Preimaging units to units I'm interested in (unity-preserving) homomorphisms $f: S \to T$ between (commutative, with-unity) rings $S$ and $T$ so that if $f(x)$ is a unit, then $x$ was a unit to start with.  For example, an inclusion of fields has this property, but a nontrivial localization like $\mathbb{Z} \to \mathbb{Q}$ does not; it sends $2 \in \mathbb{Z}$, which is not a unit, to $2 \in \mathbb{Q}$, which is.  I'm interested in answers to either of the following two questions:


*

*Is there a name for such $f$ that do preserve units through preimages?

*If not, is there a class of maps broadly recognized as useful that enjoy this property?


I'm actually only interested in the case of a surjection $R \twoheadrightarrow R_0$ whose kernel is an ideal $I$ satisfying $I^2 = 0$ (i.e., a square-zero extension of $R_0$ by $I$).  But, if this property has a name in general or if square-zero extensions occur as special types of some broadly-recognized class of maps with this property, I'd like to know so I can chat about this with other people and don't go picking a name nobody recognizes.
I'm aware that this can also be viewed as a lifting property, if that jogs anyone's memory of useful geometry words: what kind of map should $\operatorname{spec} S \to \operatorname{spec} R$ be so that for any pair of maps $\operatorname{spec} S \to \mathbb{G}_m$ and $\operatorname{spec} R \to \mathbb{A}^1$ with commuting square $$\begin{array}{ccc} \operatorname{spec} S & \to & \mathbb{G}_m \\ \downarrow & & \downarrow \\ \operatorname{spec} R & \to & \mathbb{A}^1,\end{array}$$ there exists a lift $\operatorname{spec} R \to \mathbb{G}_m$ making both triangles commute?
 A: The following result gives a characterization of when the property you ask about holds in the case of a surjective morphism.
Claim: If $f: S \to T$ is a surjective homomorphism of rings, then $f^{-1}(T^{\times}) = S^{\times}$ if and only if the kernel of $f$ is contained in the Jacobson radical of $S$.
Proof: If $f(s) \in T^{\times}$ and $f(s') = f(s)^{-1}$, then $f(s s') = 1$.
Thus the claim reduces to checking that $f^{-1}(1) \subset S^{\times}$,
i.e. that $1 + \mathrm{ker}(f) \subset S^{\times}$.  This is equivalent to asking
that $\mathrm{ker}(f)$ be contained in the Jacobson radical of $S$. QED
In general the nilradical of $S$ is contained in the Jacobson radical,
and so if $f$ is surjective and the kernel of $S$ is a nil ideal then the property you are interested in holds. (This generalizes your square-zero extension example.)
If $S$ is a Jacobson ring, e.g. a finite type algebra over a field or over $\mathbb Z$,
then the Jacobson radical of $A$ coincides with the nilradical, and so for such
$S$ your condition holds (for $f$ surjective) precisely when the kernel of $f$ is a nil ideal.
A: This isn't an answer to the exact question I asked, but it will do:
The morphism $\mathbb{G}_m \to \mathbb{A}^1$ is an open immersion, hence étale, hence formally étale.  This exactly means that maps in from morphisms $\operatorname{spec} R/I \to \operatorname{spec} R$, where $I$ is a nilpotent ideal in $R$, lift.
