Free group on 2 elements is coproduct Z*Z

Conversation started Jun 4, 2015 at 9:55.

Soham Chowdhury

Jun 4, 2015 09:55

How is $F_2 \cong \Bbb Z \ast \Bbb Z$?

The hint provided is that "with due care, the universal property for one turns into the universal property for the other."

Tobias Kildetoft

@SohamChowdhury Well, what are those universal properties?

Soham Chowdhury

For the coproduct, some unique morphism f must make this diagram commute.

Tobias Kildetoft

@SohamChowdhury Ok, so you want to get that from having a similar property on a set with two elements, right?

Soham Chowdhury

Yes.

And the universal property for the coproduct in Grp is this:

Tobias Kildetoft

@SohamChowdhury You mean for the free group

Soham Chowdhury

Jun 4, 2015 10:02

There is a unique group hom $\varphi$ making this commute.

Yeah, sorry.

The free group on A.

Tobias Kildetoft

@SohamChowdhury Ok, so you have a map from each copy of $\mathbb{Z}$ to some group, and you want to show that this gives a unique map from the free group on two objects to that group

Soham Chowdhury

Right.

Tobias Kildetoft

now, what data do you need to get such a unique map?

Soham Chowdhury

I think I'd need the two maps $\Bbb Z \to F(\{x,y\})$.

Tobias Kildetoft

@SohamChowdhury No, what data do you need to ensure a unique map to the group from the free group on two generators?

Soham Chowdhury

Jun 4, 2015 10:05

I'm not sure what you mean by "data" there.

Tobias Kildetoft

@SohamChowdhury Nothing precise. What sort of thing do you need

Soham Chowdhury

Wait.

Let me think for a bit.

skill patrol

By "data" do you mean just "information"? @TobiasKildetoft

Tobias Kildetoft

@skillpatrol Something like that. In category theory, it is often referred to as "data"

Soham Chowdhury

Not really sure. Don't I need the rest of the maps in the diagram? Then I could make a unique map that would make the diagram commute.

Tobias Kildetoft

Jun 4, 2015 10:09

@SohamChowdhury Well, for the free group, the map from $A$ to the free group is part of the free group (in some sense)

so all you need is a map from $A$ to the target group

Soham Chowdhury

Yes.

So these are the "diagonals" in the first diagram?

Tobias Kildetoft

@SohamChowdhury Now, given a map from each copy of $\mathbb{Z}$, do you see how to get a map from $A$ (which has two elements)?

Soham Chowdhury

from the two copies of Z to?

I mean, give it a name.

Tobias Kildetoft

@SohamChowdhury Well, you are trying to show that the free group on two generators is the coproduct of two copies of $\mathbb{Z}$

So let's assume we have a map from each copy to some group $G$

Soham Chowdhury

G, yes.

Tobias Kildetoft

Jun 4, 2015 10:12

Now, we want to show we get a unique map from the free group $F$ on two generators to $G$

Soham Chowdhury

So I need to find a map $A\to G$?

Tobias Kildetoft

(making everythig commute)

right

Soham Chowdhury

or $F(A) \to G$?

which one?

Tobias Kildetoft

@SohamChowdhury $A\to G$ (which is what will ensure a unique map $F\to A$)

uniqye homomorphism I mean

Soham Chowdhury

I'm having a little trouble visualizing all these maps. Are the two diagrams supposed to "join together" in some way?

How does A enter the first diagram?

Tobias Kildetoft

Jun 4, 2015 10:17

@SohamChowdhury Well, that is pretty much the entire idea here, to realize that having a homomorphism from each of two copies of $\mathbb{Z}$ corresponds to having a map from $A$

Soham Chowdhury

to G?

Tobias Kildetoft

@SohamChowdhury Right

Soham Chowdhury

@TobiasKildetoft I was confused when you said "unique map $F\to A$". That's not in the free group diagram. Did you mean $A\to F$ or $F\to G$?

Tobias Kildetoft

@SohamChowdhury I don't think I mentioned any unique maps to $A$

@SohamChowdhury Ohh, woops I did

Soham Chowdhury

I linked the comment.

Tobias Kildetoft

Jun 4, 2015 10:19

I means to $G$

Soham Chowdhury

Ahh.

Now I can think.

So what was the question?

Finding a map $A\to G$?

Tobias Kildetoft

@SohamChowdhury Yes

Andrew Thompson

@MikeMiller A summerschool. I also noted that this can be taken in a somewhat algebraic direction; I want to see where that goes.

Soham Chowdhury

@Tobias nothing's coming to mind.

Tobias Kildetoft

@SohamChowdhury Ok, let's try a simpler example. If you have a homomorphism from $\mathbb{Z}$ to a group $G$, can you find a map from a one-point set to $G$ (in some "natural" way)?

Soham Chowdhury

Jun 4, 2015 10:24

Map that single point to the identity in G?

whatever 0 maps to via the homomorphism, that is.

Tobias Kildetoft

@SohamChowdhury $0$ maps to the identity.

Soham Chowdhury

Yes.

Tobias Kildetoft

So that would be natural, but probably a bad idea as it will not depend on the homomorphism

Soham Chowdhury

What's the problem with that?

Tobias Kildetoft

@SohamChowdhury Well, there would not be much chance of making the various diagrams commute then

Soham Chowdhury

Jun 4, 2015 10:27

What diagrams? I thought this was a completely separate example.

Or are you talking about the case of $F(\{\bullet\})$?

Tobias Kildetoft

@SohamChowdhury It was meant to lead into the original example

Soham Chowdhury

Oh.

Well, I could map the single point to any $g\in G$.

Not sure how "natural" that is.

OrI could map it to whatever element I need to to make the free group diagram commute.

Is that what you're looking for?

Tobias Kildetoft

@SohamChowdhury Well, could you map it to something such that if you only knew the map from the one point to $G$ (i.e. that one point in $G$), you could figure out what the homomorphism from $\mathbb{Z}$ was?

Soham Chowdhury

Well, do I know the map $A\to \Bbb Z$?

Tobias Kildetoft

@SohamChowdhury Yes

felipa

Jun 4, 2015 10:32

@robjohn I see someone liked my question and added a bounty :)

the prob one we were discussing yesterday

do you think your previous method can be applied to it?

Soham Chowdhury

@TobiasKildetoft I'm not sure. If I know the map $A\to Z$, I know that there's a "special element" $n$ in $\Bbb Z$ that $\bullet$ maps to.

Tobias Kildetoft

@SohamChowdhury Right

@SohamChowdhury You do know that $\mathbb{Z}$ is the free group on one generator, right?

Soham Chowdhury

Yes.

Oh.

So that n is just the word "$\bullet$", in a way.

Tobias Kildetoft

right (and it should probably be $1$ or $-1$)

Soham Chowdhury

so if $f(\bullet) = g\in G$, the homomorphism is just $\varphi(n) = g^n$.

if $\bullet$ goes to 1.

because $\varphi(1) = \varphi(j(\bullet)) = g$, and then the fact that $\varphi$ is a hom implies what I wrote above.

(I'm using the notation in the second image I posted)

Ping me when you get back, please.

Tobias Kildetoft

Jun 4, 2015 10:42

@SohamChowdhury Still here

Soham Chowdhury

So was that correct?

Tobias Kildetoft

@SohamChowdhury Yeah

Soham Chowdhury

Okay. :)

How do we get to the two-generator case?

Tobias Kildetoft

@SohamChowdhury Well, now you have two maps from $\mathbb{Z}$, and you want two points

Soham Chowdhury

Oh.

I think I see what to do.

So the two copies of Z take the place of the singleton set, I guess?

Tobias Kildetoft

Jun 4, 2015 10:45

Yeah

Soham Chowdhury

@TobiasKildetoft What two points are you talking about?

Tobias Kildetoft

@SohamChowdhury Well, you want a map from a set with two elements, so that means finding two points

Soham Chowdhury

Okay. $A\to G$, right.

So let $f(x) = $ whatever $0 \in Z_1$ maps to, and $f(y) =$ whatever $0 \in Z_2$ maps to?

Masacroso

I reached the evil level... see my profile xD

Tobias Kildetoft

Not $0$, $1$

Soham Chowdhury

Jun 4, 2015 10:50

Oh, that's the generator.

Right.

skill patrol

The mark of the beast @Masacroso :O

Soham Chowdhury

Still there, @Tobias?

Tobias Kildetoft

@SohamChowdhury Yeah

Soham Chowdhury

So? What next?

Tobias Kildetoft

@SohamChowdhury So we now have a map from $A$ to $G$, right?

Soham Chowdhury

Jun 4, 2015 10:55

Yeah.

Tobias Kildetoft

@SohamChowdhury Which gives us a unique map from $F$ to $G$

Soham Chowdhury

Yes.

And we're done?

Tobias Kildetoft

Yes, because then $F$ satisfies the universal property of the coproduct

Soham Chowdhury

Okay. I'll give myself some time to digest that. Thanks. :)

Conversation ended Jun 4, 2015 at 10:56.

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