Let $(R, I)$ be a Henselian pair, with $I$ a finitely generated ideal.


We know that for any smooth $R/I$-algebra $A\_0$, there exists a smooth $R$-algebra $A$ such that $A/I\simeq A\_0$.


We also know that for any map $A\_0\to B\_0$ of smooth $R/I$-algebras, there exist $R$-smooth algebras $A$ and $B$, and a map $A\to B$ that lifts $A\_0\to B\_0$.



> 
> Suppose $A\_0\to B\_0$ is surjective. Can $A\to B$ be arranged to be surjective too?
> 
> 
> 



Lifts of smooth algebras
Yes. Let $i\colon Y=\mathbf{Spec}(B)\to \mathbf{Spec}(A)=X$ be the induced map of schemes, and let $K$ be the cokernel of
$$ i^\*\colon \mathcal{O}\_X \to i\_\* \mathcal{O}\_Y. $$
This is a coherent $\mathcal{O}\_X$-module whose support does not meet $X\_0$ by assumption. Thus after replacing $X$ with an affine open neighborhood of $X\_0$ and $Y$ with its base change to this neighborhood we get $K=0$, so $i^\*$ is becomes surjective.


