Earlier today, I (essentially) asked the following:
(Faulty) Question.
Given a set $X$, write $F_\mathbf{Semi}X$ for the semigroup freely generated by $X$. Suppose $X$ is a finite set, $n$ is a natural number, and $|X|=n$.
In terms of $n$, what is the maximum length of a square-free element of $F_{\mathbf{Semi}}X$?
It turns out that the question is ill-posed: according to the wikipedia page linked above, it holds that if $|X| \geq 3$, then $F_{\mathbf{Semi}}X$ has infinitely-many square-free elements, so they must be arbitrarily long. However, this seems to be in conflict with the formula here for the cardinality of the band freely generated by a finite set (note that a band is an idempotent semigroup). The linked formula tells us the cardinality of the free band on $n$ generators, which (rather amazingly) is always finite.
Write $F_\mathbf{Band} X$ for the free band on $X$.
The issue I'm having is that naively, it seems to be the case there is a natural bijection between $F_\mathbf{Band} X$ and the square-free elements of $F_\mathbf{Semi} X.$ Here's how to construct this (obviously non-existent) bijection. There is a projection homomorphism
$$\pi : F_\mathbf{Band} X \twoheadleftarrow F_\mathbf{Semi} X$$
Now restrict this so that it is defined only one the square-free elements of $F_\mathbf{Semi} X,$ call the result $\overline{\pi}$. Then my (no-doubt faulty) intuition whispers "$\overline{\pi}$ is a bijection."
If $|X|\geq 3$ and $|X|$ is finite, then considerations of cardinality make it clear that $\overline{\pi}$ isn't usually a bijection. For example, in the $n=3$ case, we'd have a bijection between the free band on $3$ generators, which has 159 elements, and the set of square-free non-empty words on $3$ characters, which has $\aleph_0$ elements. So $159 = \aleph_0$, a contradiction.
Question. What's going on here? Why isn't $\overline{\pi}$ a bijection?