## “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.

## Algebraic infinite sum notation

On an infinite-dimensional vector space one usually defines a linear combination as a finite sum; for instance, if is a basis of the -vector space , one may write, for some finite part of :

I find this notation cumbersome, because it seems to make the sum dependent on the arbitrary choice of the finite set , while in fact it doesn’t, provided contains all the nonzero values to be added.

For instance, if you wish to add two such linear combinations written with different finite sets and , you cannot do so directly, without first arbitrarily choosing some other finite subset of containing :

while making excuses for the necessity to define the new ‘s and ‘s as zero and for the arbitrary choice of which doesn’t change the result.

I find it more practical to accept writing sums of an arbitrary collection of objects, such as:

provided we know in advance that only for a finite number of are the values nonzero.