# Finite topological spaces

One of my favourite bits of point set topology is messing around with easy topological spaces. What could be easier than finite topological spaces? The main result (below) is that the category of finite topological spaces is equivalent to the category of finite preorders.

Recall (e.g. from algebraic geometry) the following definition:

Definition. Let be a topological space. Then the specialisation preorder on (the underlying set of) is the relation if and only if .

Note that it is indeed a preorder: clearly , and if and , then , so , showing . We denote this preorder by .

Note that the relation is usually denoted in algebraic geometry, which is pronounced “ specialises to ”.

Definition. Given a preorder , the Alexandroff topology on is the topology whose opens are the cosieves, i.e. the upwards closed sets (meaning and implies ).

To see that this defines a topology, note that an arbitrary (possibly empty) union or intersection of cosieves is a cosieve. A subbase for the topology is given by the principal cosieves for any . We denote the set with its Alexandroff topology by .

Likewise, the closed sets in are the sieves (or downwards closed sets); for instance the principal sieves . The closure of is the sieve generated by ; for instance the closure of a singleton is the principal sieve .

Theorem. Let be the functor , and the functor .

1. Let be a preorder, a topological space, and a function. Then is a monotone function if and only if is a continuous function .
2. The functors and are adjoint: .
3. The composition is equal (not just isomorphic!) to the identity functor.
4. The restriction of to the category of finite topological spaces is equal to the identity functor.
5. If is a topological space, then is if and only if is a poset.
6. If is a preorder, then is a poset if and only if is .
7. The functors and give rise to adjoint equivalences

Proof. (1) Suppose is monotone, let be a closed subset, and let . Suppose and . Since is monotone and is closed, we get , i.e. . We conclude that , so is downward closed, hence closed in .

Conversely, suppose is continuous, and suppose in . Then , so by continuity we get , so .

(2) This is a restatement of (1): the map

is a bijection.

(3) Since , we conclude that if and only if , so the specialisation preorder on is the original preorder on .

(4) In general, the counit is a continuous map on the same underlying space, so is finer than . Conversely, suppose is closed, i.e. is a sieve for the specialisation preorder on . This means that if , then implies ; in other words . If and therefore is finite, there are finitely many such , so is the finite union

of closed subsets of . Thus any closed subset of is closed in , so the topologies agree.

(5) The relations and mean and . This is equivalent to the statement that a closed subset contains if and only if it contains . The result follows since a poset is a preorder where the first statement only happens if , and a space is a space where the second statement only happens if .

(6) Follows from (5) applied to since by (3).

(7) The equivalence follows from (3) and (4), and the equivalence then follows from (5) and (6).

Example. The Alexandroff topology on the poset is the Sierpiński space with topology . As explained in this post, continuous maps from a topological space to are in bijection with open subsets , where is sent to (and to the indicator function ).

Example. Let be a set with two elements. There are 4 possible topologies on , sitting in the following diagram (where vertical arrows indicate inclusion bottom to top):

These correspond to 4 possible preorder relations , sitting in the following diagram (where vertical arrows indicate inclusion top to bottom):

We see that finer topologies (more opens) have stronger relations (fewer inequalities).

Example. The statement in (4) is false for infinite topological spaces. For instance, if is the Zariski topology on a curve, then any set of closed points is downwards closed, but it is only closed if it’s finite. Or if is a Hausdorff space, then the specialisation preorder is just the equality relation , whose Alexandroff topology is the discrete topology.

I find the examples useful for remembering which way the adjunction goes: topological spaces generally have fewer opens than Alexandroff topologies on posets, so the continuous map should go .

Remark. On any topological space , we can define the naive constructible topology as the topology with a base given by locally closed sets for open and closed. In the Alexandroff topology, a base for this topology is given by the locally closed sets : indeed these sets are clearly naive constructible, and any set of the form for upward closed and downward closed has the property .

Thus, if is the Alexandroff topology on a preorder, we see that the naive constructible topology is discrete if and only if the preorder is a poset, i.e. if and only if is .

# Application of Schur orthogonality

Lemma. Let be a finite group of order , and write for the set of irreducible characters of . Then

1.

2.

Proof. First consider the case . This is just an example; it could also be something much better. Then the second statement is obvious, and the first is left as an exercise to the reader. The general case is similar.

Here is a trivial consequence:

Corollary. Let be a positive integer, and let . Then

Proof 1. Without loss of generality, has exact order . Set , let , and note that

Part 1 of the lemma gives the result.

Proof 2. Set as before, let be the homomorphism , and the homomorphism . Then part 1 of the lemma does not give the result, but part 2 does.

In fact, the corollary also implies the lemma, because both are true ().

# Graph colourings and Hedetniemi’s conjecture II: universal colouring

In my previous post, I stated the recently disproved Hedetniemi’s conjecture on colourings of product graphs (see this post for my conventions on graphs). In the next few posts, I will explain some of the ideas of the proof from an algebraic geometer’s perspective.

Lemma. Let be a graph. Then there exists an -colouring on such that for every graph and every -colouring on , there is a unique morphism such that .

Proof. By this post, we have the adjunction

(1)

In particular, the identity gives an -colouring under this adjunction. If is any other graph, (1) gives a bijection between morphisms and -colourings of , which by naturality of (1) is given by .

Corollary. To prove Hedetniemi’s conjecture, it suffices to treat the ‘universal’ case , for every and every loopless graph .

Proof. Suppose by contradiction that there is a counterexample , i.e. there are loopless graphs and such that

(2)

Then there exists an -colouring , so the lemma gives a map such that . This forces since an -colouring on induces an -colouring on by pullback. Thus, (2) implies

showing that is a counterexample as well.

Corollary. Hedetniemi’s conjecture is equivalent to the statement that for any loopless graph and any , either or admits an -colouring.

Example. By the final example of my previous post and the proof of the first corollary above, the cases are trivially true. We can also check this by hand:

• If does not have a -colouring, then it has an edge. Then has no edges by construction, since has no edges. See also Example 2 of this post.
• If does not have a -colouring, then it has an odd cycle . We need to produce a -colouring on . Choose identifications and with adjacencies . Consider the map

To show this is a graph homomorphism, we must show that for adjacent we have . If two maps are adjacent, then for adjacent we have . Taking shows that , so

since is odd.

The case is treated in [EZS85], which seems to be one of the first places where the internal Hom of graphs appears (in the specific setting of ).

References.

[EZS85] M. El-Zahar and N. Sauer, The chromatic number of the product of two 4-chromatic graphs is 4. Combinatorica 5.2, p. 121–126 (1985).

# Internal Hom in the category of graphs

In this earlier post, I described what products in the category of graphs look like. In my previous post, I gave some basic examples of internal Hom. Today we will combine these and describe the internal Hom in the category of graphs.

Definition. Let and be graphs. Then the graph has vertices , and an edge from to if and only if implies (where we allow as usual).

Lemma. If , , and are graphs, then there is a natural isomorphism

In other words, is the internal Hom in the symmetric monoidal category .

Proof. There is a bijection

So it suffices to show that is a graph homomorphism if and only if is. The condition that is a graph homomorphism means that for any , the functions have the property that implies . This is equivalent to for all and all . By the construction of the product graph , this is exactly the condition that is a graph homomorphism.

Because the symmetric monoidal structure on is given by the categorical product, it is customary to refer to the internal as the exponential graph .

Example 1. Let be the discrete graph on a set . Then is the complete graph with loops on the set . Indeed, the condition for two functions to be adjacent is vacuous since has no edges.

In particular, any function is a graph homomorphism. Under the adjunction above, this corresponds to the fact that any function is a graph homomorphism, since is a discrete graph.

Example 2. Conversely, is discrete as soon as has an edge, and complete with loops otherwise. Indeed, the condition

can only be satisfied if , and in that case is true for all and .

In particular, a function is a graph homomorphism if and only if either or has no edges. Under the adjunction above, this corresponds to the fact that a function is a graph homomorphism to if and only if has no edges, which means either or has no edges.

Example 3. Let be the discrete graph on a set with loops at every point. Then is the -fold power of . Indeed, the condition that two functions are adjacent is that for all , which means exactly that for each of the projections .

In particular, graph homomorphisms correspond to giving graph homomorphisms . Under the adjunction above, this corresponds to the fact that a graph homomorphism is the same thing as graph homomorphisms , since is the -fold disjoint union of .

Example 4. Let be the terminal graph consisting of a single point with a loop (note that we used instead for in this earlier post). The observation above that also works the other way around: . Then the adjunction gives

This is actually true in any symmetric monoidal category with internal hom and identity object . We conclude that a function is a graph homomorphism if and only if has a loop at . This is also immediately seen from the definition: has a loop at if and only if implies .

Example 5. Let and be the complete graphs on and vertices respectively. Then has as vertices all -tuples , and an edge from to if and only if when . For example, for we get an edge between and if and only if and .

# Internal Hom

This is an introductory post about some easy examples of internal Hom.

Definition. Let be a symmetric monoidal category, i.e. a category with a functor that is associative, unital, and commutative up to natural isomorphism. Then an internal Hom in is a functor

such that is a left adjoint to for any , i.e. there are functorial isomorphisms

Remark. In the easiest examples, we typically think of as ‘upgrading to an object of ‘:

Example. Let be a commutative ring, and let be the category of -modules, with the tensor product. Then with its natural -module structure is an internal Hom, by the usual tensor-Hom adjunction:

The same is true when is the category of -bimodules for a not necessarily commutative ring .

However, we cannot do this for left -modules over a noncommutative ring, because there is no natural -module structure on for left -modules and . In general, the tensor product takes an -bimodule and a -bimodule and produces an -bimodule . Taking gives a way to tensor a right -module with a left -module, but there is no standard way to tensor two left -modules, let alone equip it with the structure of a left -module.

Example. Let . Then is naturally a set, making it into an internal Hom for :

When is the categorical product , the internal (if it exists) is usually called an exponential object, in analogy with the case above.

Example. Another example of exponential objects is from topology. Let be the category of locally compact Hausdorff topological spaces. Then the compact-open topology makes into an internal Hom of topological spaces. (There are mild generalisations of this beyond the compact Hausdorff case, but for an arbitrary topological space the functor does not preserve colimits and hence cannot admit a right adjoint.)

Example. An example of a slightly different nature is chain complexes: let be a commutative ring, and let be the category of cochain complexes

of -modules (meaning each is an -module, and the are -linear maps satisfying ). Homomorphisms are commutative diagrams

and the tensor product is given by the direct sum totalisation of the double complex of componentwise tensor products.

There isn’t a natural way to ‘endow with the structure of a chain complex’, but there is an internal Hom given by

with differentials given by

Then we get for example

since a morphism is given by an element such that , i.e. , meaning that is a morphism of cochain complexes.

Example. The final example for today is presheaves and sheaves. If is a topological space, then the category of abelian sheaves on has an internal Hom given by

with the obvious transition maps for inclusions of open sets. This is usually called the sheaf Hom. A similar statement holds for presheaves.

# Limits in the category of graphs

This is a first post about some categorical properties of graphs (there might be a few more).

Definition. For us, a graph is a pair where is a set and is a collection of subsets of of size or . An element with is called an edge from to , and a singleton is a loop at (or sometimes an edge from to itself). If , it is customary to write and .

A morphism of graphs is a map such that for all . The category of graphs will be denoted , and will be called the forgetful functor.

Example. The complete graph on vertices is the graph where and is the set of -element subsets of . In other words, there is an edge from to if and only if .

Then a morphism is exactly an -colouring of : the condition for forces whenever and are adjacent. Conversely, a morphism to a graph without loops is exactly an -clique in : the condition that has no loops forces for .

Lemma. The category has and the forgetful functor preserves all small limits.

Proof. Let be a functor from a small category , and let be the limit of the underlying sets, with cone maps . We will equip with a graph structure such that the maps for are morphisms and then show that the constructed is a limit of in .

To equip with an edge set , simply let be the set of of size or such that for all . Then this clearly makes into a graph such that the are graph morphisms for all . Moreover, these maps make into the limit cone over : for any other cone , the underlying maps factor uniquely through by the definition of , and the construction of shows that is actually a morphism of graphs .

Remark. Note however that does not create limits. On top of the construction above, this would mean that there is a unique graph structure on such that is a cone over . However, there are many such structures on , because we can remove edges all we want (on the same vertex set ).

Example. As an example, we explicitly describe the product of two graphs and : by the lemma its vertex set is . The ‘largest graph structure’ such that both projections and are graph morphisms is given by if and only if and and . This corresponds to the structure found in the proof of the lemma.

For a very concrete example, note that the product of two intervals/edges is a disjoint union of two intervals, corresponding to the diagonals in . This is the local model to keep in mind.

The literature also contains other types of product graphs, which all have the underlying set . Some authors use the notation for the categorical product or tensor product we described. The Cartesian product is defined by , so that the product of two intervals is a box. The strong product is the union of the two, so that the product of two intervals is a box with diagonals. There are numerous other notions of products of graphs.

Remark. Analogously, we can also show that has and preserves all small colimits: just equip the set-theoretic colimit with the edges coming from one of the graphs in the diagram.

Example. For a concrete example of a colimit, let’s carry out an edge contraction. Let be a graph, and let be an edge. The only way to contract in our category is to create a loop: let be the one-point graph without edges, and let be the maps sending to and respectively. Then the coequaliser of the parallel pair is the graph whose vertices are , where is the equivalence relation if and only if or , and whose edges are exactly the images of edges in . In particular, the edge gives a loop at the image .

Remark. Note that the preservation of limits also follows since has a left adjoint: to a set we can associate the discrete graph with vertex set and no edges. Then a morphism to any graph is just a set map .

Similarly, the complete graph with loops gives a right adjoint to , showing that all colimits that exist in must be preserved by . However, these considerations do not actually tell us which limits or colimits exist.

# 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 we want to construct maps to some group that differ on but agree on . In the case of abelian groups this is relatively easy, because we can take to be the cokernel, the quotient map, and the zero map. But in general the cokernel only exists if the image is normal, so a different argument is needed.

Lemma. Let be a group homomorphism. Then is an epimorphism if and only if is surjective.

Proof. We already saw that surjections are epimorphisms. Conversely, let be an epimorphism of groups. We may replace by its image in , since the map is still an epimorphism. Let be the coset space, viewed as a pointed set with distinguished element . Let be the set “ with the distinguished point doubled”, and write and for these distinguished points.

Let be the symmetric group on , and define homomorphisms by letting act naturally on the copy of in (for ). Since the action of on fixes the trivial coset , we see that the maps agree. Since is an epimorphism, this forces . But then

showing that is surjective (and a fortiori ).

Note however that the result is not true in every algebraic category. For example, the map is an epimorphism of (commutative) rings that is not surjective. More generally, every localisation is an epimorphism, by the universal property of localisation; these maps are rarely surjective.

# Concrete categories and monomorphisms

This post serves to collect some background on concrete categories for my next post.

Concrete categories are categories in which objects have an underlying set:

Definition. A concrete category is a pair of a category with a faithful functor . In cases where is understood, we will simply say is a concrete category.

Example. The categories of groups, of topological spaces, of rings, and of -modules are concrete in an obvious way. The category of sheaves on a site with enough points is concrete by mapping a sheaf to the disjoint union of its stalks (the same holds for any Grothendieck topos, but a different argument is needed). Similarly, the category of schemes can be concretised by sending to , where is the contravariant power set functor.

Today we will study the relationship between monomorphisms and injections in :

Lemma. Let be a concrete category, and let be a morphism in . If is a monomorphism (resp. epimorphism), then so is .

Proof. A morphism in is a monomorphism if and only if the induced map is injective. Faithfulness implies that the vertical maps in the commutative diagram

are injective, hence if the bottom map is injective so is the top. The statement about epimorphisms follows dually.

For example, this says that any injection of groups is a monomorphism, and any surjection of rings is an epimorphism, since the monomorphisms (epimorphisms) in are exactly the injections (surjections).

In some concrete categories, these are the only monomorphisms and epimorphisms. For example:

Lemma. Let be a concrete category such that the forgetful functor admits a left (right) adjoint. Then every monomorphism (epimorphism) in is injective (surjective).

Proof. If is a right adjoint, it preserves limits. But is a monomorphism if and only if the square

is a pullback. Thus, preserves monomorphisms if it preserves limits. The statement about epimorphisms is dual.

For example, the forgetful functors on algebraic categories like , , and have left adjoints (a free functor), so all monomorphisms are injective.

The forgetful functor has adjoints on both sides: the left adjoint is given by the discrete topology, and the right adjoint by the indiscrete topology. Thus, monomorphisms and epimorphisms in are exactly injections and surjections, respectively.

On the other hand, in the category of Hausdorff topological spaces, the inclusion is an epimorphism that is not surjective. Indeed, a map to a Hausdorff space is determined by its values on .