? For monoids $(M,b,u)$ and $(M',b',u')$, show by an example that a morphism $f\colon (M,b)\to (M',b')$ of binary operations need not be a morphism of units.
This is a homework problem I got. I think it doesn't make sense, because if $u$ is the unit of the binary operation $b$ on $M$, then by definition of morphism, for all $x$ in $M$, $f(x\,b\,u)=f(x)$ and $f(x\,b\, u)=f(x) \,b'\, f(u)$, so $f(x)=f(x)\, b'\, f(u)$ for all $x$, then by definition of unit, $f(u)$ is the unit of the binary operation $b'$ on $M'$.
So this means that all morphisms of binary operations have to be morphisms of units at the same time. I don't understand why this problem wants me to think of an example that it's not so.
Have I misread this problem, or is it just wrong?
By the way, is it OK if I just find an $f(x)$ that's not a surjection, so that $f(x) \,b'\, f(u)=f(x)$ does not work for all the elements in $M'$?