An object Q is free on the set X if there is a map h taking X into Q with the following property. For every object R, and for every function f(X) into R, there is a unique morphism g from Q to R such that g(h(X)) = f(X).
Note that f and h need not be morphisms. They are merely functions on sets.
This definition is inspired by free groups and free modules. Let Q be Z3 in the category of abelian groups. Let h map the set {1,2,3} to the three generators of this group. Now let R be another abelian group and let f map {1,2,3} into R. Let g map the 3 generators of Z3 to the same 3 images in R. The rest of g is defined by these three images in R. There is no inconsistency, because all the elements in Z3 are distinct, and they can land anywhere in R without producing a contradiction.
You can see that Zn is not a free object, by mapping it into Z. When g(1) = 1, g(2) = 2, and so on, until g(n) = n, which means g(0) = n, which violates the definition of a group homomorphism.
In contrast, a free group can map anywhere, into or onto any other group R, according to the image of its generators. And h(X) tells you what the generators are. It is often convenient to make X the set of generators in Q, whence h(X) is the identity map. Now every map f from X into R extends to a unique morphism from Q into R.
The word unique is important here. consider our example of Z3 in the category of abelian groups. The set {1,2} maps to two generators of Z3, and g(h(X)) = f(X), but there are many morphisms that make the diagram commute, since the third generator can map to anything in R.
The above proof breaks down in the category of sets of cardinality 1, i.e. each set contains one element. Every object is free, on any set, of any cardinality. All of X maps to the single element in Q, which carries over to the single element of R.
Using a bijection between the two sets, we can say Q1 and Q2 are free on the same set X without loss of generality. Thus X maps onto the generators of Q1, and the generators of Q2, and by transitivity these generators map onto each other.
A unique morphism carries the generators of Q1 onto the generators of Q2, consistent with the aforementioned map, and another unique morphism carries the generators of Q2 back to the generators of Q1. The composition of these morphisms produces the identity map on the generators of Q1. What about the rest of Q1?
The composition of the two morphisms implements a function f from Q1 into itself, that fixes the generators. Since Q1 is free on these generators, there is a unique morphism g from Q1 into itself that fixes these generators. Since g is unique, g = f. And what is this unique morphism? The identity morphism, which must exist for every object, fixes the generators of Q1. Therefore the composition of morphisms between Q1 and Q2 gives the identity morphism. This holds in the other direction as well, hence Q1 and Q2 are equivalent.
In summary, there is at most one free object, up to equivalence, for every cardinality.
Consider the category of vector spaces over a given field, where linear transformations act as morphisms. Every space is a free object. Build a basis for that space, and the image of the basis elements into R completely defines the linear transformation. Every vector space has a basis, and is free.
If Q is free on X, the rank of Q is the cardinality of X. Rank is not well defined for every category. There may be sets of different cardinalities, with Q free on both sets. However, in our example of vector spaces, rank is well defined. It is the dimension of the space.