Universal Problems

After calculating the fundamental group of the circle, and deducing lot’s of results, where one of them was really imortant, it’s time to gain some new skills. Untill now, we had two main tools for calculating the fundamental group of a topological space:

Check if the space is contractible or simply connected. If it is, it’s fundamental group is trivial
Check if the circle is a deformation retract of the space. If it is, it’s fundamental group is $\mathbb{Z}$ .

However, this is clearly not enough. With those tools, we can only find fundamental groups of really simple spaces.

Therefore, before we continue discussing about topology, we first must exapnd or knowledge in group theory.

What’s the plan?

The next 6 posts (including this one) are going to be all about group theory, we are going to work in a specific pattern:

Defining a universal problem.
Define a new group that will solve this universal problem
Show that this group is indeed a valid solution

Wait… What is even a universal problem? Well, the best way to explain it is to dive right in to the first problem, and we’ll see right away what universal means and what’s the problem.

In this post I am going to present one universal problem, similarly, each of the next posts will present one universal problem, so we’ll see a total of 5.

Trust me, I know that this may not sound so great, since we are here to learn topology where all the cool stuff happens like turing a coffee mug into a donut. But believe me, it is so worth it for two main reasons:

Once we are equipped with all those solutions to the universal problems, you won’t believe how many things we will be able to do! We will be able to calculate the fundamental groups of some pretty complicated spaces, and find some really non-trivial fundamental groups. That’s great since this is our main goal here – calculate fundamental groups in order to classify topological spaces.
This part of group theory is pretty great to be honest. We will see some nice results and special groups that are pretty useful.

Ok, after all this introduction, let’s begin with the first universal problem, and yes, this is the easiet one since you probably already familiar with it’s solution…

The first universal property

All the problems have similar structure. We are playing a game here, every game is different but the goal is similar: Find a group that fits the rules of the game.

Here are the rules for this game:

Let $G$ and $H$ be two arbitrary groups. For every group $K$ , we would like to find two homomorphism – one to $G$ and one to $H$ :

\varphi:K\to G \\ \psi:K\to H

But that’s not really a problem, we can always pick those homomorphism to be trivial and it is a valid solution.

The rules are a bit more comlicated though. I want that all the possible homomorphism from all of the groups will ‘go through’ one universal specific group that we will call $U$ . What do I mean by that? Let’s look at this diagram:

What do we have here? There is the group $K$ with it’s maps, but there is something else. $U$ has it’s own homomorphism to $G$ and $H$ . Moreover, there exist a unique map $L$ (that’s what $\exists !$ stands for) from $K$ to $U$ .

Ok, but how exactly the homomorphism $\varphi,\psi$ are ‘going through’ $U$ ? According to the diagram, they satisfy:

\varphi=g\circ L \\ \psi = h\circ L

So intuitivley, a homomorphism from $K$ to $G$ (or $H$ ) is in fact ‘going through’ $U$ , and from $U$ , it’s going to $G$ . You can think of $U$ as some place the homomorphism must visit first – like $U$ is some sort of a ‘barrier‘.

Note that $K$ can be any group, but $U$ and it’s homomorphisms $g,h$ are the same for any situation – they are universal.

Solving the problem

A solution for the problem is:

Find what $U$ is.
Prove that $U$ is unique
Figure out what the homomorphisms $g,h$ are.
Find the map $L$ for each group $K$ .
Prove that for each $K$ , the map $L$ is unique.

It might look like a lot of work, but in most cases, once we find what $U$ is, the rest will follow immediately – except one thing: the uniqueness of $U$ .

However, the uniqueness is pretty easy to prove and we will prove it in the exact same way in all of the universal problem. So let’s start that:

Uniqueness of the group

Suppose that we have found two groups that satisfy all of the above. Let’s call them $U$ and $U^\prime$ . So for any group $K$ , we have two diagrams:

But we can pick $K$ to be whatever group we want, so let’s pick $K=\color{red} U^\prime$ in the left diagram, and $K=\color{blue} U$ in the right diagram. Now we get:

Ok, now we can use the left diagram to get:

\color{red}g^\prime\color{black}=\color{blue}g\color{black}\circ\color{blue}L \\ \color{red}h^\prime\color{black}=\color{blue}h\color{black}\circ\color{blue}L

Similarly, we can now use the right diagram to get:

\color{blue}g\color{black}=\color{red}g^\prime\color{black}\circ\color{red}L^\prime \\ \color{blue}h\color{black}=\color{red}h^\prime\color{black}\circ\color{red}L^\prime

From all those four equations, we can get two equations:

\color{red}g^\prime\color{black}=(\color{red}g^\prime\color{black}\circ\color{red}L^\prime\color{black})\circ\color{blue}L\color{black} = \color{red}g^\prime\color{black}\circ(\color{red}L^\prime\color{black}\circ\color{blue}L\color{black}) \\ \color{blue}h\color{black}=(\color{blue}h\color{black}\circ\color{blue}L\color{black})\circ\color{red}L^\prime\color{black} = \color{blue}h\color{black}\circ(\color{blue}L\color{black}\circ\color{red}L^\prime\color{black})

From those equations we deduce that:

\color{red}L^\prime\color{black}\circ\color{blue}L\color{black}=Id_{\color{red}U^\prime} \\ \color{blue}L\color{black}\circ\color{red}L^\prime\color{black}=Id_{\color{blue}U}

Therefore, $\color{blue} L$ is an isomorphism between those groups and we get that:

\color{blue}U\color{black}\cong\color{red}U^\prime

As we wanted. But we’ve even got more – we’ve seen that $\color{blue}L$ shows the exact connection between the homomorphisms:

\color{red}g^\prime\color{black}=\color{blue}g\color{black}\circ\color{blue}L \\ \color{red}h^\prime\color{black}=\color{blue}h\color{black}\circ\color{blue}L

The solution

After showing the uniqueness of the solution, we only need to find one. If you think about it, the solution should be pretty clear, we’ll pick $U$ to be the cross product of the groups $G,H$ . That is:

\color{blue}U\color{black}= G\times H

The homomorphisms from it will be the projections on each group:

P_G:G\times H\to G \\ (g,h)\mapsto g

And similarly:

P_H:G\times H\to H \\ (g,h)\mapsto h

Now the diagram looks like:

The only thing left to figure out is what $\color{blue} L$ is? Let’s see what are the options: First, it must satisfy:

\varphi=\color{blue}P_G\color{black}\circ\color{blue}L \\ \color{black}\psi=\color{blue}P_H\color{black}\circ\color{blue}L

Since $L:K\to G\times H$ , we can denote it as:

L=(L_1,L_2)

Now, for every $x\in K$ :

\varphi(x)=\color{blue}P_G\color{black}\circ\color{blue}L\color{black}(x)=\color{blue}P_G\color{black}(L_1(x),L_2(x))=L_1 \\ \psi(x)=\color{blue}P_H\color{black}\circ\color{blue}L\color{black}(x)=\color{blue}P_H\color{black}(L_1(x),L_2(x))=L_2 \\

So we’ve found exactly what $L$ is, and now, of course it is unique – the only choise for $\color{blue}L$ is:

\color{blue}L\color{black}(x)=(\varphi(x),\psi(x))

And that’s it, we’ve found the universal solution! Now we now that every time we are mapping a group to two different groups, those maps have to ‘go through’ the cross product.

Summary

Great! We’ve just solved our first universal problem! We still have 3 more left before we’ll go back to talk about topology. However, the problems are getting harder and harder.

Ok, so that’s it for this post, in the next one, thing will start to get interesting….

What’s the plan?