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

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.

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.