A question about definition of a category

[Arian.D]

I know it may sound idiotic to ask questions about definition of something, but I'm going to do that now. I've seen the definition of categories in several different contexts, in all of them categories consisted of objects like groups, rings, R-modules (in particular, vector spaces), topological spaces and many other mathematical objects that I don't know and the arrows were group homomorphisms, ring homomorphisms, R-homomorphisms, linear transformations, continuous maps, etc that have the following properties:

1. The composition of arrows gives us a third.
2. There exists an identity arrow that if it's composed with another arrow it gives it back.
3. The composition of arrows is associative (which sounds a little bit obvious to me because composition of functions is associative and we think of arrows as functions acting on mathematical objects that preserve some structure. Right?).

Now my book adds an additional condition on categories. I need you to clarify things a bit please:

A category $\mathbb{C}$ is a class of mathematical objects that we usually denote them by A,B,C,D,... with following properties:
1) For any two objects A and B, a set $Hom_{\mathbb{C}}(A,B)$ is associated to them with the property that for any four objects A,B,C and D that $(A,B)\neq (C,D)$, $Hom_{\mathbb{C}}(A,B) \cap Hom_{\mathbb{C}}(C,D) = \emptyset$
Other conditions are understandable for me.

Why they have put this condition here? What does it want to tell me? Isn't the trivial homomorphism that sends A to B by f(A)=0 in the intersection?
I mean suppose that we have (A,B)≠(A,C) (which is true because one of the components of the ordered pairs is different). Then if we define f(A)=0, isn't this homomorphism in the intersection?

 

[micromass]

This is following the convention that functions should be determined by their domain, their codomain and their mapping properties.

That is, the following functions should be regarded as different
1) $f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x^2$
2) $f:\mathbb{R}\rightarrow \mathbb{R}^+:x\rightarrow x^2$
3) $f:\mathbb{R}^+\rightarrow \mathbb{R}:x\rightarrow x^2$

They might do the same thing, but their domain and codomain are different, so they are different functions.

In your case, the two functions
1) $f:A\rightarrow B:x\rightarrow 0$
2) $f:A\rightarrow C:x\rightarrow 0$

should be regarded as different if B is different from C.

Note that this is just a technical condition. We introduce it in order for each function to have a unique domain and a unique codomain. That is, we wish the the functions
$dom(f:A\rightarrow B)=A,~codom(f:A\rightarrow B)=B$
are well-defined.
That this is just a minor, technical condition is clear since it is easy to force the condition to be true. Indeed, one can easily replace f in Hom(A,B) by (A,f,B). This does satisfy the property.

 

[Arian.D]

This is following the convention that functions should be determined by their domain, their codomain and their mapping properties.

That is, the following functions should be regarded as different
1) $f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow x^2$
2) $f:\mathbb{R}\rightarrow \mathbb{R}^+:x\rightarrow x^2$
3) $f:\mathbb{R}^+\rightarrow \mathbb{R}:x\rightarrow x^2$

They might do the same thing, but their domain and codomain are different, so they are different functions.

In your case, the two functions
1) $f:A\rightarrow B:x\rightarrow 0$
2) $f:A\rightarrow C:x\rightarrow 0$

should be regarded as different if B is different from C.

Note that this is just a technical condition. We introduce it in order for each function to have a unique domain and a unique codomain. That is, we wish the the functions
$dom(f:A\rightarrow B)=A,~codom(f:A\rightarrow B)=B$
are well-defined.
That this is just a minor, technical condition is clear since it is easy to force the condition to be true. Indeed, one can easily replace f in Hom(A,B) by (A,f,B). This does satisfy the property.

Thanks. It makes sense now.

 

[Arian.D]

OK. Another question. My book uses very odd definitions for categories, I don't know why. I like the book because it has defined tensor product in a very neat way that I could understand all theorems easily, but it has very weird definitions for categories I guess. Anyway, I don't understand this definition:

Definition. We call a category 'tangible!' if there exists a function $\sigma$ such that for any object like $A$ it corresponds a set $\sigma(A)$ with following properties:

1. Any morphism $f: A \to B$ is a function from the set $\sigma(A)$ to the set $\sigma(B)$,
2. For any object like $A$, the morphism $1_A$ is the identity function on $\sigma(A)$,
3. The composition of two morphisms $f: A \to B$ and $g: B \to C$ is the composition of them as functions from $\sigma(A)$ to $\sigma(B)$ and $\sigma(B)$ to $\sigma(C)$.

Then it claims that the categories $\mathbb{Set}$, $\mathbb{Grp}$, $\mathbb{Ring}$, $\mathbb{_RM}$, $\mathbb{M_R}$ are tangible while the following category is not:

Suppose that (G,*) is a group. Consider $\mathbb{C}$ to be the set $\{G\}$ and also consider the set $Hom_{\mathbb{C}}(G,G)$ to be the set of all elements of G. Morphisms in this category are elements of G and composition of morphisms is defined in the following way:

$Hom_{\mathbb{C}}(G,G) \times Hom_{\mathbb{C}}(G,G) \to Hom_{\mathbb{C}}(G,G)$
$(a, b) \mapsto a*b$
​

I don't understand why this category is not tangible while the other ones that I mentioned above are tangible. Does it have anything to do with being small or large?

Thanks in advance

 

[micromass]

That reminds of being a "concrete category", but it's not quite the same.

A category is tangible if it's morphisms are actually functions between sets. For example, a morphism between groups is a function f satisfying f(xy)=f(x)f(y). What matters is that this is actually a function between sets.

The category you list comes from a group G. It's morphisms are elements of G. These morphisms are not functions between sets but rather elements of the group.

 

[Arian.D]

That reminds of being a "concrete category", but it's not quite the same.

A category is tangible if it's morphisms are actually functions between sets. For example, a morphism between groups is a function f satisfying f(xy)=f(x)f(y). What matters is that this is actually a function between sets.

The category you list comes from a group G. It's morphisms are elements of G. These morphisms are not functions between sets but rather elements of the group.

Yea, that's what I guessed too but then I couldn't realize why the definition is so weird like this. How does this definition imply what you said and why these two things are equivalent? Could you create more examples please?
I don't understand the definition yet. One more question, why the identity function doesn't work in the definition as sigma? Can't we always set sigma=identity function?

 

[micromass]

The reason why we don't set sigma = identity function is probably the following:

Remember that a group is a couple (G,*). But a morphism between groups (G,*) and (H,*) is a function between G and H, not between the couples (G,*) and (H,*)!

So we define $\sigma(G,*)=G$, then we can say that a morphism between groups is exactly a function between $\sigma(G)$ and $\sigma(H)$.

This is likely the reason for introducing the $\sigma$.

Other examples are easy to be found. For example, a ring is a triple (R,+,*). Set $\sigma(R,+,*)=R$.

 

[Arian.D]

The reason why we don't set sigma = identity function is probably the following:

Remember that a group is a couple (G,*). But a morphism between groups (G,*) and (H,*) is a function between G and H, not between the couples (G,*) and (H,*)!

So we define $\sigma(G,*)=G$, then we can say that a morphism between groups is exactly a function between $\sigma(G)$ and $\sigma(H)$.

This is likely the reason for introducing the $\sigma$.

Other examples are easy to be found. For example, a ring is a triple (R,+,*). Set $\sigma(R,+,*)=R$.

Aren't your notations conflicting now?
I guess you meant this:


So we define $\sigma(G,*)=G$, then we can say that a morphism between groups is exactly a function between $\sigma(G,*)$ and $\sigma(H,*)$.

So you say because the object that we map A to it might have a different mathematical structure the identity map doesn't work. For example in this case sigma acts on an ordered pair which represents a group but the output is a set which is obviously not the same thing as an ordered pair. Is that what you mean?

OK. Fine. Then why that example is not tangible? Which condition of the definition make it fail?

 

[micromass]

So we define $\sigma(G,*)=G$, then we can say that a morphism between groups is exactly a function between $\sigma(G,*)$ and $\sigma(H,*)$.

Yes, I meant that.

So you say because the object that we map A to it might have a different mathematical structure the identity map doesn't work. For example in this case sigma acts on an ordered pair which represents a group but the output is a set which is obviously not the same thing as an ordered pair. Is that what you mean?

Yes.

OK. Fine. Then why that example is not tangible? Which condition of the definition make it fail?

What example is not tangible?? The case where the morphisms are the elements of the group?? In that case, the morphisms aren't really functions.

 

[Arian.D]

What example is not tangible?? The case where the morphisms are the elements of the group?? In that case, the morphisms aren't really functions.

Yea, but why morphisms need to be functions? This is what I didn't fully understand.

A question which is a bit off-topic now. Later in the book it claims that there exists only the trivial homomorphism between $\mathbb{Q}$ and $S_3$. I'm trying to figure it out why. Well, $\mathbb{Q}$ is abelian, and any non-identity element in it has an infinite order, while S₃ is non-abelian and every element in it is of finite order. So the structure of the two groups is radically different and there could be no structure-preserving maps between them except the trivial homomorphism. Is my explanation good enough or I need to reason in some way else?

 

[Arian.D]

Aaaah! How naive I am. It explicitly says in the definition that any morphism must be a function from sigma(A) to sigma(B). so being a function is required in the definition.

 

[micromass]

A question which is a bit off-topic now. Later in the book it claims that there exists only the trivial homomorphism between $\mathbb{Q}$ and $S_3$. I'm trying to figure it out why. Well, $\mathbb{Q}$ is abelian, and any non-identity element in it has an infinite order, while S₃ is non-abelian and every element in it is of finite order. So the structure of the two groups is radically different and there could be no structure-preserving maps between them except the trivial homomorphism. Is my explanation good enough or I need to reason in some way else?

No, you need a better explanation.
That $\mathbb{Q}$ is abelian only means that the image under the homomorphism is an abelian subgroup of $S_3$. Such abelian subgroups certainly exist.
That every nonzero element has infinite order in $\mathbb{Q}$ doesn't mean anything as elements of infinite orders can be mapped to elements of finite order.

So you need a new reasoning. Let's prove this by contradiction. Assume that there exists a nonzero element $q\in \mathbb{Q}$ such that f(q) is not the identity. Then f(q) can have order 2 or order 3.
Now, think about f(q/2) and f(q/3). We have that q/2 + q/2 = q, thus f(q/2)+f(q/2)=f(q). What must f(q/2) map to then?? Try to finish the proof.

What I essentialy used here is that $\mathbb{Q}$ is a divisble group and $\mathbb{S}_3$ is not.

 

[Arian.D]

No, you need a better explanation.
That $\mathbb{Q}$ is abelian only means that the image under the homomorphism is an abelian subgroup of $S_3$. Such abelian subgroups certainly exist.
That every nonzero element has infinite order in $\mathbb{Q}$ doesn't mean anything as elements of infinite orders can be mapped to elements of finite order.

So you need a new reasoning. Let's prove this by contradiction. Assume that there exists a nonzero element $q\in \mathbb{Q}$ such that f(q) is not the identity. Then f(q) can have order 2 or order 3.
Now, think about f(q/2) and f(q/3). We have that q/2 + q/2 = q, thus f(q/2)+f(q/2)=f(q). What must f(q/2) map to then?? Try to finish the proof.

What I essentialy used here is that $\mathbb{Q}$ is a divisble group and $\mathbb{S}_3$ is not.

I don't know what a divisible group is and I have no idea how to finish the proof yet. :(

 

[mathwonk]

that first example of a non tangible category is kind of a fraud, because you could consider the set sigma(G) associated to G to be the set of elements of G, and each element of G then defines a function from the set sigma(G) to itself by multiplication by that element. Then composition of two elements is indeed multiplication by their product so it corresponds exactly to their definition of composition. moreover multiplication by the identity element does define the identity mapping. so that's a fake example.

in fact its hard to think of a real example, i.e. one not essentially equiovalent to functions.

e.g. if you take the category to be the integers i.e. the objects to be integers, and define the set of morphisms HOM(m,n) from one integer m to another integer n, to be non empty with exactly one morphism if and only if m divides n.

then you could still consider the set sigma(m) to be the one point set {m}, and the morphism from m to n, to be the unique function from {m} to {n}. still if m = n, the unique map from {m} to {m} is indeed the identity map.

so its kind of a fake, but maybe i can give a little better example.

lets take the category to be topological spaces, but the morphisms from X to Y are the equivalence classes of homotopic maps from X to Y. Now these are really not functions, since as far as I can see anyway, there is no way to choose a particular representative of a given homotopy class.

 

[dcpo]

that first example of a non tangible category is kind of a fraud, because you could consider the set sigma(G) associated to G to be the set of elements of G, and each element of G then defines a function from the set sigma(G) to itself by multiplication by that element. Then composition of two elements is indeed multiplication by their product so it corresponds exactly to their definition of composition. moreover multiplication by the identity element does define the identity mapping. so that's a fake example.

in fact its hard to think of a real example, i.e. one not essentially equiovalent to functions.

e.g. if you take the category to be the integers i.e. the objects to be integers, and define the set of morphisms HOM(m,n) from one integer m to another integer n, to be non empty with exactly one morphism if and only if m divides n.

then you could still consider the set sigma(m) to be the one point set {m}, and the morphism from m to n, to be the unique function from {m} to {n}. still if m = n, the unique map from {m} to {m} is indeed the identity map.

so its kind of a fake, but maybe i can give a little better example.

lets take the category to be topological spaces, but the morphisms from X to Y are the equivalence classes of homotopic maps from X to Y. Now these are really not functions, since as far as I can see anyway, there is no way to choose a particular representative of a given homotopy class.

I was thinking something similar. I like your homotopy class example.

 

[mathwonk]

The notion of concrete categories C as defined here is a bit misleading to me. It says that there is a way of assigning to each object A in C, a set s(A) such that each morphism in Hom(A,B) “is” a function from s(A) to s(B). What does “is” mean? Technically a morphism from A to B “is” not a function from s(A) to s(B). It is a morphism from A to B nothing else. But the definition seems to mean that it considers only categories where the objects A,B ”are” in fact already given as sets with structure. E.g. a group is usually defined as a set plus an operation, and a homomorphism is defined as a function with certain properties. Then the “assignment” of a set s(A) to a group A is done by merely forgetting the structure.

But if this is so, why does the definition allow some freedom in the “assignment” of the set s(A) to the object A? I.e. having allowed us to “assign” a set s(A) to A, subject to the requirements that the identity morphism “is” the identity function, and compositions of morphisms “are” compositions of functions, why not allow us the same flexibility in “assigning” the functions s(f) to the morphisms f:A→B?

Note that if we just change the word “is”, or rather make the definition more precise, the definition would say that we assign a set s(A) to each object A, and also assign a function s(f):s(A)→s(B) to each morphism f:A→B, such that the same conditions on identities and compositions hold. This would be what is called a “functor” from the category C to the category S of sets. The difference would be that now every category admits such a functor. (Just send every object to the same set and every morphism to the identity function.) Hence every category would be concrete!

Presumably we want different morphisms to correspond to different functions, (although this is not true in “scheme theoretic” algebraic geometry, where the function on the underlying set does not always determine the algebraic part of the morphism.) Thus we could strengthen the definition and ask that the induced map from Hom(A,B) to Fun(s(A),s(B)) is injective. Such a functor is called an embedding.

So we could say that a “concrete category” is a category equipped with an embedding functor to the category S of sets. In this sense the examples I called fakes the other day do have such embeddings in the category of sets, as we discussed. All we needed to do was change the phrase “is a function” to the word “determines a function”. Moreover we get that the function also determines the morphism back again.

Then it turns out that, except for set theoretical paradoxes caused by large sets (“proper classes”), in fact every category is essentially concrete. I.e. every morphism can be viewed as determining and determined by, at least a family of functions on sets.

I.e. If C is a category, and S is the category of sets, then for every object A in C we can assign s(A) to be the collection of families of morphisms Hom(X,A) for all X in C. I.e. let Fun(C,S) be the collection of functions (actually functors) from C to S. Then given an object A, it determines a functor s(A) from C to S, sending X to Hom(X,A).

The only set theoretic problem here is that the functor can be represented as a collection of ordered pairs, i.e. as a subcollection of CxS, but C and S are too big to be sets, and the subcollection s(A) is also too big to be a set. But if C is a “small” category. i.e. if the collection of all objects in C is itself a set, then everything is ok.

I.e. a morphism from A to B, defines a function from Hom(X,A) to Hom(X,B) by composing, hence also a function from the full collection {Hom(X,A)} to the collection {Hom(X,B)}. And identities and compositions behave as required.

The theorem then is that, in our sense, every small category is a concrete category, i.e. has an embedding in the category of sets.

In fact if you do the exercise, you will see that this realizes, in the case of the category above with one element, the group G, and such that Hom(G,G) = G, the same functor embedding that I described.
I.e. the object G goes to the set Hom(G,G) = G, and morphisms G-->G, i.e. elements a of G, define via "composition" i.e. multiplication, the same maps from Hom(G,G) = G-->G = Hom(G,G) that I described, namely multiplication.

There is another notion meant to mimic the category of abelian groups, called an “abelian category”, where there exists a zero object, and morphisms have kernels and cokernels, and all monic or epic morphisms occur as kernels or cokernels. Then the theorem is that all small abelian categories embed in the category of abelian groups. In fact in that case one can make the map Hom(A,B)→Hom(s(A),s(B)) a bijection. These theorems are due to several people, including Peter Freyd and Barry Mitchell.

I can only conclude that books which give the misleading kind of definition above of "concrete" categories, are making a (to me misguided) effort to avoid defining functors. I say misguided because functors are far more important than categories. Categories are only a necessary precursor to defining functors.