Is the failure of MaxSpec to be functorial due to homomorphisms which take non-units to units? I suspect this is easy, but that I'm missing something obvious: 

If a ring homomorphism $f:R\rightarrow S$ is such that a maximal ideal $\mathfrak m$ in $S$ does not have $f^{-1}(\mathfrak m)$ maximal in $R$, must there be some non-unit element $x\in R$ whose image under $f$ is a unit in $S$?

My intuition is just that some "localization" must have occurred. In the classic example of $\mathbb{Z}\hookrightarrow\mathbb{Q}$, the zero ideal becomes maximal because 2, 3, 5, ... all become units.
 A: Maybe these remarks will be useful:
The map $R/f^{-1}(m) \to S/m$ is a map from an integral domain to a field, and if the
source is not also a field (i.e. if $f^{-1}(m)$ is not itself maximal), then there are certainly
non-units in $R/f^{-1}(m)$ that map to units in $S/m$. 
One interpretation of your question is whether, in this context, there must necessarily be a non-unit of $R$ that maps to a unit in $S$?
The answer is "no" in general. Here is a counterexample:
Let $R$ be a domain which is not a field, and let $I$ be any subset of $R\setminus \{0\}$ which generates $R\setminus \{0\}$ as a monoid under multiplication.  Let $S$ be the polynomial ring $R[\{x_i\}_{i \in I}].$
If $Q$ denotes the fraction field of $R$, then there is a natural surjection
$S \to Q$ given by mapping $x_i$ to $1/i$.  (This is where we use the assumption that
$I$ generates $R\setminus \{0\}$, so as to get a surjection.)
The kernel of this surjection is a maximal ideal $m$ of $S$, whose preimage in $R$ is the
zero ideal (hence not maximal).
On the other hand, since $S$ is a polynomial ring over $R$, any non-unit in $R$ remains a non-unit in $S$.
Some additional comments: if $R$ is Jacobson (e.g. $\mathbb Z$, or a polynomial ring over a field), then any map between finite type $R$-algebras preserves the corresponding MaxSpecs.
On the other hand, for such domains, the set $I$ is necessarily infinite, and so $S$ is necessarily infinitely generated as an $R$-algebra (and hence no contradiction ensues!).
But we can get finite type examples by taking $R$ to be e.g. a DVR.  Then $I$ can be taken
to consist of a single element (the uniformizer), and so $S = R[x]$ is a polynomial ring
in a single variable.  For example, if $R = \mathbb Z_p$ (the $p$-adic integers) then
we have 
$$\mathbb Z_p \hookrightarrow \mathbb Z_p[x] \mapsto \mathbb Z_p[x]/(p x - 1) = \mathbb Q_p,$$
and so $p x - 1$ generates a maximal ideal in $\mathbb Z_p[x]$ whose preimage is not maximal, but every non-unit in $\mathbb Z_p$ remains a non-unit in $\mathbb Z_p[x]$.
A: HINT Consider the image of a maximal ideal containing $f^{-1}(m)$.
