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Here is my failed attempt. Please, see if it is far from being fixable or not.

Suppose $\mathcal{A}$ is a concrete category over $\mathcal{X}$, i.e. there is a faithful functor $U \colon \mathcal{A} \longrightarrow \mathcal{X}$ and for $X \in \mathcal{X}$, $\mathcal{F}^{X}$ is the following category:

Objects of $\mathcal{F}^{X}$ are pairs $\left(i,A\right)$, such that $i:X \longrightarrow A$ is a morphism! (1)

Morphisms of $\mathcal{F}^{X}$ are $f:\left(i_1,A_2\right) \longrightarrow \left(i_2,A_2\right)$ such that $f\circ i_1 = i_2$

Assuming that $U:$ Grp $ \longrightarrow$ Set is the underlying functor, the initial object of $\mathcal{F}^{X}$ for $X \in$ Set is the free group generated by $X$.

  1. Here is the catch, $i$'s are supposed to imitate the function from sets to the groups, in definition of free groups but obviously it is not clear where they live. I guess,they must be characterized in terms of $U$, the underlying functor.
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You need to be more clear about what category your $i$ is a morphism in. Hint: There is no such thing as a morphism from ane element of $\mathcal X$ to an element of $\mathcal A$. – Thomas Andrews Nov 19 '12 at 16:15
If you're just thinking of Groups and Sets for the time being, the $i$-morphisms have no choice but to be morphisms of sets (after all, $X$ is a set, and $A$ could be viewed as a set). The $f$ morphisms, however, should be morphisms of groups. You can see this by trying to prove the universal property for a free group from what you have above, and look at which conditions start to become necessary. – Thomas Belulovich Nov 19 '12 at 16:22
I'm a complete noob in these matters, but aren't free objects the thing that you're looking for? – Dan Shved Nov 19 '12 at 16:23
Thanks a lot. I guess I have heard the term free objects. I read it and now and I am satisfied with that. – Hooman Nov 19 '12 at 16:36
up vote 2 down vote accepted

This all can be made precise:

First of all, note that you are basically looking for the left adjoint of the "concretization functor" $U$.

Secondly, we can consider a category that (disjointly) contains both $\mathcal X$ and $\mathcal A$, and arrows accordingly: let the $f:X\to A$ arrows be the $f:X\to U(A)$ arrows in $\mathcal X$. This way we get the collage of the profunctor $U^*:\mathcal X^{op}\times\mathcal A\to\mathcal{Set}$ which maps $\langle X,A\rangle\mapsto \hom_{\mathcal X}(X,U(A))$.

Within this collage, which itself is a category, each $A$ is coreflected by $U(A)$, by definition, and the free objects you look for are the reflections of the $X$'s.

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