What is an example of a non-zero "ring pseudo-homomorphism"? By "pseudo ring homomorphism", I mean a map $f: R \to S$ satisfying all ring homomorphism axioms except for $f(1_R)=f(1_S)$.
Even if we let this last condition drop, there are only two ring pseudo-homomorphisms from $\Bbb Z$ to $\Bbb Z[\omega]$ where $\omega$ is any root of unity, for example. They are the identity homomorphism and the "$0$ pseudo-homomorphism".
I couldn't find a non-zero example. It is easy to conclude that $0=f(a)(1_S-f(1_R))$, so I know that the counterexample will have to be some $S$ with zero-divisors.

The (conceptually) simplest example I know is $f:\Bbb Z\to\Bbb Z\times\Bbb Z$ defined by $f(n)=(n,0)$. The ring identity of $\Bbb Z\times\Bbb Z$ is $(1,1)$ but $1$ is instead mapped to $(1,0)$. The same example shows that a subset of a ring that is a ring is not necessarily a subring if you don't require it to contain the ring unit.
