While on holiday, I started reading a couple of on-line texts on algebraic topology on my Kindle (Hatcher 2002), (May 1999). Kindle’s aren’t great for reading maths but they save having to carry around heavy books of which one might only read a fraction. Along with the usual holiday activities (ok not that usual as we did cycle into Burma), I made it as far as calculating the fundamental group of the circle, .

The idea is that two continuous functions are **homotopic** if they can be deformed continuously into each other. All functions from now on are considered continuous unless otherwise specified.

More formally if are homotopic then there exists a function such that and .

A **path** is a map where is the unit interval . A homotopy of paths is a function such that and . We write . The two paths and are homotopic and we write . Homotopy of paths is an equivalence relation, a fact which we do not prove here. The equivalence class of a path is denoted .

Where the end point of one path is the start point of another path, we can join these together to form new paths. This joining operation respects homotopy classes as the diagram below illustrates.

A **loop** is a path with .

We can join loops based at the same point together to form new loops giving a group operation on equivalence classes under homotopy. In more detail, suppose that and are loops then we can form the new loop which is the obvious loop formed by first traversing and then traversing . The inverse of the loop is the loop and the identity is the constant loop . The way we have defined homotopy ensures that this is all well defined (see either of the on-line books for the details).

The group of loops under these group operations is called the fundamental group or the first homotopy group and is denoted .

**Theorem** where the latter denotes the integers with the group structure being given by addition.

**Proof** Define

by

that is for each integer we define an equivalence class of loops. In more standard notation (not using as a way of introducing anonymous functions)

where

Further we have and clearly . Thus making a homomorphism.

We wish to show that is an isomorphism. To do this we are going to take a loop on and lift it to a path on so that the map which wraps around the circle projects this lifted map back to the original map: . Since we must have that for some integer (this will be the number of times the loop winds around the origin). Let us call the map that creates this integer from a loop on the circle

Note that we also have so the lifted path could start at any integer. For the sake of definiteness let us start it at .

We now construct this map, , show that it is well defined and is a homomorphism.

We know that is a smooth manifold. Let us give it an atlas.

We define the co-ordinate maps as . The transition maps are then just a translation around the circle. For example

And these are clearly smooth.

By continuity of , every point in is contained in some open interval whose map under is contained in one of the charts. This gives an open cover of . By the compactness of (it is closed and bounded), we therefore have a finite subcover. From this we can construct such that for some and .

Since we could to define . But how do we define beyond ?

We know that

where is a homeomorphism from each to .

So we can equivalently define

where has been chosen to contain .

Given our specific atlas we have

Suppose we have the loop , then we could define

So that

And now we know how to continue. There must be a such that .

So we can define

where has been chosen to contain .

Given our specific atlas we have

With our specific example we could define

So that (again)

Continuing in this fashion, after a finite number of steps we have defined on the entirety of . Note that this construction gives a *unique* path as at each point the value of on is uniquely determined by its value at (and of course by itself).

We still have a problem that if is homotopic to then might not equal .

Assume we have a homotopy then since is compact we can proceed as above and choose a partition of of rectangles

with , , and such that for some .

Thus we can define on as

where has been chosen to contain .

We can continue as before; we know there must be a such that so we can define

where has been chosen to contain for all using the previously defined .

Eventually we will have defined on the whole of . But now we can start the same process for and by choosing to be the value of the previously defined and by the uniquess of lifts of paths we can define on the whole of .

Carrying on in this way we will have defined on the whole of .

and must be constant paths because for all (constant lifted paths get projected onto these and they are unique).

Since we require that , we must have that .

By the uniquess of lifted paths we must then have

Since is constant we must have that . Thus is well defined on homotopy classes.

Recall that so that so that is a homomorphism.

Suppose that then with a homotopy of loops. This lifts to a unique homotopy of paths with and . Since is a homotopy of paths, the end point is fixed. Thus and is injective.

Suppose that is a loop starting at 1 then there is a lift starting at . We also have that by the linear homotopy . Thus and is surjective.

# Bibliography

Hatcher, A. 2002. *Algebraic Topology*. Tsinghua University. http://books.google.co.uk/books?id=xsIiEhRfwuIC.

May, J.P. 1999. *A Concise Course in Algebraic Topology*. Chicago Lectures in Mathematics. University of Chicago Press. http://books.google.co.uk/books?id=g8SG03R1bpgC.