 23:14
• Shouldn't one take only affine $Y$ in the definition of formal smoothness, see [EGA IV_4, 17.1.1]? Due to nonuniqueness, the extensions on an affine cover need not glue. Dec 17, 2013 at 23:14
• Kestutis is correct: there are cohomological obstructions to lifting maps without the affineness hypothesis on $Y$. On the other hand, for affine $Y$ the lifting problem can be checked Zariski-locally since the relevant obstruction to globalizing lives in a higher quasi-coherent cohomology group on $Y$ that vanishes since $Y$ is affine. Dec 17, 2013 at 23:21

Certainly not for a fixed morphism $Y \to X$. Perhaps if you choose the morphism $Y \to X$ it might work. Counterexample:
$Y_0 =Y_0'= X'= \operatorname{Spec} k$. $Y= Y' = \operatorname{Spec} k [x]/x^2$. $X = \operatorname{Spec} k[x]$. The first two commutative diagrams are the obvious ones. Then the obvious morphism $Y \to X$ does not extend to a commutative diagram.
Geometrically, we have a point $X'$ inside a line $X$. $Y_0'$ and $Y_0$ are the same point, and $Y$ and $Y'$ are tangent vectors. We can extend the point on the line to a tangent vector in more than one way. If the tangent vector is nonzero and exits the point, than it does not form a commutative diagram. Only if the morphism $Y \to X$ is chosen to be the zero tangent vector does it live inside the point and form a commutative diagram.