## 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.

## One universal enveloping algebra

In my search for less abstract or trivial examples of a universal enveloping algebra, I have come upon this one.

Let be a vector space and a free associative algebra on . The mapping is a vector space morphism .

can also be viewed as a Lie algebra. is a part of ; let be the sub-Lie algebra of generated by .

(It appears that, except in trivial cases, is strictly smaller than ; for instance, if we take as the free associative algebra on constructed in chapter 18, , in it appears that all the components are antisymmetric.)

We have a natural Lie algebra morphism . That is, is the canonical injection from to .

Proposition: The associative algebra is the universal enveloping algebra of the Lie algebra .

Proof:

Let us first note that since is the Lie subalgebra of generated by , we have . Let us define , that is, is with its codomain restricted to . We can then write .

We must prove that given any associative algebra and any Lie algebra morphism , there exists one and only one associative algebra morphism such that .

So let be an associative algebra and a Lie algebra morphism ; that is, a linear mapping such that .

Then we have vector space morphism . Since is a free associative algebra on the vector space , there exists a unique associative algebra morphism such that , that is, .

Thus and agree on all elements of , which is also , which generates the sub Lie algebra of . We wish to show that they actually agree on all elements of .

Let be the set of all elements of on which and agree. Since and are both linear mappings, is a linear mapping too, even if it may not be an associative algebra or Lie algebra morphism. is its kernel, and is a sub vector space of . Furthermore, if and agree on vectors and of , then . Since both and are Lie algebra morphisms, this is equal to , and, since and agree on and , this expression reduces to zero. Thus too is an element of . This completes the proof that is a sub Lie algebra of . Since includes , and since is the smallest sub Lie algebra of itself including , it follows that , that is, that and agree on all . Since is their common domain, they are equal: .

Hence there exists at least one associative algebra , namely , such that .

Let us show that such a is unique.

Let be an associative algebra morphism such that .

Then , that is, .

But, because is a free associative algebra on the vector space , we know that there is a unique associative algebra morphism having such a property, namely .

Hence there is one and only one associative algebra morphism such that .

This shows that is the free associative algebra on the Lie algebra .

## No maverick elements in a free space

This keeps cropping up, and every time it takes me some effort to rediscover the proof. It happened first with free groups, then with free vector spaces and again with free Lie algebras.

I’ll formulate the issue with groups, but it is easy to carry it over to other cases.

A free group on set is defined by its universal property; furthermore, a particular specimen can be constructed. This construction is essential for proving the existence of a free group, but that should be its only role.

The question is: Is the subgroup of generated by necessarily itself? Can’t there be any “maverick” elements in , that are not in that subgroup? It seems that there shouldn’t be any, because they would be “too free”; when applying the universal property, the value given to them by the supposedly unique group morphism from to the “test group” might not be uniquely determined. But how do we prove this?

If we examine the constructed specimen ‑ in the case of free groups, the set of “strings” of elements of considered as “letters” plus their “primed versions”, with a certain composition rule ‑ the answer is that indeed, the whole of is generated by . And since all free groups on a given set are isomorphic, this answers our question.

But having to make use of a particular specimen doesn’t seem right to me. I want a proof with the universal property alone! Here it is, for my own repeated reference.

Let be a set, and any free group on . Let be the subgroup of generated by .

Now since, by definition of a generated subgroup, , is also a mapping from to ; or rather, we will call it as the codomain has changed. Thus is the mapping .

We thus have a group and a mapping ; the universal property of free groups satisfied by implies the existence of a group morphism such that .

Now can also be seen as a mapping from to ; or rather, since again we have changed the codomain, we define the new mapping . Since is a group morphism, so is .

Our equality  can be rewritten with the larger codomain , yielding . So is a group morphism from to such that . We know that there is only one such group morphism. Since we also have , necessarily .

For any , we have . Hence .

## “Free Object Functors”

Chapter 12 (Functors) gives as an example of a functor the “free group functor” from the category of sets () to the category of groups (). We will call this functor , and its construction, as described, is the following:

For any set , the object is the free group on .

Let be a -morphism, that is, any mapping, between sets and . Let and be the respective free groups, with and the associated mappings.

Then, by virtue of the universal property of free groups, there exists a unique -morphism such that, as mappings, .

So we decide that will be this . Of course, for this to be a functor, we must check that the composition of morphisms are preserved, as are identities.

Later in the chapter, the general concept of a free object in category on an object in via a functor from to is defined, as an extension of the notion of a free group; a free group thus becomes the free object in category on an object in category following the forgetful functor from to . We are called to consider the “free object functor” from  to (hence in the reverse direction respective to ), based as above on the mapping of objects in to their respective free objects in and of -morphisms to the corresponding -morphisms following the above construction.

This set-up at first appears clear, but left me feeling uneasy. After some thought I managed to pinpoint two issues:

1. Free objects, when they exist, are unique up to an isomorphism, which means that they aren’t «unique, period». For a functor to map a set to its free group, one must have a rule that says what exactly that group is; you can’t just say what it is up to an isomorphism.
2. «Free objects» are not actually objects in the target category; they are ordered pairs , where is an object of the target category, that is, in our example, groups, and is a morphism between objects of the source category. If I build a free group functor retaining only the target object, in our case , the part is lost.

The first issue can be solved easily for groups, and, I suspect, in most or all other cases. The existence of a free object is usually (always?) demonstrated by building some particular instance; one can decide that the functor will map the source object to that specific free object, that we will call the canonical free object.

For the second issue, one must note that the associated morphisms is not complete. We need these to build from . Thus the ‘s remain at least partly included in the definition of the free group functor.

I don’t have the impression that if we are given the functor , without the knowledge of the canonical free groups that it is built upon, we can reconstruct the morphisms that are part of the definition of thise canonical free objects. The picture is rather this:

• We define a particular rule for building a free group on any set .
• Based on this rule, we define the mappings from objects in to objects in , and from morphisms in to morphisms in .
• The entire above procedure (including the rule for building a specific free group) constitutes our free group functor, that is our rule for mapping to and morphisms in to morphisms in .

In practice, it doesn’t matter much what particular rule we take for building our free objects. It does matter, however, that it be conceived of as well specified, and that we understand that not only the part but also the in the free object serve to define the free object functor.

## Functors Between Real and Complex Vector Spaces

(Unfinished)

In Chapter 12, R.G. describes three possible transformations “from real to complex vector spaces and back”. Inspired by Chapter 17 (Functors), I have examined these transformations from the functor point of view.

We can distinguish not two, but three categories here, the first and third of which are equivalent:

1. : Complex vector spaces with their usual morphisms and composition law.
2. : Real vector spaces with their usual morphisms and composition law.
3. : The category of pairs , where is a real vector space and a vector space morphism such that (with the identity morphism on ). A morphism in this category is any linear mapping from to subject to the condition: . These morphisms are composed as mappings. It is simple to check that this does make up a category.

## Equivalence of  and

The equivalence of these two categories stems from the procedures described in Chapter 12, procedures that we will describe as isofunctors.

### Functor  from to

Action on objects:

If is a complex vector space, is such that:

• is the real vector space obtained from by “forgetting” multiplication by imaginaries, that is the one based on the same set as with the same addition, the multiplication by a scalar being simply restricted to real scalars.
• is the linear mapping that sends a vector to , the scalar multiplication here being that in the complex vector space . One immediately checks that we do have .

Action on morphisms:

A morphism between complex vector spaces is also, as a mapping, a morphism between the corresponding real vector space obtained by “forgetting” multiplication by imaginaries; thus, for a -morphism , we can define as the same mapping. It is simple to check that the condition on -morphisms is satisfied.

### Functor  from to

The second construction described by R.G. takes a pair with real vector space and morphism such that to a complex vector space. We can make this into a functor from to by defining what happens to morphisms: they are left as they are, as mappings; the above condition on morphisms in ensures that the resulting mapping is a -morphism.

These two functors are isofunctors

It is easy to check (R.G. should at least mention it – criticism) that these two operations are the inverse one of the other, which makes  an isofunctor with  its inverse, and and equivalent categories.

Taking free objects between these categories yields nothing new, as we saw in the post on isofunctors.

The main interest in this category is that it allows us to work with real vector spaces only, avoiding in particular the need to consider complex and real versions of a given vector space, with the scalar multiplication in the former that «knows how» to multiply by complex numbers, while it does not in the latter. In the rest of this post I will talk of complex vector spaces and these «real plus» vector spaces as if they were the same thing.

## Forgetful functor from complex to real vector spaces

One can make a forgetful functor from to simply by forgetting the part. The functor leaves morphisms unchanged. This is the first construction described by R.G. in Chapter 12, from to .

## A sort of inverse construction of

The functor forgets the , which cannot be retrieved. It can, however, be reinvented, though in general there is a necessary arbitrariness in the process.

Let be a real vector space. Suppose that it is possible to find in two subspaces, and , that are complementary and isomorphic. There will in general be an infinite number of ways of choosing the pair , and, this pair being chosen, an infinite number of ways of choosing and isomorphism . (One can, for instance, arbitrarily choose bases on and , and create from an arbitrary bijection between these bases.)

is a mapping between two specific subspaces of . We can, however, create a full mapping by posing:

It is easy to check that this is linear, and that . Hence is an object of . Applying the functor defined above, we obtain a complex vector space.

The necessary and sufficient condition for it to be possible to find in two complementary and isomorphic subspaces is that be either finite and even-dimensional, or infinite-dimensional (exercise 84).

Providing this condition is satisfied, this procedure involves the arbitrary step of choosing the subspaces, and the second arbitrary step of choosing the isomophism between them. For this reason, one cannot make it into a functor; for a functor must consist of a rule for mapping objects and morphisms. We don’t have an inverse functor for .

This arbitrariness is present even in the simplest non-trivial case, that of a two-dimensional real vector space. This is another formulation of the fact that in a complex vector space there is no non-arbitrary notion of “real” and “imaginary” vectors (as noted in Chapter 12).

## Free objects following the forgetful functor from complex to real vector spaces

We do, however, have a functor from to , but one that implies doubling the initial vector space. It uses the third construction described in Chapter 12.

It is simpler to talk of this construction with rather than ; we have seen that these categories are equivalent. Given an object in , that is, a real vector space, we take as the real vector space direct sum . We define the mapping by . It is easy to check that is linear, and that . Hence is an object in . We thus have a rule that maps objects of to objects of . Now let us consider a morphism in , between real vector spaces and . Let and . We can form the mapping . It is linear, and furthermore ; hence is a morphism in category , between  and . We write . We can check that this rule preserves composition and identities. It is thus a functor from to .

Now let us examine if this cannot be viewed as the — or better, a — free object functor following functor .

Let be a real vector space, and , that is and . and are objects in . Let be the -morphism defined by . We wish to check that is a free object on following and that the transformation of morphisms follows suit.

Let be another object of and a -morphism from to , that is a linear mapping .

We wish to find a -morphism such that . But is just the mapping itself. For , . So must map any to . Since must also be a -morphism, we must have .

In other words, necessarily, our must be such that .

Now we must check that thus defined is indeed a -morphism and that  (for up to now we have shown what must be; we must check that the result is indeed adequate). This checking is unproblematic.

Thus our is a free object on the real vector space following the functor .

But it’s not quite over! We must also check that, taking

## 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.