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Learning about free groups introduced me to free objects. At the bottom are my attempts at explaining free objects, which I have questions about here:

  1. What $F(g) \circ i$ means in the universal property definition in (2). This is a bit confusing b/c $F(A)$ and $F(B)$ plugs in objects, while this is plugging in a morphism and ~composing~ it (not sure) with $i$.
  2. An intuition on how to describe the universal property definition. The way I wrote it in (1) doesn't explain it much.
  3. Where the generators / relators are in the definition in (2). Or if they're just part of the creation process.
  4. If the general technique for creating a free object works for all free object types.

(1) Suppose we have a category, a faithful functor, a basis set, and an injection. A category object is a free object iff, for any other category object and corresponding injection, there is a unique morphism such that [question 1]. That is, it satisfies the universal property.

(2) Formally, suppose we have a concrete category $(C, F)$ -- where $F : C \to \mathbf{Set}$ is a faithful functor -- an object $A \in C$, a set $X$ called the basis, and a canonical injection $i : X \to F(A)$. Then we call $A$ the free object on $X$ with respect to $i$ iff it satisfies the universal property. The universal property says that $\forall B \in C,\ f : X \to F(B),\ \exists!g : A \to B\ \ s.t.\ \ f = F(g) \circ i$. That is, the following diagram commutes:

$${\begin{array}{c}X{\xrightarrow {\quad i\quad }}F(A)\\{}_{f}\searrow \quad \swarrow {}_{F(g)}\\F(B)\quad \\\end{array}}$$

The creation of free objects happens in two steps:

  1. Define all collections of words over an alphabet. For non-associative algebraic relations, use parentheses.
  2. Impose defining equivalence relations on the words.

The free object then consists of the set of equivalence classes.

From Wikipedia:

The most general setting for a free object is in category theory, where one defines a functor, the free functor, that is the left adjoint to the forgetful functor.

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  • $\begingroup$ I don't understand your first point/question. $F$ is a functor, so it accepts morphisms and objects. $F(g)\circ i$ is thus a perfectly valid function between sets. The universal property then implies $\mathbf{Set}(X,FB)\cong \mathcal{C}(A,B)$. $\endgroup$ – Jo Be Jun 9 '18 at 17:37
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A way to characterize the existence of free objects in a concrete category $(C,U)$ is to say $U$ is a right adjoint, that is, there exists a functor $F \colon Set \to C$ and natural transformations $\eta \colon id_{Set} \to UF$ and $\epsilon \colon FU \to id_{C}$ such that

  • For every morphism $f \colon F(X) \to Y$ there exists an unique $g \colon X \to U(Y)$ such that the diagram below commutes.
  • For every morphism $g \colon X \to U(Y)$ there exists an unique $f \colon F(X) \to Y$ such that the diagram below commutes.

enter image description here

With this in mind, let's address your questions:

  1. g is a morphism in the domain of $F$, that is, in $C$. Since $F(g) \colon F(A) \to F(B)$, we have $g \colon A \to B$ (in $C$). Functors take objects to objects and morphisms to morphisms. Also the composition $F(g) \circ i$ is well defined (the codomain of $i$ is equal to the domain of $F(g)$).

  2. The definition I gave is more general and complete. For intuition you can think of how a linear transformation is determined by its action on the basis of the domain. Using the diagrams above, $X$ is a set, $Y$ is a vector space, $F(X)$ is the vector space with basis $X$ and $U(Y)$ is the underlying set of $Y$.

  3. In order to have presentations of objects by generators and relations you actually need two properties in your concrete category: The existence of free objects (that is, $U$ being a right adjoint) and for every relation in the underlying set of a free object a canonical choice of a congruence (in the category of groups, for example, you can't quotient an object by an arbitrary relation. In fact, you can't even quotient an object by an arbitrary equivalence relation!). The former is what is being discussed in this question, the latter is a topic for another time.

  4. That is unfortunately not true. For example, as I hinted above, every vector space is a free object in the category of vector spaces. If you consider the base field as $\mathbb{R}$ and a singleton $\{*\}$, the free vector space with basis $\{*\}$ is $\mathbb{R}$ itself (up to isomorphism), which is uncountable! (hence can't be obtained by taking words in the alphabet $\{*\}$). Another example is the category of topological spaces: its free objects are the discrete spaces. It seems pointless to consider the set of words of an alphabet and partition it in a way to be in bijection with the set you had originally (just take the set you had initially and equip it with the discrete topology directly!).

Feel free to comment if the answer is incomplete.

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