Semigroup homomorphism which isn't a monoid homomorphism

Question

I am trying to find monoids $G$ and $H$ and a semigroup homomorphism $f:G\to H$ that is not a monoid homomorphism.

I know that I need $f(1_G)\neq 1_H$.

Is this a valid example:

$f:(\mathbb{Z},\times)\to (\mathbb{Z},\times)$, $f(x)=0$. In this example, $f(xy)=f(x)f(y)$ for every $x,y\in \mathbb{Z}$, but $f(1)\neq 1$.

The reason I'm not sure about this example is that the image of $f$ is $\{0\}$, so it seems like the group $H$ should just be the trivial group, instead of $(\mathbb{Z},\times)$.

Does my example work? Or if not, can someone give me a hint about how to do this?

Answer 1

To add to the answers:

Any semigroup morphism $f:M\to S$ with $M$ a monoid, induces a monoid structure on the image $f(M)\subseteq S$, with $f(1)$ being its 'local' unit element.
However, if $f$ is not surjective, $f(1)$ needs not be a unit in the whole semigroup $S$, as your example shows.

Answer 2

Your example is correct. The image of $f$ is $\{0\}$, but the codomain is $(\mathbb{Z},\times)$, and it's the codomain you use to determine whether it is a monoid homomorphism.

Answer 3

Your example is correct. $0\colon (\mathbb{Z},\times)\to (\mathbb{Z}, \times)$ is a semigroup homomorphism but not a monoid homomorphism.

Observe that $\{0\}$ is not a submonoid of $(\mathbb{Z},\times)$. You could choose the codomain to be $(\{0,1\},\times)$, though.

Answer 4

your example, can have a monoid homomorphism, because \begin{align*} 1.f(x)=f(x)=f(1.x)=f(1).f(x) \qquad \forall x\in \mathbb{Z} \end{align*} Since $(\mathbb{Z}, \times)$ is cancelable, we have \begin{align*} 1=f(1). \end{align*}