In both my textbook (Hungerford's Algebra), and in class, it is claimed that Monoid Homomorphisms are not required to preserve the identity. Interestingly enough, the Wikipedia page for Monoids requires Monoid Homomorphisms to preserve the identity element: https://en.wikipedia.org/wiki/Monoid#Monoid_homomorphisms. I haven't found an example of the former, so I thought I'd prove the opposite statement.
I believe that I have proved the opposite assertion, based on the proof that I used to show that Group Homomorphisms preserve the identity. Since I don't use any information stating that elements are invertible, I think my proof is still valid.
Let $M, N$ be monoids, and let $f:M\rightarrow N $ be a homomorphism of monoids.
Let $m,e_{M} \in M$ be an arbitrary element and the identity in $M$ respectively .
Then:
$$f(m) = f(m\cdot e_{M}) = f(m)\cdot f(e_{M})$$
$$f(m) = f(e_{M} \cdot m) = f(e_{M}) \cdot f(m)$$
Thus: $$f(m)\cdot f(e_{M}) = f(e_{M}) \cdot f(m) = f(m), \forall m \in M $$
This seems to imply my assertion.
Is there anything wrong with my proof?
Best Answer
As 57Jimmy points out in their comment, you have not proved that the "identity" you have found is the identity of the whole monoid.
Let make this all formal:
As a counter-example, take your favourite monoid $A$ and then attach an identity to obtain a new monoid, $B$. Then the embedding map $A\hookrightarrow B$ is a semigroup homomorphism, but the image of the identity isn't the identity of $B$. For example, take $A=\{e\}$ such that $e^2=e$ and attach an identity $1$ to obtain a new monoid $B$, so $B=\{e, 1\}$ where $1\cdot e=e=e\cdot 1$ and $1^2=1$. Then clearly the monoid $A$ embeds into $B$, but $e$ is not the identity of $B$ (it is in fact the zero, as $1\cdot e=e$, etc.)