TITLE: definition of a free object in a category
QUESTION [8 upvotes]: My book has this definition:

Let $ A $ be an object of a concrete category $ \scr{C} $, $ X $ a nonempty set, and $ i: X \to A $ a map (of sets). We say that $ A $ is free on a set $ X $ provided that for any object $ B $ of $ \scr{C} $ and a map (of sets) $ j: X \to B $, there exists a unique morphism $ \phi \in {\text{Hom}_{\scr{C}}}(A,B) $ such that $ \phi \circ i = j $ (composition of functions).

Question 1: Wikipedia's definition says that the map $ i $ is a canonical injection. The one above allows the map $ i $ to be not injective. I feel that a good definition must include the injectivity of $ i $; am I correct?
Question 2: Can the definition of free objects be extended to categories other than concrete categories? One definition I thought of is: Let $ \scr{C} $ and $ \scr{D} $ be two categories, and let $ F := (F_{0},F_{1}) $ be a functor from $ \scr{C} $ to $ \scr{D} $. Let $ A $ and $ X $ be objects in $ \scr{C} $ and $ \scr{D} $ respectively. (Notation: Here, $ F_{0} $ is the map between the objects of $ \scr{C} $ and $ \scr{D} $, and $ F_{1} $ is the map between the morphisms of $ \scr{C} $ and $ \scr{D} $). Let $ i \in {\text{Hom}_{\scr{D}}}(X,{F_{0}}(A)) $. We say that $ A $ is free on $ X $ iff for every object $ B $ in $ \scr{C} $ and morphism $ j \in {\text{Hom}_{\scr{D}}}(X,{F_{0}}(B)) $, there exists a unique morphism $ \phi \in {\text{Hom}_{\scr{C}}}(A,B) $ such that $ {F_{1}}(\phi) \circ i = j $. Again, I don't know what condition to add on the morphism $ i $ (if there is any to add).
As I am new to category theory, I don't know if it is interesting to create a definition for free objects in non-concrete categories (I just think it is).
Thank you all.

REPLY [5 votes]: Let's answer question 2 first. What you have written down is essentially the local universal property of a left adjoint. Specifically, suppose we have a functor $U : \mathcal{A} \to \mathcal{S}$. Let $X$ be an object in $\mathcal{S}$. A "free object on $X$ relative to $U$" is a pair $(F X, \eta_X)$ consisting of an object $F X$ in $\mathcal{A}$ and a morphism $\eta_X : X \to U F X$ in $\mathcal{S}$ such that, for every object $A$ in $\mathcal{A}$ and every morphism $f : X \to U A$ in $\mathcal{S}$, there exists a unique morphism $\overline{f} : F X \to A$ in $\mathcal{A}$ such that $f = U \overline{f} \circ \eta_X$. If free objects exist for all objects in $\mathcal{S}$, then these can be assembled into a functor $F : \mathcal{S} \to \mathcal{A}$ that is left adjoint to $U : \mathcal{A} \to \mathcal{S}$.
Now, question 1. As Hurkyl has indicated, the "insertion of generators" $\eta_X : X \to U F X$ can fail to be monic. There is a complete classification of when this happens when $\mathcal{S} = \textbf{Set}$ and $U : \mathcal{A} \to \textbf{Set}$ has a left adjoint $F : \textbf{Set} \to \mathcal{A}$:

$U F X = 1$ for all sets $X$. This corresponds to the algebraic theory (or "variety" in the sense of universal algebra) with one constant $a$ and the axiom $x = a$.
$U F X = \emptyset$ when $X = \emptyset$ and $U F X = 1$ otherwise. This corresponds to the algebraic theory with no constants and the axiom $x = y$.

Now, why is this? Suppose $\eta_X : X \to U F X$ fails to be injective for some set $X$. Obviously, $X$ must have at least two elements for this to happen: so suppose $\eta_X (x_1) = \eta_X (x_2)$ with $x_1 \ne x_2$. Now take any $A$ in $\mathcal{A}$. If $U A$ is empty then there is nothing to do; otherwise, suppose $a_1$ and $a_2$ are elements of $U A$. We can construct a map $f : X \to U A$ by defining
$$f(x) = \begin{cases}
a_1 & \text{if } x = x_1 \\
a_2 & \text{otherwise}
\end{cases}$$
and by the universal property of $F X$ there exists a morphism $\overline{f} : F X \to A$ in $\mathcal{A}$ such that $U \overline{f} \circ \eta_X = f$. But then
$$a_1 = f(x_1) = U \overline{f} (\eta_X (x_1)) = U \overline{f} (\eta_X (x_2)) = f(x_2) = a_2$$
so we are forced to conclude that $U A$ has only one element. Thus, if $\eta_X : X \to U F X$ is not injective, then $U A$ is either empty or a singleton for all objects $A$ in $\mathcal{A}$. On the other hand, the fact that there is a map $\eta_Y : Y \to U F Y$ for all sets $Y$ means that $U F Y$ is empty if and only if $Y$ is empty. Thus the only possibilities are the two examples listed above.