Let $f:S\to T$ be a surjective mapping of monoids that holds the homomorphic property: $f(xy)=f(x)f(y)$. I want to show that $f$ is a homomorphism and also find an example of a mapping that holds the homomorphic property, but not having $f(1_S) = 1_T$.
My attempt is in the answer below.
On
The mapping is homomorphic, as all that is required for homomorpism is that the monoid is properly defined, and that the homomorphic property is satisfied.
An example of a homomorphic mapping $f$ that doesn't have $f(1_S)=1_T$ is $f:S\to T$, $f(x)=e^x$. Clearly $f(x+y)=e^{x+y}=e^x e^y=f(x)\cdot f(y)$. If we take $f:S\to T$ with $S=(\Bbb R, \cdot), T=(\Bbb R, +)$, we have $1_S=1,1_T=0$ and thus $f(1)=e\ne 0$
Hint: Try the map $f\colon S \to T$ defined by $f(x) = 1_T$ for all $x$.
For a less trivial example consider the ring $A = k[x]/(x^2 - x)$ and let $S = (A, \cdot)$. Then the map $f\colon S \to S$ defined by $f(a) = ax$ is a homomorphism.