Free objects on same set are isomorphic Let $(\mathbf{C}, \mathcal{F})$ be a concrete category.
Let $S$ be a set, we define $F \in C$ together with $i : S \rightarrow \mathcal{F}(F)$ as a free object on $S$ if it follow universal property :
for any $Y \in C$ and any map $\varphi : S \rightarrow \mathcal{F}(Y)$ there exists a unique morphism $\Phi : F \rightarrow Y$ such that $\varphi = \mathcal{F}(\Phi) \circ i$.
My question is : how can we show that free objects on $S$ are essentially the sames (i.e. they are isomorphic) ? I don't see where to start. I know how to do it in the similar case of free groups when $i$ is one-to-one.
 A: If there exist two free objects $F_1$ and $F_2$ then there are maps $\Phi_1:F_1\rightarrow F_2$ and $\Phi_2:F_2\rightarrow F_1$. Both composite maps $\Phi_1\Phi_2$ and $\Phi_2\Phi_1$ must by uniqueness be identity maps.
A: Let $i\colon S\to F(X)$ and $j\colon S\to F(Y)$ be free objects on $S$. By the universal property for the first, there is a suitable morphism $X\to Y$. By the universal property for the second, there is a suitable morphism $Y\to X$. The suitability of these morphisms implies that their composition (in both directions) is the identity (by using the uniqueness in the universal property for each of the assumed free objects).
The most conceptual way of understanding this situation is that free objects can be used to construct a functor $L\colon \mathbf{Set}\to \mathbf{C}$, and this is left adjoint to your $\mathcal F$, but concreteness of the category has nothing to do with this property. Different constructions of free objects thus yield two left adjoints for the same functor. But left adjoint are uniquely determined up to a natural isomorphism, and so their values at any particular set (or any object in the relevant category) are isomorphic objects in $\mathbf C$.
