## Union of transitively overlapping connected spaces

Exercise 225 (Chapter 32, “Connectedness”):

Prove the following generalization of theorem 40. Let ( in ) be a collection of connected subsets of topological space X, and let the equivalence relation on this collection of sets generated by if have just one equivalence class. Then is connected.

For clarity, rather than a family I will consider a set of connected subsets of . The condition is that the equivalence relation in generated by the above relation has only one class.

Let .

Let and be eventual open subsets of such that and . To show that is connected, we must show that either or is empty.

First, let be an eventual element of . Since , the two subsets of , and are disjoint, and their union is . Furthermore, they are open in . Since is connected, this implies that one or the other is empty. If is empty, then ; if instead is empty, then . Hence in all cases an element of is entirely in or entirely in .

Let us consider the relation on defined by ; in other words, if and only if both are subsets of , or both are subsets of . Clearly, this is an equivalence relation.

Furthermore, if , that is, , we cannot have and , for and are disjoint. Hence and are either both in or both in ; that is, . Hence is an equivalence relation greater than the equivalence relation generated by .

If had two or more classes, there would be some element of in one class and another in another; these two elements would not be equivalent following . Since is greater than the equivalence relation generated by , they would not be equivalent either following this latter relation; which contradicts the fact that this has only one class. Hence too can have only one class, which means that for all and in , either both are in or both are in . But this implies that all are in or all are in , which in turn implies that their union is a subset of either or , hence that either or is empty.

## Direct products/sums: when the associated morphisms are epi/mono

Exercise 7 (Chapter 2, “Categories”):

In the category of sets, the two morphisms ( and ) in a direct product are monomorphisms and the two morphisms in a direct sum are epimorphisms. Is this true in every category?

There are two mistakes in this wording.

• The author has exchanged “monomorphisms” and “epimorphisms”. In a direct product of sets, the morphisms are generally not monomorphisms (injective), and in a direct sum of sets, they are generally not epimorphisms (surjective).
• In a direct product of sets, the morphisms are almost always epimorphisms. If one of the sets is empty, the product is empty, and the morphism associated with the other set (an empty mapping) is not an epimorphism; unless this other set is empty too.

The general theorem I will prove is this:

Let and be objects of a category , and a -direct product of and . If is not empty then is an epimorphism.

We consider the above conditions satisfied. There exists a morphism .

Let be the identity morphism and the morphism . Applying the universal property of the direct product  to the triplet , we obtain that there exists a unique such that both and .

The first of these two relations says that is the identity on . If we have an object and two morphisms and from to such that , then, composing this relation on the right with and eliminating the resulting identity morphisms, we get . Hence is an epimorphism.

We obtain, of course, a similar result exchanging the roles of and .

Reversing the arrows, we get the corresponding result for direct sums:

Let and be objects of a category , and a -direct sum of and . If is not empty then is a monomorphism.

Plus, of course, the similar result obtained by exchanging and .

It so happens that in the category of sets, in the case of a direct product, both and are always monomorphisms, even if , respectively , are empty.

## Direct sum of Hausdorff topological spaces

Exercise 185 (Chapter 28, “The category of topological spaces”):

Prove that both the direct product and the direct sum of two Hausdorff topological spaces is Hausdorff.

We will consider Hausdorff topological spaces and and their direct sum (in the first part) or product (in the second part) .

We will proceed without reference to the explicit constructions of direct topological sums and products, using instead the results in “Theorems about the direct product and sum of two topological spaces”.

## Direct sum

Let and be two eventual points of , with . The first result about direct sums in the above page implies that . The second result implies that . Two cases are possible:

• One of and is in and the other in .
• Both  and are in , or both are in .

We will consider successively the case where and , and the case where both are in . The other possible cases are similar and will lead to similar results.

First case ( and ):

We have and , with and .

itself is an open neighbourhood of , and hence (last result about direct sums on the above page) is an open neighbourhood of in . Similarly, is an open neighbourhood of in . Since  and  are disjoint, we have found disjoint neighbourhoods respectively of and .

Second case ():

We have and , with . Since , we have . Since is Hausdorff, there exist and disjoint open neighbourhoods in respectively of and .

//TODO (we need injective)

## Direct product

Let and be two eventual points of , with .

The first result about direct products in the above page implies that   or .

Let us suppose that .

Since is Hausdorff and and are distinct points of , there exist open sets and of such that , and .

Then and  are disjoint (the image by of a common element would be both in and , which are disjoint). They are open (since and are open and is continuous). Since , we have ; similarly, . Thus there exist two disjoint open neighbourhoods respectively of and in .

The same would result if we had .

This being the case for all with , the topological space is Hausdorff.

## Universal definition of the indiscrete topology

Exercise 182 (Chapter 29, “The category of topological spaces”):

Give a universal definition which leads to the introduction of the indiscrete topology on a set.

On a set , the indiscrete topology is the unique topology such that for any set , any mapping is continuous.

This isn’t really a universal definition. We have seen that the discrete topology can be defined as the unique topology that makes a free topological space on the set . We will give the indiscrete topology, which is at the other end of the spectrum of topologies, a similar definition.

For any category , we can define a “reversed arrows” category , which has the same objects as and the same morphisms, except that the sources and targets of the morphisms are reversed. Composition too is, obviously, reversed; the identities are the same. It is easily checked that this does lead to a proper category, and also that the arrows-reversed category of an arrows-reversed category leads back to the original category.

If we have categories and and a covariant functor from to , then we can consider the three modified functors from to , from to and from to . These functors will map both objects and morphisms exactly as the original does. Those with only one reversal are contravariant; the one with a reversal in both categories is covariant, and is the one that will interest us here.

We consider the classical category of sets ; in its arrows-reversed version, , the morphisms from set to set are the mappings from to . Similarly, we have the category of topological spaces and its arrows-reversed version in which morphisms from topological space to topological space are the continuous mappings from to.

For a given set , object of , we can ask if we can find a free object in category following the forgetful functor from to . This mens finding a topological space and a -morphism such that for any topological space and any -morphism , there exists a unique -morphism such that (as -morphisms) .

Factoring in the arrows-reversals, this means that we wish to find a topological space and a mapping such that for any topological space and any mapping , there exists a unique continuous mapping such that (as mappings) .

We propose that the topological space with the indiscrete topology on , together with the mapping , is a free object on following the functor .

Proof:

Let be a topological space and a mapping . If is to be a continuous mapping such that , it must be a mapping such that ; that is, we must have . This is indeed continuous, since is the indiscrete topology. Hence it is the unique continuous mapping such that .

Thus we have shown that with the indiscrete topology, together with the identity mapping on , is a free topological space following the arrows-reverse categorical definition of free objects.

Free objects are unique up to an isomorphism. However, we wish to show more specifically that given the set , the indiscrete topology on is the unique topology such that be a free object on following the above definition. We could use the “unique up to an isomorphism” card, but more simply: If is to be a free object on following the above definition, in particular, taking and , we need there to exist a continuous , that is, such that , that is such that . In other words, we need to be continuous from to , which implies , that is, that must be coarser than ; which is possible only if .

Hence the indiscrete topology on is the unique topology such that be a free object in category following the functor from category to category .

## Identity functors, isofunctors and equivalent categories

Exercise 105 (Chapter 17, “Functors”):

Define the identity functor from a category to that same category. What do you suppose is meant by equivalent categories?

## Identity functors

We have encountered two flavors of identity definitions. Identity mappings on sets are defined by the way they act on set elements, namely that they don’t change them. Identities in abstract categories, on the other hand, are defined by the way they behave respective to the composition of morphisms.

Categories are a bit like sets, in that they have “elements”, or rather two kinds of such: objects and morphisms. Inspired by this, we can define identity functors in a category as the functor from itself to itself that maps each object to itself and each morphism to itself. Such a functor always exists, and, since its effect is completely specified, it is unique.

But since we also have composition of functors, we can also try to define the identity functor on category as a functor, if it exists, from to such that for any category , for any functor from to , and for any functor from to , .

It is immediate that the functor from the first definition satisfies the second. Since the former always exists, so does the latter. Furthermore, if there were two identity functors following the second definition, and , we would have both and , hence .

Thus the two definitions are equivalent; one and only one identity functor exists for each category.

## Isofunctors and equivalent categories

The same exercise asks us to define the notion of equivalent categories.

We may first want to define an isofunctor between two categories:

A functor from category to category is an isofunctor if and only if there exists a functor from to such that is the identity functor on and is the identity functor on .

We can then, of course, define equivalent categories as categories between which there exists at least one isofunctor.

## Free objects following an isofunctor

If and are equivalent categories and is an isofunctor from to with its inverse isofunctor from to , then we have:

If is an object of and , then (with the identity morphism on ) is a free object on following the functor .

The proof is easy. If is any object of and a -morphism , then for a -morphism to be such that , necessarily , that is (applying to each side) . We can check that this indeed satisfies . Hence there is one and only one -morphism such that . This being the case for any object  of and any -morphism , the pair is a free  object on following .

The “free object functor” in this case thus simply maps objects of to the same as does. It is easy to check that the effect of this free object functor on morphisms of is too just the same as that of . Thus, the free object functor in the case of equivalent categories following an isomorphism between them is just the inverse of that isomorphism.