Proving Brouwer's Fixed Point Theorem

[MUSIC PLAYING] There's a strong interplay between geometry and algebra.

Graphs and shapes in a coordinate plane correspond to algebraic equations, and algebraic equations correspond to geometric features.

It's almost as if there's a bridge or a portal between geometry and algebra.

Well, today I'd like to tell you about an actual mathematical portal.

[MUSIC PLAYING] Analogous to the relationship between geometry and algebra, there is a mathematical portal from a looser version of geometry, topology, to a more sophisticated version of algebra.

This portal can take problems that are very difficult to solve topologically, and recast them in an algebraic light, where the answers may become easier.

For example, take Brouwer's Fixed Point Theorem, a well-known result from topology.

In two dimensions, the theorem is often illustrated with maps.

If you have two copies of a map of the room you're in and you lay one flat, but crumple the second and lay it on top of the first, there's always one point on the crumpled map that sits directly above that corresponding point on the map below.

More formally, the theorem says that every continuous function from a disc to itself has a fixed point.

A point, x, is called a fixed point of a function, g, if g assigns x to itself-- if g of x equals x.

So no matter how you rotate, bend, twist, stretch, or deform a disc or any shape that's topologically equivalent to a disc, like a rectangular map, there's always at least one point that ends up in its original location.

Michael from "Vsauce" has a great video about fixed points, where he explains Brouwer's Fixed Point Theorem in more detail.

But what does this have to do with mathematical portals?

Well, you might wonder why Brouwer's Fixed Point Theorem is true.

How would you prove it?

It's not an obvious fact.

Fortunately, a portal from topology to algebra provides an easy answer, and that's what I'd like to talk about today.

Let's prove Brouwer's Fixed Point Theorem using this portal.

To start, I'll tell you what the portal does, then we'll use it to prove the theorem.

First, notice that there are not one, but two topological spaces in the statement of Brouwer's Fixed Point Theorem.

There is a disc, and then there's the boundary of the disc, which is a circle.

The portal from topology to algebra assigns to each of these shapes an algebraic gadget that keeps track of the loops within the shape.

The set of integers is assigned to the circle because a circle is like a loop, which we can wind around 0, 1, 2, 3 times clockwise or counterclockwise.

And the number 0 is assigned to the disc because any loop within a disc can be shrunk to a point.

Topologically, there are no fundamental loops in a disc.

So to a topologist, a disc is like the number 0 and a circle is like the integers.

By the way, I'm referring to the integers and 0 as algebraic gadgets because they're actually groups.

But don't worry if you're not familiar with group theory.

We won't need it for this episode.

OK, here's the key feature of these assignments.

Any scenario that we can construct between a circle and a disc must correspond to an identical scenario between the integers and the number 0.

In short, whatever happens in the land of topology should be mirrored in the land of algebra.

That's the key.

So let's use this mirror relationship between topology and algebra to prove Brouwer's Fixed Point Theorem.

We'll do it using a proof by contradiction.

In other words, we will not prove the theorem directly.

So we won't identify the fixed point of every possible deformation of a disk.

Instead, we'll prove the theorem indirectly.

If we assume the theorem is false and discover a contradiction, we'll know our assumption must have been wrong, and therefore, the theorem is true.

Here's an outline of the next few minutes.

We'll start by assuming the theorem is false.

So we'll assume there's a continuous function from the disc to itself with no fixed points.

Then we'll use that assumption to construct a scenario between the circle and a disc.

Then, we'll use the portal from topology to algebra to get an analogous scenario between the integers and zero.

Finally, we'll discover that the two scenarios do not match.

This will contradict the key feature of our portal.

Whatever happens in one realm should be mirrored in the other.

All right, let's start.

Suppose Brouwer's Fixed Point Theorem is not true.

That means there is some transformation-- some bending or stretching or twisting of the disc so that no point ends up where it was before the transformation.

In other words, there is a continuous function, g, from the disc to itself so that any point in the disc, x, and the location of that point after the transformation, g of x, are never the same.

Therefore, we can draw an arrow from g of x to x that touches a point on the outside circle.

This assignment defines a new function.

Let's call it h. To every point, x, in the disc, h assigns a point, h of x, on the boundary circle.

And notice if x is already on the circle, then h of x equals x.

Now what happens if we compose the function, h, with another function?

Start with the boring function that assigns a point on the boundary circle to itself.

Then undo the process.

That is, send that point back to itself, this time using the new function, h. This composition defines a function from the circle to the circle, with this special feature that every point is assigned to itself.

This is our topological scenario.

Next, let's use the portal we introduced earlier to obtain the algebraic analog.

We can replace the circles by the integers and the disc by the number 0.

Now here's a very simple observation.

Any function that assigns to each integer the number 0-- so 1 goes to 0, 2 goes to 0, and so on-- composed with a function that assigns 0 to exactly one integer, like minus 3, clearly does not assign each integer to itself.

For example, 1 will go to minus 3, which is not 1.

It's almost like 0 acts as a black hole.

We can't send all the integers to a single number and expect to recover the full set.

We can't undo the process.

So here we have a composition of functions in which each integer is not assigned to itself.

This is our algebraic scenario.

Finally, let's compare it with the scenario from topology.

We started by constructing a composition of functions in which every point on the circle is assigned to itself.

And yet, when we look at the algebraic analog of this same composition where we assign each circle with the integers, there is no way every integer can be assigned to itself.

But a key feature of the portal from topology to algebra is that the two scenarios must match.

But they don't.

And there's the contradiction.

Conclusion-- Brouwer's Fixed Point Theorem must be true.

What's amazing is that the portal from topology to algebra is, itself, mathematics.

I've been calling it a portal, but it goes by a much more sophisticated name.

Number one, it assigns groups like the integers to shapes like circles.

And number two, it preserves the relationships or functions between them.

So instead of calling this assignment or portal a function, it's called a functor.

And now we have entered into a branch of math called category theory.

I won't go into any detail, but category theory, whatever that is, opens up more portals, or functors, between a myriad of other mathematical worlds.

It's very exciting mathematics.

But we'll have to talk about that in a future episode.