Image of the complement of an ideal - when $\phi\left(R\setminus I\right)=\phi\left(I\right)\setminus \phi\left(I\right)$? In the following all rings are considered commutative and unital.
Now during my repetition of our introduction to fibers, I stumbled over the following
Hypothesis.
Let $P$ be a prime in $R$, and $\phi:R\to S$ a homomorphism of rings. Then
either $\phi\left(R\setminus P\right)=\phi\left(R\right)\setminus \phi\left(P\right)$ or the ideal generated by $\phi\left(P\right)$ is equal to $S$.
I suppose $"\supseteq"$ follows from set theory.
I would prefer not to write down my surely wrong proof, since i suppose this can be easily proven or disproven, which would take me ages tho.
 A: Before producing some counterexamples, let's look at your first condition a little more closely, by proving the following:

Let $I$ an ideal of $R$ and $\phi: R \rightarrow S$ a ring homomorphism.  Then $\phi(R \setminus I) = \phi(R) \setminus \phi(I)$ iff $\ker(\phi) \subseteq I$. 

Proof: Note that the $\supseteq$ containment always holds, so the equality holds iff the $\subseteq$ relation holds.  This in turn holds iff $\phi(R \setminus I) \cap \phi(I) = \emptyset$.  We claim $\phi(R \setminus I) \cap \phi(I) = \emptyset$ iff $0 \notin \phi(R \setminus I)$ (equivalently $\ker(\phi) \subseteq I)$.  This is necessary because always $0 \in \phi(0) \in \phi(I)$.  It is sufficient because if $a = \phi(b) = \phi(c)$ with $b \in R \setminus I$ and $c \in I$, then $\phi(b-c) = 0$ and $b - c \in R \setminus I$. $\square$
(Notice that primeness didn't play any role here).  
The observation that we just proved will make it easy to produce counterexamples where the homomorphism $\phi$ is even surjective.  

Let $R$ be a ring containing an ideal $I$ and a prime $P$ such that: (1) $I + P \not= R$ and (2) $I \nsubseteq P$.  Then $P$ and $\phi: R \twoheadrightarrow R/I$ constitute a counterexample to your hypothesis.

Indeed, since $I$ is the kernel of $R \twoheadrightarrow R/I$, the assumption that $I \nsubseteq P$ ensures that $\phi(R \setminus P) \not= \phi(R) \setminus \phi(P)$.  And since $I + P \not= R$, $\phi(P)$ doesn't generate $R/I$.  
Some really trivial counterexamples immediately present themselves.  If $D$ is any integral domain that is not a field, so that $(0)$ is a prime ideal of $D$ and $D$ has a nontrivial ideal $I$, then $(0)$ and $D/I$ are a counterexample. Or generally if $R$ has a nonmaximal prime ideal, you can get a counterexample in the same way.  
Of course we don't need to restrict ourselves to $P \subseteq I$ in the above.  For example, we could consider taking $R = D[x,y]$, $P = (x)$ and $I = (y)$.  
