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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.

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2 Answers 2

up vote 4 down vote accepted

Let $\mathscr{C}$ be the category with one object (which I will call "*") and only the identity morphism, with the forgetful functor that sends the object to the one-element set (which I will call "1").

By this definition, every map $i : X \to *$ makes $*$ free on $X$.

This is good, because it coincides with the usual notion of free from algebra; $\mathscr{C}$ is, for example, equivalent to the category of algebras for the universal algebra with no operations, one constant symbol $c$ and one relation $$x = c$$

The unique algebra for this theory is clearly the free algebra on any number of generators.

One good generalization of the notion of "free object" along the lines of universal algebra comes from the notion of a monad, which corresponds to a special sort of adjunction.

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Apart from this example, is it necessary to insure that $i$ is injective? –  Amr Dec 30 '12 at 2:56
    
For universal algebras, I think that is the only non-injective example. –  Hurkyl Dec 30 '12 at 3:18
    
I see. Do you have an idea why wikipedia said "canonical injection". –  Amr Dec 30 '12 at 3:29
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@Amr: IMO, the most likely reasons are (from most to least) ignorance, error, rejection of degeneracy, and sloppiness. A good term I've heard for this is "insertion of generators", or just the "insertion map". –  Hurkyl Dec 30 '12 at 4:11
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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.

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