Free objects; the way they are defined in terms of commutative diagrams Wikipedia defines free objects as follows:

Let $(\mathcal{C},F)$ be a concrete category (i.e. $F : \mathcal{C} \to {\rm \bf{Set}}$ is a faithful functor), let $X$ be a set (called basis), $A \in \mathcal{C}$ an object, and $i: X \to F(A)$ a map between sets (called canonical injection). We say that $A$ is the free object on $X$ (with respect to $i$) if and only if they satisfy this universal property:
for any object $B$ and any map between sets $f : X \to F(B)$, there exists a unique morphism $\tilde{f} : A \to B$ such that $f = F(\tilde{f})\circ i$. That is, the following diagram commutes:


I just can't seem to understand the categorical definition of free objects in terms of commutative diagrams, so I would appreciate any help in grasping the intuition behind the way free objects are defined in categorical setting.
 A: There's a more sophisticated way to think about free objects that involves fewer diagrams. Recall that the forgetful functor has a left adjoint, which produces free objects. From this perspective, forming a free object (generated by some set $S$) is equivalent to finding the canonical object containing $S$.
For example, consider the forgetful functor from groups to sets. Then the free group generated by $\{a,b\}$ is the canonical group which contains the symbols $a,b$. Hence it will have no relations. On the other hand, when you take the forgetful functor from abelian groups to sets, it will be the simplest abelian group containing $a,b$, which in this case would be $\mathbb{Z}[a,b]$.
A: You have to think of this as the following: To uniquely determine a map $A \to B$, we just have give a map of sets $f:X \to B$. This has to be denoted by $X \to F(B)$ since $X$ and $B$ do not come from the same category. Given concrete categories like groups, modules, vector spaces, you should think of $F(B)$ as the set $B$. Map of sets means that there are no restrictions at all in the sense that $f(x_1)$ and $f(x_2)$ for $x_1 \neq x_2 \in X$ are nowhere related.
This is exactly what you are used to: To give a homomorphism from a vector space $V$ to another vector space $W$, you have to say where a basis maps to. And choosing the image of the basis vectors, you have all freedoms. This is the same as giving a map of sets from the basis to the vector space $W$.
