How does this commutative triangle arises in the idea of freely generated things?

Question

I know that in category theory you normally abstract ideas. So I guess the Universal Mapping Property comes from other fields. In fact, I've already studied free modules and it was there.

I'm trying to figure out how this for every map $f$ there is a map $\overline{f}$ such that $\overline{f}\circ i = f$ comes from. I'm on page 17 of Steven Awodey's book and it talks about the Alphabet being freely generated.

The closet answer I could find was https://math.stackexchange.com/a/2469950/166180. It kinda talks about other map existing but I can't see the connection with this composition/commutative triangle thing. In this example the alphabet is also used. I understand it very well, but I still don't see the connection.

Answer 1

As Derek Elkins well commented, you miss to give information about 'where do $i,f,\bar f$ live'.
The title of the question suggests that it's about the universal property of the free objects.

So, for a given set $X$, $\ i$ will be the embedding $X\to F(X)$ where $F(X)$ denotes the free whatever structure generated by $X$.
Now, freeness of $F(X)$ says that

For any function $f:X\to A$ which goes to [the underlying set of] some object $A$ of whatever structure, there is a unique morphism $\bar f:F(X)\to A$ (preserving the whatever structure) such that $\bar f\circ i=f$.

Once we know that $i$ is an embedding, $\bar f\circ i=f$ means nothing but that $\bar f$ extends $f$.
In other words, given an 'evaluation' $f(x)\in A$ for all the 'variables' $x\in X$, there is a unique way to extend this to the free object $F(X)$ to obtain a morphism $F(X)\to A$.

To make it precise, one usually states this triangle in $\mathcal Set$ (because $X$ is not assumed to have any structure), and in our formulas we might want to use the 'underlying set' functor $U$ to make things clearer. (Note that $U$ will map a homomorphism to the exact same mapping, but viewed as only a pure function.)
With that, we should rather write $U\bar f\circ i=f$: $$\matrix{X & \overset i\to & UF(X) \\ &{}_f\!\!\searrow & \ \ \downarrow U\bar f \\ &&U(A) } $$

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If you want to work out a specific example, I suggest semigroups (or monoids), as a free semigroup over a set ('alphabet') $X$ is just the set of finite 'words' (sequences) built from elements of $X$, with concatenation as the associative operation. (Free monoids include the empty word as unit element.)