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 f(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).

Cauchy’s theorem

This is a really classical result, but this proof is just so magical that I have to include it.

Lemma. Let G be a finite group, and let p be a prime dividing \# G. Then G has an element of order p.

Proof. Consider the action of \Z/p\Z = \langle \sigma \rangle on

    \[ X = \{(g_1, \ldots, g_p) \in G^p\ |\ g_1 \cdots g_p = 1\} \]

given by

    \[ \sigma \cdot (g_1, \ldots, g_p) = (g_2, \ldots, g_p, g_1). \]

Then \# (X^{\Z/p\Z}) \equiv \# X \pmod{p}, since the orbits of size > 1 all have size p (and the fixed points X^{\Z/p\Z} are exactly the union of the orbits of size 1). The right hand side is 0 \pmod{p} as p \mid \# G. Thus,

    \[ \# (X^{\Z/p\Z}) \equiv 0 \pmod{p}. \]

But there is a bijection

    \begin{align*} \{g \in G\ |\ g^p = 1\} &\rA X^{\Z/p\Z}\\ g &\rM (g, \ldots, g). \end{align*}

The former trivially contains the element 1, hence (since its size is divisible by p) it has to contain some other element g. This element will then have (exact) order p. \qedsymbol

Remark. Of course, the result also follows from the much stronger Sylow theorems (of which we only need the existence statement).

Separation properties for topological groups

Although this is quite a classical result, I really like it.

Lemma. Let G be a topological group. Then G is T_1 if and only if G is Hausdorff.

Proof. One implication is clear. Conversely, suppose G is T_1. Then the identity element is closed. The map

    \begin{align*} G\times G &\rA G\\ (g,h) &\rM gh^{-1} \end{align*}

is continuous. Hence, the inverse image of the identity is closed. But this is the diagonal, hence G is Hausdorff. \qedsymbol

Exercise. Prove that Hausdorff is in fact equivalent to T_0.