@[Anindya](https://forum.azimuthproject.org/discussion/comment/18231/#Comment_18231): If we have a monoid homomorphism \$$f : X \to Y\$$, then we have \$$f(1_X) \otimes f(x) = f(1_X \otimes x) = f(x) = f(x \otimes_X 1_X) = f(x) \otimes_Y f(1_X)\$$. So if nothing else, \$$f(1_X)\$$ is a unit for the image \$$f(X) \subseteq Y\$$. And as you say, if \$$1_Y\$$ is mapped to, then \$$1_X\$$ must be one of the things that map to it. (Nice proof!)

Unfortunately, it's entirely possible for something that isn't the unit of the larger monoid to nonetheless behave like a unit for a subset of the monoid. Consider the monoid \$$\langle \mathcal{P}(S), \cup, \emptyset \rangle \$$ of subsets of a set \$$S\$$, with union as product and the empty set as unit. Let \$$x \in S\$$, and consider the upset \$$\operatorname{\uparrow}\\{x\\}\$$ consisting of the collection of all sets containing \$$x\$$. Then \$$\\{x\\}\$$ acts as a unit for this collection, because every subset in this collection already contains \$$x\$$.

One reason to require that a homomorphism preserve the unit is because this lets us classify _submonoids_ (and [subalgebras](http://planetmath.org/homomorphismbetweenalgebraicsystems) in general). A submonoid is a subset of a monoid where the unit and multiplication are the same as the containing monoid. With monoid homomorphisms as defined, every submonoid is just the image of a homomorphism into the containing monoid. (This can be as simple as an injection, like \$$i : \mathbb{N} \to \mathbb{Z}\$$ defined by \$$i(x) = x\$$.)