Epimorphisms of groups

In my previous post, we saw that injections (surjections) in concrete categories are always monomorphisms (epimorphisms), and in some cases the converse holds.

We now wish to classify all epimorphisms of groups. To show that all epimorphisms are surjective, for any strict subgroup H \subseteq G we want to construct maps f_1, f_2 \colon G \to G' to some group G' that differ on G but agree on H. In the case of abelian groups this is relatively easy, because we can take G' to be the cokernel, f_1 the quotient map, and f_2 the zero map. But in general the cokernel only exists if the image is normal, so a different argument is needed.

Lemma. Let f \colon H \to G be a group homomorphism. Then f is an epimorphism if and only if f is surjective.

Proof. We already saw that surjections are epimorphisms. Conversely, let f \colon H \to G be an epimorphism of groups. We may replace H by its image in G, since the map \im(f) \to G is still an epimorphism. Let X = G/H be the coset space, viewed as a pointed set with distinguished element * = H. Let Y = X \amalg_{X\setminus *} X be the set “X with the distinguished point doubled”, and write *_1 and *_2 for these distinguished points.

Let S(Y) be the symmetric group on Y, and define homomorphisms f_i \colon G \to S(Y) by letting G act naturally on the i^{\text{th}} copy of X in Y (for i \in \{1,2\}). Since the action of H on X = G/H fixes the trivial coset *, we see that the maps f_i|_H agree. Since f is an epimorphism, this forces f_1 = f_2. But then

    \[H = \Stab_{f_1}(*_1) = \Stab_{f_2}(*_1) = G,\]

showing that f is surjective (and a fortiori X = \{*\}). \qedsymbol

Note however that the result is not true in every algebraic category. For example, the map \mathbf Z \to \mathbf Q is an epimorphism of (commutative) rings that is not surjective. More generally, every localisation R \to R[S^{-1}] is an epimorphism, by the universal property of localisation; these maps are rarely surjective.

Rings are abelian

In this post, we prove the following well-known lemma:

Lemma. Let (R,+,\times,0,1) satisfy all axioms of a ring, except possibly the commutativity a + b = b + a. Then (R,+) is abelian.

That is, additive commutativity of a ring is implied by the other axioms.

Proof. By distributivity, we have 2(a+b) = 2a + 2b, so multiplication by 2 is a homomorphism. By our previous post, this implies R is abelian. \qedsymbol

Criteria for groups to be abelian

This is a review of some elementary criteria for a group to be abelian.

Lemma. Let G be a group. Then the following are equivalent:

  1. G is abelian,
  2. The map G \to G given by g \mapsto g^2 is a group homomorphism;
  3. The map G \to G given by g \mapsto g^{-1} is a group homomorphism;
  4. The diagonal G \subseteq G \times G is normal.

Proof. We prove that each criterion is equivalent to (1).

For (2), note that (gh)^2 = ghgh, which equals gghh if and only if gh = hg.

For (3), note that (gh)^{-1} = h^{-1}g^{-1}, which equals g^{-1}h^{-1} if and only if gh = hg.

For (4), clearly \Delta_G \colon G \hookrightarrow G \times G is normal if G is abelian. Conversely, note that (e,h)(g,g)(e,h^{-1}) = (g,hgh^{-1}), which is in the diagonal if and only if gh = hg. \qedsymbol

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).

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.