I'm not sure about this since if you have two free monoid homomorphisms: $f, g : M \to M$, then $f \circ g = h$, $h$ hiding the information about the two maps, in other words we don't always have to represent elements in the space of homs as a string as we do in a free monoid. Confused! Thanks.
2026-05-10 12:34:27.1778416467
Is the space of monoid homomorphisms on a free monoid free?
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I assume the question is whether the monoid of endomorphisms of a free monoid is free. The answer is no: for $n \ge 2$, the free monoid on $n$ generators has endomorphisms of finite order (namely permutations of the generators), which free monoids do not. For $n = 1$, the endomorphism monoid is the multiplicative monoid of $\mathbb{N}$, which is a free commutative monoid (but not a free monoid) on countably many generators, one for each prime.