# Homework Help: The definition of a free object

1. May 16, 2007

As the title suggests, I'm confused with the definition of a free object, and would be thankful if someone could present some illustrative examples and motivation for understanding the definition.

Let F be an object in a concrete category C, X a nonempty set, and i : X --> F a map (of sets). F is free on the set X provided that for any object A of C and map (of sets) f : X --> A, there exists a unique morphism of C, f* : F --> A, such that f*i = f.

Ok, first of all, since concrete categories are mentioned, one is dealing with mappings between underlying sets. So, what does the deifnition actually tell? Is the emphasis on the fact that for every mapping f, we can "simulate" this very mapping with f*i? I believe it is a bit hard to understand such definitions without examples (the *one* in the book didn't really help me) and greater experience in mathematical abstraction.

Last edited: May 17, 2007
2. May 16, 2007

### StatusX

Here's a rough way to understand these definitions of universal properties in terms of the existence of maps. I'll try to explain the situation for a groups, and hopefully this will help you understand the general case.

Say f:G->H is a group homomorphism, which we may assume is surjective by restricting to its image. Then H is a quotient of G. This is another way of saying that H is just G, where we've identified some elements with zero (and extended the equivalence relation so the structure is still a group). Another way still is to say H is G with extra relations added, where a relation is just a true equation involving elements in the group. Note that if $a_1 a_2 ... a_n=e$ then $f(a_1 a_2 ... a_n)=f(a_1) f(a_2) ... f(a_n)=f(e)=e$, so any relations that hold in G must also hold in H, where we identify elements with their images. A rough way of summarizing this is that if a (surjective) homomorphism exists from G to H, then H has at least as many relations as G, or equivalently G has no more relations than H.

So when we assert that an object U have the universal property that there is always a morphism from it to some other object with certain properties, we are effectively saying that it is the most general object with those properties, in the sense that it has no more relations than are absolutely necessary to give it those properties. If it did, there would be some object A with the properties and with fewer relations, and it would be impossible to have a morphism from U into A. Note these definitions use maps from the universal object, not to it, which makes sense given the last sentence of the previous paragraph. For example, there is always a homorphism to the trivial group, because it has, in a sense, every relation.

In the example of a free group, we are saying it is the most general group containing the set X. That is, given any other group A containing X (where we speak of two different groups containing the same subset by using the maps f:X->A and i:X->A), there is a morphism from F to A taking X to itself.

For example, if X has a single element, F is just the group of integers, Z. This is reflected in the fact that every element in a group A (ie, every map from X to A) generates a cyclic subgroup, the image of Z in A under f*, isomorphic to a quotient of Z. More generally, the subgroup generated by a subset is the image of the homomorphism from the free group on that set into the group.

This is all just a sketch of an argument motivating the definition, and there are details to be filled in (like how we use maps to identify elements in different objects). The best way to get a feel for these defintions is to look at some examples. For example, try to understand in what sense the tensor product encodes the most general bilinear map from two vector spaces into another. There's a really good page on this http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html".

Last edited by a moderator: Apr 22, 2017
3. May 16, 2007

### Hurkyl

Staff Emeritus
You can try working out examples. It's instructive.

Let's start with an easy one: for any set X, can you find a free set on X? (i.e. let C be the category of sets)

What about other categories you know well? Can you find a free group on X? A free abelian group on X? A free topological space on X? A free complex vector space on X?

Here's a very useful one: let R be a commutative ring. Can you find a free [commutative R-algebra with unity] on the one point set {*}? What about a finite set? Or an arbitrary set? (the maps in this category must preserve the unit)

Incidentally, the category of commutative R-algebras with unity is an example of a slice category (or coslice category, or comma category, depending on your notation). It is equivalent to the category defined by:

Objects are ring homomorphisms R ---> X for commutative rings X
A morphism from R ---> X to R ---> Y is a ring homomorphism X ---> Y such that
R ---> X ---> Y = R ---> Y.​
(i.e. the morphisms are commutative triangles)

They might even be isomorphic!

(eep! I hope I have that right)

Last edited: May 16, 2007
4. May 16, 2007

### mathwonk

a vector space is free on any basis. Z is free on the element 1.

free means it is easy to define maps out of it. al you have to do is specify them on the basis, and you can do that biowever you want.

5. May 16, 2007

### mathwonk

bioever? you like?

6. May 16, 2007

### Hurkyl

Staff Emeritus
For the sake of precision:
Z is a free group on {1}.
Z is a free abelian group on {1}.
Z is a free ring on {}.
Z is not a free ring on {1}.

7. May 17, 2007

First of all, thanks to all for your help.

Well, since the identity exists in any category, we have i : X --> X. If we take any set A and mapping f : X --> A, there exists a unique mapping f* : X --> A, such that f*i = f, so for the mapping f* we simply take f. Hence, X is free on X, for any set X. (I hope I got that one right.)

If Z is a free group on {1}, then for every group A from the category of groups, and for every map f : {1} --> A, there must exist a unique map f* : Z --> A, such that f*i = f, where i : {1} --> Z. Ok, let's define f with f(1) = a^k. Further on, let i(1) = k. If we define f* with f*(x) = a^x, then f*i = f, and hence Z is free on {1}.

I hope what I wrote is correct.

But I'm still pretty much foggy about the term "free object". I guess I'll have to read more and get a feeling what it means. The problem is I didn't yet study rings and modules, so the number of examples I can process is small.

8. May 17, 2007

### matt grime

Free means, essentially, 'no relations', or at least 'with no extraneous relations. It says that any map of F is defined by the restriction to X, and that it is universal with this property. So, firstly, F has to be generated by X in whatever appropriate sense.

Secondly, let's suppose that we have some putative F that is generated by X, and we want to examine maps from F to any object. Any map can be viewed as a surjection if we restrict to its image. So the image of every map from F is essentially, F modulo some relations, i.e. a quotient of F by something (the kernel). So to be universal, any relations on X as a subset of F must be a subset of all possible relations if F is to be free. Which sounds like a mouthful but isn't really.

It's easiest with groups.

Let X be a set. What is a free group on X? Call this F[X]. Well, we must send each element of x to some element of F[X]. Let's associate g_x with x. So, we can just write down a group:

F[X]:=<g_x : x in X>

the group with generators g_x, and no relations. This looks good. Clearly any map from X to some group with generators indexed by x factors through it.

So what are we saying in general? Well, if we have f:X-->A any map from X to an object A, it generates some subobject of A. This subobject is just what ever f(x) for x in X generates, modulo some relations that A imposes on it. The only thing F that can fit into the diagram X-->F-->A must then be the object F that has as relations those relations that are a subset of every set of relations for all objects A.

As we saw for GRP, this meant the free group with generators indexed by X. For ABGRP, the category of abelian groups, we have to have generators g_x again, but the minimall set of relations is 'all g_x, g_y commute', so a free object in ABGRP is isomorphic to Z^|X|, i.e. the product of Z with itself |X| times.

Free, for rings is somewhat more complicated. A free module for the ring R, is a copy of direct sums of R.

Last edited: May 17, 2007
9. May 17, 2007

### Office_Shredder

Staff Emeritus
Nothing's free in life.

Sorry, it had to be said

10. May 20, 2007

So, if I'm looking at the category of vector spaces, and linear operators between them, and if I take a particular vector space V, and a basis B for it, how do I show, by strictly using the definition of a free object (in the category of vector spaces), that V is free on B? I have tried, but there are still things confusing me. Any push is appreciated.

11. May 20, 2007

### Hurkyl

Staff Emeritus
Well, start one step at a time.

If you have a set map B --> |W|, can you construct a linear transformation V --> W?

(|W| is the underlying set of W)

Last edited: May 20, 2007
12. May 20, 2007

### matt grime

It is easiest to see it is free on its basis in the sense you wrote by thinking instead of X as some set of cardinality n. I.E. the set over which it is free is just a set of cardinality n. A free object on that set is then a vector space of dimension n. Send the elements {1,..,n} to e_1,..e_n.

If you just think about the any map from the set {1,..,n} to a vector space, then it can only be universal if the image set is a linearly independent spanning set, i.e. a basis. You can take each of the two conditions separately. If the set e_1,..,e_n don't span, then you can extend linear maps out of the vector space in different ways. And if they're not linearly independent, then you can not necessarily write down a map which takes those n elements to n arbitrary elements in another vector space (e.g. an linearly independent set in another vector space, for instance).

13. May 20, 2007

Okay, so the first step is simply to take a mapping i : {1, ..., n} --> {v1, ..., vn}, where {v1, ..., vn} is a basis of a vector space V. Next, we consider the "any" mapping f : {1, ..., n} --> W, where W is some vector space. We want to show that there exists a unique mapping f* such that f*i = f. Let {w1, ..., wk} be a basis of W. If we define i(j) = vj, j = 1, ..., n, and f(j) = wj, j = 1, ..., k, then the map f* : V --> W defined with f*(vj) = wj satisfies f*i = f. This holds for k less or equal to n. If k > n, the the image of f doesn't span W, and I guess I come to the spot matt mentioned, about extending linear maps out of vector spaces in different ways (?).

Then again, I'm confused since this looks more like showing that the basis of a vector space is free on {1, ..., n}. Is it equivalent to saying that a vector space is free on {1, ..., n}?

14. May 20, 2007

### Hurkyl

Staff Emeritus
You can't do that! You said
consider the "any" mapping f : {1, ..., n} --> W​
so you have to use the f under consideration: you can't just ignore it and define an entirely new f!

15. May 20, 2007

I know I can't, and I don't know why I did it.

So, suppose some vector space W and some mapping f : {1, ..., n} --> W are given. Frankly, I can't see how to prove anything.

16. May 20, 2007

OK, here's another try. For any vector space W and mapping f : {1, ..., n} --> W, Im(f) is a subset of W, say Im(f) = {w1, ..., wk}, where k is less or equal to n, so there exists a mapping f* : {v1, ..., vn} --> Im(f). Further on, if we let f*(vj) = f(j), then f* satisfies f*i = f, since (f*i)(j) = f*(i(j)) = f*(vj) = f(j). If this works, then I should show the uniqueness of f*, which I can't figure out right now.

Edit: and I assume I should show that f* is a linear operator, since the definition insists f* to be a morphism?

17. May 21, 2007

### matt grime

What does f* do? It sends a basis element v_i to some element w_j in V. Well, that is obviously a linear map - why assume anything? Any other map that could be a different factoring must do the same thing to the basis, hence f* must be unique (if two linear maps agree on a basis they are equal).

Let's try this from the beginning.

I claim that a vector space V of dimension n is free for any set with n elements. I may as well let the set of n elements be X={1,...,n}. Free means we have to specify a map

i:X-->V

so I'll define i(k) to be e_k for some basis of V.

Now, does this satisfy the definition of free?

Let f be any map from {1,..,n} to W. Thus f(j)=w_j for some elements w_1,..,w_n (with possible duplicates - so dont' make a set out of them for no reason). Does f factor through i? Yes. The map f*(v_j)=w_j is a linear map that does the job. Since v_j form a basis, this map is completely defined, and is unique: any other g* must send v_j to w_j as well, but the v_k are a basis. Hence V is free on {1,..,n} with the map i picking out any basis.

Thus we know that i and {1,..,n} above imply free. But what about the converse?

Suppose that X={1,..,m} is any set and V any vector space (dimension unimportant). Let i(k)=v_k be some choice of map from X to V. If v_k are not a basis can this be free? No. If the v_k don't span then any maps defined on v_k can be completed in infinitely many different ways to maps defined on the whole of V. Secondly, if they are not linearly independent, then there is a linear combination of them that is zero. But now pick some linearly independent set in W of cardinality m w_1.,,.w_m. Any notional factorisation through V must send v_k to w_k by definition. But then this f* sends a linearly dependent set to a linearly independent set, which is impossible.

18. May 21, 2007

Thanks a lot for the detailed explanation, everything is clear now!

19. May 23, 2007