free semigroup with involution

Let $X,X^{‡}$ be two disjoint sets in bijective correspondence given by the map ${}^{‡}:X\to X^{‡}$. Denote by $Y=X\coprod X^{‡}$ (here we use $\coprod$ instead of $\cup$ to remind that the union is actually a disjoint union) and by $Y^{+}$ the free semigroup on $Y$. We can extend the map ${}^{‡}$ to an involution ${}^{‡}:Y^{+}\to Y^{+}$ on $Y^{+}$ in the following way: given $w\in Y^{+}$, we have $w=w_1{w}_2\mathrm{\dots }{w}_k$ for some letters $w_i\in Y$; then we define

$$w^{‡}=w_k^{‡}{w}_{k-1}^{‡}\mathrm{\dots }{w}_2^{‡}{w}_1^{‡}.$$

It is easily verified that this is the unique way to extend ${}^{‡}$ to an involution on $Y$. Thus, the semigroup ${(X\coprod X^{‡})}^{+}$ with the involution $‡$ is a semigroup with involution. Moreover, it is the free semigroup with involution on $X$, in the sense that it solves the following universal problem: given a semigroup with involution $S$ and a map $\mathrm{\Phi }:X\to S$, a semigroup homomorphism $\overline{\mathrm{\Phi }}:{(X\coprod X^{‡})}^{+}\to S$ exists such that the following diagram commutes:

where $\iota :X\to {(X\coprod X^{‡})}^{+}$ is the inclusion map. It is well known from universal algebra that ${(X\coprod X^{‡})}^{+}$ is unique up to isomorphisms.

If we use $Y^{*}$ instead of $Y^{+}$, where $Y^{*}=Y^{+}\cup \{\epsilon \}$ and $\epsilon$ is the empty word (i.e. the identity of the monoid $Y^{*}$), we obtain a monoid with involution ${(X\coprod X^{‡})}^{*}$ that is the free monoid with involution on $X$.