We have (A, ∗) as a semigroup and the set M.We can say that (A, ∗) is a free semigroup on set M if there is a function f: M → A such that for any semigroup (B,⊗) and any function g : M → B there is a unique morphism of semigroups φ : (A, ∗) → (B, ⊗) such that g = φ ◦ f .
It also says that if (A, ∗) is a free semigroup on set M then the function f: M → A is injective.As a result,the set M can be identified with it's own image f(M) ⊆ A.
I don't know what a free semigroup on a set looks like and what is it's purpose.
Can you please help me understand this definition from an abstract point of view?
I am new to this kind of algebra.
Define $A$ as the collection of finite tuples $\langle m_1,\dots,m_n\rangle$ where $n$ is a positive integer and the $m_i$ are elements of $M$.
Then define composition by:$$\langle m_1,\dots,m_n\rangle\star\langle m'_1,\dots,m'_k\rangle=\langle m_1,\dots,m_n,m'_1,\dots,m'_k\rangle$$
Then $\langle A,\star\rangle$ can be recognized as semigroup free over set $M$.
We have the inclusion map $f:M\to A$ prescribed by $m\mapsto\langle m\rangle$ which is evidently injective.
This allows us to identify $M$ with subset $f(M)=\{\langle m\rangle\mid m\in M\}\subseteq A$.
If $\langle B,\otimes\rangle$ is a semigroup and $g:M\to B$ is a function then there is a unique homomorphism $\phi:\langle A,\star\rangle\to\langle B,\otimes\rangle$ prescribed by: $$\langle m_1,\dots,m_n\rangle\mapsto g(m_1)\otimes\cdots\otimes g(m_n)$$
and such that $\phi\circ f=g$.
This is not the construction of the semigroup free over $M$ but the construction of a semigroup free over $M$.
Semigroups free over a fixed set $M$ are unique up to isomorphisms.
So a semigroup $\langle S,\odot\rangle$ is free over $M$ if and only if an isomorphism $\langle S,\odot\rangle\to\langle A,\star\rangle$ exists.