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? 



Does formal smoothness work compatibly across morphisms?
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. 


