# Fundamental group of the circle S^1

• Mar 19th 2009, 07:06 AM
math8
Fundamental group of the circle S^1
The question is to prove that the fundamental group of the circle S^1 is isomorphic to the group of integers under addition.

So I think I should show that the following map Phi is an isomorphism.

Phi: F(S^1, (1,0)) --> Z defined by Phi([f])= f*(1) where f* is the lifting path of f ( pof*=f) and f*(1) is the degree of f and p is the map
p:Reals--> S^1 defined by p(t)=(cos 2 pi t, sin 2 pi t).

I am able to show that Phi is onto, but I am having trouble showing that it is 1-1 and that it is well defined .
• Mar 19th 2009, 04:55 PM
aliceinwonderland
Quote:

Originally Posted by math8
The question is to prove that the fundamental group of the circle S^1 is isomorphic to the group of integers under addition.

So I think I should show that the following map Phi is an isomorphism.

Phi: F(S^1, (1,0)) --> Z defined by Phi([f])= f*(1) where f* is the lifting path of f ( pof*=f) and f*(1) is the degree of f and p is the map
p:Reals--> S^1 defined by p(t)=(cos 2 pi t, sin 2 pi t).

I am able to show that Phi is onto, but I am having trouble showing that it is 1-1 and that it is well defined .

Lemma. For loops f, g in S^1 with base point (1,0), [f]=[g] iff deg(f)=deg(g).

If [f]=[g], then Phi([f])=deg(f)=deg(g)=Phi([g]) by lemma. Thus, Phi is well-defined.
If Phi([f])=Phi([g]), then deg(f) = deg(g). By lemma, if deg(f)=deg(g), then [f]=[g]. Thus, Phi is one-to-one.