Operating homomorphism

Question

I'm doing this proof but I'm not sure that the yellow part in the proof is correct. I think it might be true because of the composition between homomorphism, is this correct?

Thank you!

Answer 1

Given arbitrary map $f \colon A \to B$ and subset $X \subseteq A$, we shall use the notation $f[X]=\{f(x)\}_{x \in X}$ to describe the direct image of $X$ through $f$. We introduce the direct image map associated to $f$ as follows: $$\begin{align} \hat{f} \colon \mathscr{P}(A) &\to \mathscr{P}(B)\\ \hat{f}(X)\colon&=f[X]. \end{align}$$ Consider an arbitrary semigroup $(S, \cdot)$. On the powerset $\mathscr{P}(S)$ consider the binary operation defined by: $$\begin{align} \cdot \colon \mathscr{P}(S) \times \mathscr{P}(S) &\to \mathscr{P}(S)\\ \cdot(X, Y)\colon&=XY\colon=\{xy\}_{\substack{x \in X\\y \in Y}}. \end{align}$$ Equipped with this new operation, $(\mathscr{P}(S), \cdot)$ becomes itself a semigroup, the so-called extension to the powerset of the original semigroup structure on $S$. Given an arbitrary subset $X \subseteq S$ and nonzero natural number $n \in \mathbb{N}^{\times}=\mathbb{N} \setminus \{0\}$, we shall write $X^{(n)}$ for the $n$-th power of element $X$ in the powerset semigroup $\mathscr{P}(S)$.

Consider now a second semigroup $T$ together with a semigroup morphism $f \in \mathrm{Hom}_{\mathbf{Sg}}(S, T)$. The following assertion is immediate:

Proposition. The direct image map $\hat{f} \in \mathrm{Hom}_{\mathbf{Sg}}\left(\mathscr{P}(S), \mathscr{P}(T)\right)$ is a morphism between the powerset semigroups.

Proof. Consider arbitrary subsets $X, Y \subseteq S$. We shall establish the relation $f[XY]=f[X]f[Y]$ by double inclusion. Let $w \in f[XY]$ be arbitrary; this means there exists $z \in XY$ such that $w=f(z)$ and by definition of the subset product there exist $x \in X$ and $y \in Y$ such that $z=xy$; it follows that $w=f(z)=f(xy)=f(x)f(y) \in f[X]f[Y]$, since obviously $f(x) \in f[X]$ and $f(y) \in f[Y]$. Conversely, let $w \in f[X]f[Y]$; again, by definition of the subset product there exist $u \in f[X]$ and $v \in f[Y]$ such that $w=uv$ and furthermore there exist $x \in X$ such that $u=f(x)$ respectively $y \in Y$ such that $v=f(y)$. We have $w=uv=f(x)f(y)=f(xy) \in f[XY]$, since clearly $xy \in XY$. $\Box$

In general, given a semigroup morphism $g \in \mathrm{Hom}_{\mathbf{Sg}}(P, Q)$ and a nonempty finite family $x \in P^I$ (this simply means the index set $I$ is nonempty and finite) such that $(I, T)$ is a totally ordered set (in other words, the object $T$ is a total order on $I$), we have the general relation $g\left(\displaystyle\prod_{\substack{i \in I\\T}}x_i\right)=\displaystyle\prod_{\substack{i \in I\\T}}g(x_i)$. As a particular instance of this, for every $n \in \mathbb{N}^{\times}$ and $x \in P$ we have $g\left(x^n\right)=g(x)^n$.

This applies in your particular instance to the end of deriving the relation $\hat{f}\left(X^{(n)}\right)=\hat{f}(X)^{(n)}$ or more explicitly $f\left[X^{(n)}\right]=f[X]^{(n)}$, for every subset $X \subseteq S$ and every nonzero natural number $n \in \mathbb{N}^{\times}$.

When working with monoids, the above relations are also valid for null exponents, as the powerset semigroup of a monoid $M$ is itself a monoid -- of unity given by $1_{\mathscr{P}(M)}=\left\{1_M\right\}$ -- and in monoids the notion of "product of a finite family" is naturally defined for empty families as well.