A fun example of a representable functor

This post is about representable functors:

Definition. Let $F \colon \mathscr C \to \Set$ be a functor. Then $F$ is representable if it is isomorphic to $\Hom(A,-)$ for some $A \in \ob \mathscr C$ . In this case, we say that $A$ represents $F$ .

Exercise. If such $A$ exists, then it is unique up to unique isomorphism.

Really one should encode the isomorphism $\Hom(A,-) \stackrel\sim\to F$ as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation $\Hom(A,-) \to F$ is uniquely determined by the element of $F(A)$ corresponding to the identity of $A$ .

When $\Hom(A,-) \to F$ is a natural isomorphism, the corresponding element $a \in F(A)$ is called the universal object of $F$ . It has the property that for every $B \in \mathscr C$ and any $b \in F(B)$ , there exists a unique morphism $f \colon A \to B$ such that $(Ff)(a) = b$ .

Example. The forgetful functor $\Ab \to \Set$ is represented by $\Z$ . Indeed, the natural map

$\begin{align*} \Hom(\Z,M) &\to M\\ f &\mapsto f(1) \end{align*}$

is an isomorphism. The universal element is $1 \in \Z$ .

Example. Similarly, the forgetful functor $\Ring \to \Set$ is represented by $\Z[x]$ . The universal element is $x$ .

A fun exercise (for the rest of your life!) is to see whether functors you encounter in your work are representable. See for example this post about some more geometric examples.

The main example for today is the following:

Lemma. The functor $\Top\op \to \Set$ that associates to a topological space $(X,\mathcal T_X)$ its topology $\mathcal T_X$ is representable.

Proof. Consider the topological space $Y = \{0,1\}$ with topology $\{\varnothing, \{1\},\{0,1\}\}$ . Then there is a natural map

$\begin{align*} \Hom(X,Y) &\to \mathcal T_X\\ f &\mapsto f^{-1}(\{1\}). \end{align*}$

Conversely, given an open set $U$ , we can associate the characteristic function $\mathbb I_U$ . This gives an inverse of the map above. $\qedsymbol$

The space $Y$ we constructed is called the Sierpiński space. The universal open set is $\{1\}$ .

Remark. The space $Y^I$ represents the data of open sets $U_i$ for $i \in I$ : for any continuous map $f \colon X \to Y^I$ , we have $U_i = f^{-1}(Y_i)$ , where $Y_i = \pi_i^{-1}(\{1\}) \subseteq Y^I$ . If $Z_i$ denotes the complementary open, then the $U_i$ form a cover of $X$ if and only if $\bigcap_{i \in I} Z_i = \varnothing$ . This corresponds to the statement that $f$ lands in $Y^I\setminus\{(0,0,\ldots)\}$ .

Thus, the open cover $Y^I\setminus\{0\} = \bigcup_{i \in I} Y_i$ is the universal open cover, i.e. for every open covering $X = \bigcup U_i$ there exists a unique continuous map $f \colon X \to Y^I\setminus\{0\}$ such that $U_i = f^{-1}(Y_i)$ .

2 thoughts on “A fun example of a representable functor”

Martin on 8 May 2019 at 01:40 said:

Great post, thanks for sharing. I think in your definition of a universal object, it should be:

… It has the property that for every $B \in \mathscr C$ and any $b \in F(B)$ , there exists a unique morphism $f \colon A \to B$ such that $(F f)(a) = b$ .

Reply ↓
- Remy on 8 May 2019 at 14:00 said:
  
  Thanks, you’re absolutely right! Fixed.
  
  Reply ↓

Lovely little lemmas

Or maybe ‘amusing observations’

A fun example of a representable functor

2 thoughts on “A fun example of a representable functor”

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