Do Henselian extensions have left lifting property with respect to smooth morphisms?

Question

Consider the following commutative diagram of schemes:  $\require{AMScd}$ \begin{CD} T @>{}>> X\\ @VVV @VVV\\ T' @>{}>> Y \end{CD}

where  $T\hookrightarrow T'$ is a Henselian extension and $X\to Y$ is a smooth morphism. Can one always find a morphism $T'\to X$ such that the following diagram is commutative:

$$\begin{array}{ccc} T & \rightarrow & X \\ \downarrow & \nearrow & \downarrow \\ T' & \rightarrow & Y \end{array} $$

If $T\hookrightarrow T'$ is a first order thickening , this is just the claim that smooth implies formally smooth. It follows that the claim is true if we replace $T'$ by its completion along T.  What about the general case?

We are mainly interested in the case when $T$ is a spectrum of a field $F$. In this case, $T'$ is a spectrum of Henselian ring with residue field $F$.

In fact, we did not find the notion of Henselian extensions in the literature (though it looks rather natural to have). we will be grateful for a reference for this notion too.

Answer 1

In the local case this is a well known property of henselian rings. It is standard when $X \to Y$ is étale; in the general case it follows from the fact that any smooth maps is Zariski-locally a composite $X \to \mathbb{A}^n_Y \to Y$, where $X \to \mathbb{A}^n_Y$ is étale.

About the general case, for the lifting property to hold you need $T$ to be affine. I don't know what a "henselian extension" is; if it is of the type $\operatorname{Spec}(A/I) \subseteq \operatorname{Spec}A$, where $(A,I)$ is a henselian pair, I would suspect that the answer is still positive, but I don't know a reference.