@[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\\).)