## Open tubes

In Chapter 31, “The Compact-Open Topology”, it is asserted that “clearly” the third topology on (with and the real line) is finer than the second. The issue is more fully discussed here.

In this post I discuss the concept of open tubes, particular subsets of the Cartesian product with  a set and a metric space.

## Definition

Let be a set and a metric space, with distance .

For any and any , we note the open ball of of radius and centred on , that is set of all such that . Open balls are notoriously open subsets.

Let be any mapping , and a real .

Then the open tube of radius and centred on is the following subset of :

Equivalently:

The “open” in “open tube” refers to its being based on open balls (second expression above), but does not imply that it is itself necessarily open; indeed, we have not given any topology to , so we don’t have any particular one on and cannot speak of being open or not.

However, if is a topological space and is continuous , then  is open, as I will show.

## An open tube centred on a continuous mapping is open

If is a topological space and is a continuous mapping , then any open tube centred on is an open subset of .

Proof:

I will show that in the above circumstances is a neighbourhood of each of its points.

Let be an eventual point of .

Then .

Lettuce define . We have .

Let . Since is continuous and is an open subset of , is an open subset of . Since , .

Let , open subset of . Clearly, , so is a neighbourhood of .

Let be an eventual point of .

, hence , that is .

, hence .

Since  and , this leads to:

With:

Thus we have: .

Which implies: .

Hence .

Thus we have found a neighbourhood of that is a subset of , which implies that itself is a neighbourhood of .

Thus is an open subset of .

## Uniform convergence topology and the open set topology

Chapter 31, “The Compact-Open Topology”, describes the compact-open topology, and goes on to compare it with two other topologies on in the case where . It asserts that “clearly” the second of these three is finer than the first, and the third finer than the second.

This may well be “clear” intuitively, but is not so easy to prove. Here I will prove that the third (that the author does not name, but that I call the open set topology) is finer than the second, which is the uniform convergence topology.

I will only need to specify that is a topological space and a metric space with distance .

I define the set of continuous mappings from to .

I note the open ball in centred on and with radius . It is the set of all such that .

Furthermore, for any subset of , I note the set of all continuous mappings the graph of which is a subset of ; in other words, .

The uniform convergence topology on is such that a subset of is open if and only if for any , there exists such that for any , if , then . We can express this condition in terms of subsets of that I have named open tubes: for any mapping (not necessarily continuous) and for any , the open tube is . Then is open for the uniform convergence topology if and only if for any , there exists such that .

The third topology ‑ the “open set” topology ‑ is the topology on generated by all the for all open subsets of . (The book says the topology is this collection itself, but I have not managed to prove that it directly forms a topology ‑ as is more explicitly requested in exercise 215.)

To show that the open set topology is finer than the uniform convergence one, we consider an open set for the uniform convergence topology and endeavour to show that it is open in the open set one.

Let be an open set in the uniform convergence topology.

Let be an eventual member of .

There exists an such that .

I prove here that, being continuous, is an open subset of . It includes the graph of . Hence contains , and is by definition an open set for the open set topology on . Since it is a subset of , is a neighbourhood of for the open set topology.

Hence for any , is a neighbourhood of for the open set topology. This implies that is itself an open set in the open set topology.

## Connected subsets one of which meets the closure of the other

Exercise 224 (Chapter 32, “Connectedness”):

Let and be connected subsets of topological space such that intersects . Prove that is connected. Find an example to show that it is not enough to assume that intersects .

Let us assume that and are open subsets of such that is empty and that . To show that is connected, we need to show that either or does not intersect .

Since is connected, and  is empty and , necessarily is entirely in or in . Similarly, is entirely in or entirely in .

Let us suppose that they are not both in or both in ; for instance, that and .

If , then with open in . Since does not intersect , cannot be in . Hence cannot intersect . This contradicts our assumption.

Hence it is impossible that  and . Similarly, it is impossible that and . Thus, and are either both in or both in , which implies that either or does not intersect .

If we only assume that and intersect, it does not always follow that is connected. For instance, let be the topological plane and and two open disks that “almost touch”, that is such that the corresponding closed disks ‑ which are their closures ‑ intersect in one point. The union of and is then not connected.

## 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.

## Theorems about the direct product of two topological spaces

A lot of results concerning the direct product of two topological spaces and are trivial if you consider the usual particular construction of these entities. I find it more satisfying to derive them directly from the universal definitions.

In the following, we will always have and topological spaces and a direct product thereof.

## No duplicates

In the construction, this is just the trivial fact that if you know , you know both and .

Theorem:

For every , if and , then .

Proof:

Let us suppose that we have such that and .

Let us take a singleton topological space, that is that is a singleton (singletons exist!) and we equip it with the only possible topology, which is both the indiscrete and the discrete one.

We define and as the constant mappings, and respectively, that map the only element of to , which is also , and to , which is also , respectively.

We define and as the constant mappings, both , that map the only element of to and to respectively.

Since these four mappings are constant, or, alternatively, since the topology on is discrete, they are all continuous.

Furthermore, for both and we have:

Since is a direct product of and , there can be only one morphism satisfying these two relations. Hence necessarily , which implies .

## Existence of an antecedent

If we have and , does there exist a such that and ? Again, this is trivial in the representation: is the answer. Sticking to the universal definition of the direct product, we have:

For any and , there exists such that and .

Proof:

We take as a singleton topological space as above, and and as the mappings and respectively that map the unique element of to and to respectively. These two mappings again are continuous, since has the discrete topology.

The universal definition of as a direct product implies that there exists such that and . Taking as the image by of the unique element of , we thus have and . .

This result and the preceding one come together as:

For any and , there exists a unique such that and .

## The associated morphisms are surjective

At least, usually.

If or , then is surjective. If or , then is surjective.

This is just a consequence of the general result proved in exercise 7 (corrected): in any category, if is a direct product of and , and if , then is an epimorphism (and similarly, exchanging the roles of and , and and , respectively); plus the fact that in the category of topological spaces, epimorphisms are surjective mappings.

## Characterisation of the topology

That the topology on the the direct product space is the one generated by the collection of all and for open in and open in is a stipulation of the explicit construction of as . It can however be proven without reference to this construction.

The topology of is the topology generated in by the collection of all and for open in and open in . A subset of is open if and only if it is the union of some collection of subsets of the form with  open in and open in .

Proof:

Let be the direct product topology on and the topology on generated in by the collection of all and for open in and open in .

Since and are defined as continuous, necessarily contains all the  and for open in and open in . Hence is finer than . We wish to show that is also finer than .

The mapping , considered from to , remains continuous, since all the for open in are in . Similarly, is continuous . Hence we have a topological space and two continuous mappings and , from to and respectively. Since is a direct product of and , there exists a unique continuous mapping such that and .

This implies that any element has the same image as by both and . The first theorem above implies that . Thus .

Hence is continous ; which means that is finer than .

Hence , that is the topology of the direct product is the topology of generated by the collection of all and for open in and open in .

The open sets of the topology generated by a collection of sets are the unions of any number of finite intersections of elements of .

In this case, . A finite intersection of elements of will be of the form with the ‘s open subsets of and the ‘s open subsets of . Since intersections are well-behaved respective to inverse images, and since finite intersections of open subsets are open subsets, this can be written with open subset of and open subset of . Hence an open subset of is a union of subsets of of the form with open subset of and open subset of

## Continuity of a mapping to a product space

If is a topological space and a mapping , then is continuous if and only if both and are continuous.

If is continuous, since both and are continuous, by composition of continuous mappings we conclude that and are continuous.

Conversely, suppose that we know that  and are continuous. Let be an eventual open subset of . We wish to show that is an open subset of .

Let be an eventual point of . Then . Since is open in and (see above) “a subset of is openif and only if it is the union of some collection of subsets of the form with  open in and open in ”, there exist  open in and open in such that . Taking of this expression, we obtain . But  . Since we know that both and are continuous, this last expression is an open subset of , that contains and is included in . Thus is a neighbourhood of . This being the case for any , is an open subset of .

## 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.

## Theorems about the direct sum of two topological spaces

A lot of results concerning the direct sum of two topological spaces and are trivial if you consider the usual particular construction of these entities. I find it more satisfying to derive them directly from the universal definitions.

In the following, we will always have and topological spaces and a direct sum thereof.

## Non-overlapping images

and are disjoint.

Let and be eventual elements respectively of and .

Let and be any two distinct objects (such exist), and , equipped with any topology (the discrete one will do). Then the constant mappings and are continuous.

Hence there exists a continuous mapping such that and .

This implies in particular that and , that is, and . Since , it follows that .

Thus and cannot have a common point, which would both be the image by of a point of and by of a point of .

## The images cover the direct sum

Let with the topology induced on by that of .

Let be the mapping and the mapping .

Since an open set on is the intersection of with an open set on and since certainly contains no points of , its preimage by is equal to the preimage of by and hence is open in . Thus is continuous. Similarly, is continuous.

We can thus apply the universal property of the direct sum , which implies that there exists a continuous mapping such that and .

Let be the mapping . The preimage by of an open set of is again the same as its preimage by , hence is an open set of . Hence is continuous.

For any , ; hence . Similarly, .

But we also have and , with continuous . The uniqueness requirement of the universal property of the direct sum thus implies that .

But the image of is that of , and is included in . Thus the image of , which is , is included in . It follows that , that is, that .

## Topology on the direct sum

The topology on the direct sum is the topology generated by the collection of the and for all open in and all open in .

//TODO

## 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 .

## Representation frameworks and representations

Chapter 22 on representations is, I feel, the worst written I have come upon up to now. I haven’t yet come to wrap my head around it.

I certainly appreciate the desire to give a general definition of representations in category theory terms. This is how the author puts it:

Let and be two categories. We suppose that we are given the following two things: i) a forgetful functor from the category to the category of sets, and ii) a rule which assigns, to each object in category , an object in category and an isomorphism (in the category of sets) from set to the set .

I see two issues about this setting. I will “correct” these issues in this post, giving what I hope is a better definition, and come to the issue of direct products and sums of representations and of subrepresentations in another post.

## About the functor to the category of sets

About the above complex statement, the author reassures us:

There is an obvious forgetful functor from every category we shall consider to the category of sets, and we shall always use this one for item i). The purpose of i) is to allow us to speak of “elements” of objects in category .

The existence of this “obvious functor” in the categories we consider is very familiar; indeed, groups, vector spaces and so on are built upon sets. We are clearly used to speaking of an element of a group, though in fact we should speak of an element of the underlying set.

Postulating the existence of this forgetful functor from to is, however, not enough to ensure that we are in the “familiar situation”. There is another element of this familiar situation that is not captured by the mere existence of this functor. Namely: in the familiar situation, the morphisms between objects of are no more and no less than morphisms – i.e., mappings – between the corresponding sets. If and are objects of , then is literally a subset of . This implies in particular that is “one to one” in its action on morphisms; there cannot be two different morphisms such that .

That this is not necessarily implied by the mere existence of a functor from to can be made clear by considering this trivial fact: for any category , there is a functor from to that takes any object of to the set and any morphism in between objects and to the one and only mapping that exists from set to set . One can check that this is indeed a functor; but clearly, in general it will take many different morphisms in to just one morphism in .

To be closer to the familiar situation, we thus need (at least) the following additional condition: that the functor from to be “morphism-injective”, that is, such that, for any given objects and of and any , we have .

We will see why this condition is necessary particularly in the case of the direct products and sums of representations. In the book, it is hidden behind the notations of the author, that tend to muddle the distinction between morphisms in and the corresponding morphisms in .

## Must it really be an isomorphism?

The author postulates that there must be “ii) a rule which assigns, to each object in category , an object in category and an isomorphism (in the category of sets) from set to the set “.

It makes for a simpler formulation if instead of speaking of an isomorphism, we say that this rule must assign to the object of an object of such that .

For instance: if is the category of real vector spaces, and that of real associative algebras, we want our rule to “make the set of endomorphisms of a given real vector space into an associative algebra”, that is, is an associative algebra the underlying set of which is .

But this doesn’t work well if, instead, we want to be the category of groups. The author calls this a “special case of the definition of a representaton”, but doesn’t clearly note that in this case, we much change his definition, replacing “isomorphism” by “monomorphism”. In the case of groups, it is not the whole of that is “made into a group”, but only the set of bijective linear mappings from to itself.

So we will adopt the modified specification of ii): We need a rule which assigns, to each object in category , an object such that .

## My definition of a representation framework

A representation framework is made of:

• Two categories and .
• A morphism-injective functor from to .
• A rule that assigns to each object  in category , an object such that .

## A representation is…

Given such a representation framework, a representation of an object of is an ordered pair where is an object of and a -morphism .

This is very close to the definition given by the author. He then goes on to state:

(…) for each element of object , we must have a certain morphism, which we write , from object to itself.

This “” is misleading, in that it gives the impression that is the image of by . But  is a morphism between and . It is not (in general) a mapping at all. It is not , but that is a mapping and can map to this , element of .

Despite the awkwardness, I will often prefer the notation “” to the shorter ““, because the latter can give rise to the notion that if we happen to have, for each , a certain -morphism that we call , for instance, then necessarily this mapping forms a -morphism ; while it only means that we have a mapping of sets ‑ if indeed the ‘s are all in , which may be only a subset of ‑ and this mapping may or may not be some with a -morphism from to . This false reasoning arises precisely in the presentations by the author of direct products of representations, direct sums and subrepresentations. That will be for another post.