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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.

Guerlando OCs
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  • A typical example of such triangle are homotopy liftings. – Javi Mar 04 '18 at 14:10
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    There's a whole lot of context you're omitting that you should include. For example: What is $i$? What categories are you working in? What are the "types" of $f$ and $i$, i.e. their domains and codomains? Are there conditions on $f$ or $\overline f$? Are you sure Awodey just say "there exists" a map? – Derek Elkins left SE Mar 04 '18 at 14:51

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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 $U\bar f\circ i=f$: $$\matrix{X & \overset i\to & UF(X) \\ &{}_f\!\!\searrow & \ \ \downarrow U\bar f \\ &&U(A) } $$

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.)

Berci
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  • I completely understand the triangle. What I do not understand exactly is the following: – Guerlando OCs Mar 05 '18 at 00:34
  • Once we define an aphabet, we can multiply elements of the alphabet to come up with words. The set of all words is the free object. In order to define this set of all words, the commutative triangle appears. But why? I know it has something to do with the following: – Guerlando OCs Mar 05 '18 at 00:37
  • given another map $f$ from the set of letters (the alphabet) to another set, then $|f|$ must exist such that $|f|\circ i = f$. Why? What is the purpose of this $|f|$ existing? – Guerlando OCs Mar 05 '18 at 00:40
  • The free construction itself doesn't require the commutative triangles at all. We construct the free object in an appropriate way, then we prove it satisfies the universal property. – Berci Mar 05 '18 at 00:40
  • The purpose is that we can exchange a $\mathcal Set$ morphism targeted to a structured set for a homomorphosm with same codomain. – Berci Mar 05 '18 at 00:44