General chat with user 170039

user131753

1:45 PM

Hello @Max. I just saw your comment. Please give me some time to read it.

Max

@user170039 no problem, take all the time you need

user131753

Ok. I finished reading your comment. Just for the sake of completeness let me quote it below.

user131753

There are two ways of defining a category, one has that the $\hom(A,B)$ need to be disjoint (this is the one in JoC), in which case for groups $\hom(G,H)$ cannot be a subset of $\hom(UG,UH)$; the other one has no such requirement, in this version $\hom(G,H)$ can be defined to be a subset of $\hom(UG,UH)$, but then for any concrete category $(\mathbf{A},U)$ over $\mathbf{X}$, there is a concretely isomorphic concrete category $(\mathbf{B},V)$ where $\hom(A,B)\subset \hom(VA,VB)$, so it really isn't a crucial difference — Max 3 mins ago

Max

Ok. Do you understand what I mean or are there parts that are still unclear ?

user131753

@Max I have some doubts still.

Max

1:48 PM

What about ?

user131753

Before I say anything. Let me just say that even though in JoC they define the $\text{hom}$-sets to be disjoint while defining a category, they later say that adding that condition is for technical convenience only and hence can be easily dropped.

user131753

I have basically two questions:

user131753

(1) Why do you say that in case for groups $\text{hom}_{\mathbf{Grp}}(G,H)$ cannot be a subset of $\text{hom}_{\textbf{Set}}(UG,UH)$? What is(are) the flaw(s) in my arguments,

user131753

This can be clarified by the saying that a group homomorphisms (i.e., $\mathbf{Grp}$-morphisms) for two given groups $(G,\circ_G)$ and $(H,\circ_H)$ is precisely those functions (i.e., $\mathbf{Set}$-morphisms) $f:G\to H$ such that the following diagram commutes, $$\require{AMScd}\begin{CD} G\times G @>{f\times f}>> H\times H\\ @V{\circ_G}VV @VV{\circ_H}V \\ G @>{f}>> H \end{CD}$$ — user 170039 10 mins ago

user131753

The point being $\mathbf{Grp}$-morphisms are precisely those $\mathbf{Set}$-morphisms which satisfy some additional properties. This is not always the case if we consider faithful functors $U:\mathbf{A}\to\mathbf{X}$ where $\mathbf{X}$ can be any category. — user 170039 10 mins ago

user131753

1:53 PM

(2) You wrote, "...for any concrete category $(\mathbf{A},U)$ over $\mathbf{X}$, there is a concretely isomorphic concrete category $(\mathbf{B},V)$". Would you mind to elaborate how one can construct $(\mathbf{B},V)$ from $(\mathbf{A},U)$?

user131753

I am done @Max.

Max

Well it can't if you require the hom-sets to be disjoint, because for instance if you take a bijection $A\to B$, any group structure on $A$ induces one on $B$ and a corresponding group morphism

If you don't require that the hom sets be disjoint, then it can, but in this case you have point (2) that comes in

So if you require the hom sets to be disjoint, a group morphism (G,*)-> (H,.) can't be a map of sets G-> H satisfying certain conditions

it has to be a map of sets satisfying those conditions + the data that its domain is (G,*) (and not some other group with G as underlying set) and that its codomain is (H,.) (not some other group with H as underlying set)

Now if you don't require the hom-sets to be disjoint (which is a bit more than a technical convenience, it's interesting to define source and target, and thus to define internal categories; but also for conceptual reasons of typing, but anyway)

then start from your concrete category (A,U), now take a category B that has as objects the same as those of A, and hom_B(A,C) := U(hom_A(A,C)), and composition is defined as : if f:A-> C, g: C-> D in B, then f=U f' for f' : A-> C in A, g=Ug' for g' : C-> D in A, f' and g' are unique with these specifications, and we may define g\circ f := U(g'\circ f'). It's easy to check (using injectivity of U on hom sets) that this is associative and that Uid_A is id_A in B;

then you may define VA = UA on objects, Vf = f on arrows which gives V:B-> X; it's trivially faithful, and trivially concretely isomorphic to A; and by definition hom_B(A,C)\subset hom_X(VA,VC).

(also, about the disjoint hom-sets, note that it is really really really more convenient to require that about them)

Don't hesitate if you have more questions or if something is unclear

@user170039

user131753

Ok. I will take some time to fully understand this. I will let you know if something is unclear and/or I have more questions regarding this. Thank you very much @Max.

user131753

@Max But isn't the commutative diagram $$\require{AMScd}\begin{CD} G\times G @>{f\times f}>> H\times H\\ @V{\circ_G}VV @VV{\circ_H}V \\ G @>{f}>> H \end{CD}$$saying precisely that the operation on $G$ is $\circ_G$ and similarly for $H$ (of course if we assume that $\circ_G$ satisfies the group axioms) and hence the group homomorphism is really between $(G,\circ_G)$ and $(H,\circ_H)$?

Max

No, because it's not because f commutes with \circ_G that it can't commute with other group structures on G. Pick the stupidest example : f is the trivial map to H; then any group structure on G works

so f will be in Hom((G,*),(H,.)) for any * making G a group and any . making e_H the neutral element

user131753

2:11 PM

@Max Oh. I see. We are using a slightly different notion of the domain and codomain of a function then. If $f:A\to B$ be a function with range say $C\subseteq B$ and $g:A\to C$ is a function. Even though we normally treat them as being the same function, here we don't because the codomains of them are different.

Max

That's not even the point

Take H = {e} so there is only one group structure on H; then all the hom-sets hom((G,), (H,.)) will be *equal and non empty, notdepending on *

So they will not be disjoint, they will only depend on G, the set

user131753

Let $H=\{e\}$ and let $(H,\ast)$ be the group structure on $H$. Then you say that $\text{hom}((G,\circ),(H,\ast))$ will be equal for all group structure $\circ$ on $G$, right?

Max

Yes

user131753

And your point is since is it possible to give different group structures on $G$ but even then only one group homomorphism from $(G,\circ)$ to $(H,\ast)$ all of which has the same underlying function, so even though the $\mathbf{Grp}$-morphisms must be distinct (and hence their $\text{hom}$-sets disjoint), this isn't the case. Am I with you so far @Max?

Max

I don't see why you say "hence". They're different group structures, i.e. different groups, that's why the hom-sets must be disjoint

user131753

2:21 PM

@Max Yes. I agree. That's what I wanted to say.

Max

Then, yes, so far that's right

user131753

I see. I get your point now.

Max

Great ! Do you see how to fix this, how to change the definition of \hom_{Grp}((G,*),(H,.)) to account for this ? And to see why it can't be a subset of \hom(G,H) ? (though it can be pretty close; but then again, that's the case for any concrete category (A,U) over any category X)

Exercise : let (A,U) be a concrete category (so with disjoint hom-sets) over X. Prove that (A,U) is concretely isomorphic to a concrete category (B,V) that satisfies for all C,D, \hom_B(C,D) \subset \hom_X(VC,VD)\times \{(C,D)\}. Try to understand why this is "the best you can get". Unwrap what this means for group for instance

user131753

@Max What is $(C,D)$?

Max

just a pair, first coordinate C, second coordinate D

user131753

2:31 PM

So your $\mathbf{B}$-morphisms are of the form $(f,(C,D))$, right?

Max

Yep

user131753

I guess that the construction will be similar to what you have provided earlier.

Max

Yes, it's actually exactly the same, but you add (C,D) to ensure disjointness

user131753

I think that $\text{hom}_{\mathbf{B}}(C,D):=U(\text{hom}_{\mathbf{A}}(C,D))\times\{(C,D)\}$ works.

user131753

But I need to check it rigorously.

Max

2:40 PM

Rather than checking it rigorously (which you have to do if you're not sure) try to understand why it will work (and then write down the details if you still need to)

user131753

@Max Intuitively it seems to work (in fact that's how I came up with this) because as you said "you add $(C,D)$ to ensure disjointness" without tampering with the categorical properties.

user131753

Anyway, I need to go now. Thank you very much once again for your help (and of course, for your patience) @Max.

Max

No problem, hope I helped !

Transcript for

$Mathematics$ General chat with user 170039