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Given a function $f: A \to B$, I can construct a morphism $g : A^* \to B^*$ where $X^*$ denotes some free structure generated by $X$ (Could be monoid, group, module, etc.).

I'd like to study morphisms generated by functions a bit more, but I'm not sure where I should be looking. Do they have a specific name? Where would I look in an algebra/category theory book for more info?

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You could start by reading the Wikipedia article on free objects in a category: – Fredrik Meyer Jan 26 '13 at 18:56
@FredrikMeyer That's what inspired the question! Maybe I'm an idiot, but I don't see any functions of the form $g$ has on that page. – Mike Izbicki Jan 26 '13 at 18:57
You should make your question more precise in order to get interesting/helpful answers. What do you want to know about $g$? – Martin Brandenburg Jan 27 '13 at 12:00
@Mike: $g$ is the part of the free functor which acts on morphisms. – Qiaochu Yuan Aug 23 '13 at 23:30

Let $\mathcal C$ be some category of algebraic structures.

Define the forgetful functor $U : \mathcal C \to \mathbf {Set}$.

Construct its left adjoint $F \dashv U$.

Then $F$ takes sets to freely generated objects and morphisms between the generators to morphisms between freely generated objects.

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You should look in category theory books.

There, you would read something along these lines:

Given a set A you can build a group (or a monoid, module, ect) which is called the free group and is denoted by $F(A)$ (or even $FA$, which I find very bad, personally).

Given a function $f:A \to B$ you can build a group-morphism (or monoid, module,.., -morphism) $F(f):F(A) \to F(B)$ between the free groups generated by $A$ and $B$.

The $F$ operator is called the Free functor from $\mathcal{Set}$ - the category of sets and functions - to $\mathcal{Grp}$ -the category of groups and group-morphisms.

$$F:\mathcal{Set} \to \mathcal{Grp}$$

The interesting thing about this operator is that it can be proved that it is a functor.

Correspondence with your notation:

$F(X) \longleftrightarrow X^*$

$F(f) \longleftrightarrow g$

As was mentioned by @user58512, it turns out that $F$ is the left adjoint to the $U$ functor.

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