Showing the Fundamental Group of S^1 is isomorphic to the integers

[trmukerji14]

Hi,

I am reading J.P. May's book on "A Concise Course in Algebraic Topology" and have approached the calculation where $\pi$$_{1}$(S$^{1})$$\cong$Z

He defines a loop f$_{n}$ by e$^{2\pi ins}$

I want to show that [f$_{n}$][f$_{m}$]=[f$_{m+n}$]

I understand this as trying to find a homotopy between f$_{n}$*f$_{m}$ and f$_{m+n}$

I have some attempts some attempts which have been unsuccessful are

H(s,t)= f$_{n+mt}$*f$_{m(1-t)}$
H(s,t)={e$^{2\pi in2st}$e$^{2\pi im2s(1-t)}$ for s in [0,1/2]
{e$^{2\pi im(2s-1)t}$e$^{2\pi in(2s-1)(1-t)}$ for s in [1/2,1]

Any help would be very much appreciated on my part.

 

[rasmhop]

Your idea seems correct. To separate into cases depending on whether s is smaller or larger than 1/2 is a good idea in this problem (as you have done).

However try to think about what you are doing. $f_{m+n}$ loops with a speed of (m+n) for the whole interval [0,1]. $f_n * f_m$ instead first loops with speed 2m for [0,1/2], and then loops with speed 2n for [1/2,1].

Let us first focus on what our homotopy should be for $s \in [0, 1/2]$. In this case we start with a speed of 2m and would like to end up with a speed of m+n. In other words for the part $s\in[0,1/2]$ we would like to have
$$e^{2\pi i 2 m s}$$
when t = 0 and
$$e^{2\pi i (m+n) s}$$
when t = 1. In your case when we plug in t=0 we get
$$e^{2\pi i m 2s}$$
as we wanted, but when we plug in t=1 we get
$$e^{2\pi i n 2 s}$$
but we wanted
$$e^{2 \pi i (m+n) s}$$
Can you figure out how to correct this? The same problem happens when $s \in [1/2,1]$, so you must also correct your homotopy here.

If you think you have done this, then all that remains is to check that this is a homotopy between the correct loops and that your two expressions agree on the intersection $s = 1/2$.

As a side remark the homotopy you constructed instead shows
$$[f_n][f_m] = [f_m][f_n]$$
which will follow from what you want to prove eventually, but it is also interesting if you couldn't figure out how to prove the general result.

 

[trmukerji14]

Thank you for the help.

So far what I have is for s in [0, 1/2]

H(s,t)= e$^{2\pi i(m+n)st}$e$^{2\pi im2s(1-t))}$

and for s in [1/2, 1], I have:

H(s,t)= e$^{2\pi i(m+n)st}$e$^{2\pi in(2s-1)(1-t))}$

This satisfies our conditions for a homotopy for t=0 and for t=1.

Unfortunately, when we consider s it fails miserably. For a basepoint of 1. We have that H(0,t)=1 and that is good but when we try to plug in s=1/2 or s=1 we do not get what we want.

So I'm scratching my head at the moment trying to fix this.

As for the side remark, I understand that if we show this equivalence we can define a homomorphism from the integers to the fundamental group of the circle which will give us the abelian property. But I don't think I showed the abelian property with the homotopy I had since that one fails also when s=1.

Thanks for the help again.

 

[rasmhop]

There are basically still 3 issues with your homotopy:
1) You forgot an extra coefficient of 2 in front of n in your H for s >= 1/2.
2) It doesn't line up on s=1/2, i.e. the two expressions you gave are different when s=1/2
3) It isn't a homotopy of loops (i.e. not a based homotopy).
1 is easily fixable and I assume you can easily do this. If you solve 2, then 3 should be solved automatically since your speeds gives you whole revolutions (this is obviously not a rigorous argument and you will need to check, but intuition never hurts).

Fixing 1 for you I assume you wrote down:
$$H(s,t) = \begin{cases} e^{2\pi i (m+n) st}e^{2\pi i m 2 s (1-t)} & \textrm{for }s \in [0,1/2] \\ e^{2\pi i(m+n)st}e^{2\pi i n 2 (2s-1)(1-t)} & \textrm{for }s \in [1/2,1] \end{cases}$$
Let us focus on fixing the fact that this is not well-defined for s=1/2. Ideally we would like some continuous function $\alpha : [0,1] \to \mathbb{R}$ such that
$$H(s,t) = \begin{cases} e^{2\pi i (m+n) st}e^{2\pi i m 2 s (1-t)} & \textrm{for }s \in [0,1/2] \\ e^{2\pi i(m+n)st}e^{2\pi i n 2 (2s-1)(1-t)}e^{2\pi i \alpha(t)} & \textrm{for }s \in [1/2,1] \end{cases}$$
would work as a homotopy. Note that this does not change the "speeds" we work with, but just where around the circle the second case takes over when s reaches 1/2, so as long as $\alpha$ is well-chosen this should let us fix the problem at s=1/2. Clearly for this to work you must have (up to addition by an integer)
$$(m+n)\frac{1}{2}t + 2m\frac{1}{2}(1-t) = (m+n)\frac{1}{2} t + 2(1-t) + \alpha(t)$$
This should give you an expression for $\alpha$ which will let you fix your homotopy, and then you should get a well-defined homotopy of loops.

 

[trmukerji14]

I got it! This is awesome.

I got $\alpha(t)$=m(1-t) which gave the well defined homotopy

$$H(s,t) = \begin{cases} e^{2\pi i (m+n) st}e^{2\pi i m 2 s (1-t)} & \textrm{for }s \in [0,1/2] \\ e^{2\pi i(m+n)st}e^{2\pi i n 2 (2s-1)(1-t)}e^{2\pi i m(1-t)} & \textrm{for }s \in [1/2,1] \end{cases}$$

This almost worked except when s=1, we did not get our desired result of 1. So if we use this:

$$H(s,t) = \begin{cases} e^{2\pi i (m+n) st}e^{2\pi i m 2 s (1-t)} & \textrm{for }s \in [0,1/2] \\ e^{2\pi i(m+n)st}e^{2\pi i n (2s-1)(1-t)}e^{2\pi i \alpha(t)} & \textrm{for }s \in [1/2,1] \end{cases}$$

We have our homotopy of loops.

Thanks again for the help.I was really stuck out there.