# Fundamental group of the disk is trivial

• Apr 1st 2009, 01:49 PM
math8
Fundamental group of the disk is trivial
How do we show that the fundamental group of the disk D^2={(x,y) in RxR: x^2 +y^2< or eq. to 1} is trivial?

I know how to show that the fundamental group of the circle is isomorphic to the group of the integers under addition, but for some reason, I don't see a way to show that the fundamental group of the disk is trivial.
• Apr 1st 2009, 07:09 PM
aliceinwonderland
Quote:

Originally Posted by math8
How do we show that the fundamental group of the disk D^2={(x,y) in RxR: x^2 +y^2< or eq. to 1} is trivial?

I know how to show that the fundamental group of the circle is isomorphic to the group of the integers under addition, but for some reason, I don't see a way to show that the fundamental group of the disk is trivial.

Lemma 1. A convex subset A in R^n is contractible to each point x_0 in A.

$\displaystyle D^2$ is a convex subset of $\displaystyle R^2$ and is contractible to each point $\displaystyle x_0$ in $\displaystyle D^2$ by lemma 1.

Let X be $\displaystyle D^2$ and $\displaystyle x_0 \in D^2$; let $\displaystyle F:X \times I \rightarrow X$ be a contraction such that

$\displaystyle F(x, 0) = x, F(x, 1)= x_0, F(x_0, t)=x_0, x \in X, t \in I$.

For [a] in pi_1(X, x_0), define a homotopy $\displaystyle H:I \times I \rightarrow X$ by

$\displaystyle H(t,s) = F(a(t), s), (t,s) \in I \times I$.

A contraction F ensures that H is a homotopy between H( . , 0) = a and H(. , 1) = c, which is a constant loop at x_0. Thus, [a] = [c]. We conclude that pi_1(X, x_0) is a trivial group.