Math4202 Topology II (Lecture 15)

Algebraic Topology

Fundamental group of the circle

Recall from previous lecture, we have p:\mathbb{R}\to S^1 by x\mapsto e^{2\pi ix}.

We want to study the relation between the paths in \mathbb{R} starting at 0 and the loops in S^1 at 1.

Definition for lift

Let p:E\to B be a map. If f is a continuous map from X\to B, a lifting of f is a map \tilde{f}:X\to E such that p\circ \tilde{f}=f

A natural question is, whether lifting always exists? and how many of them (up to homotopy)?

Back to the circle example, we have f:I\to S^1, representing a loop, and p:\mathbb{R}\to S^1, by p(x)=e^{2\pi ix}.

Lemma for unique lifting for covering map

Let p: E\to B be a covering map, and e_0\in E and p(e_0)=b_0. Any path f:I\to B beginning at b_0, has a unique lifting to a path starting at e_0.

Back to the circle example, it means that there exists a unique correspondence between a loop starting at (1,0) in S^1 and a path in \mathbb{R} starting at 0, ending in \mathbb{Z}.

Idea for Proof

Starting at b_0, by the covering map property, there exist some open neighborhood U_0 of b_0 such that V_0=p^{-1}(U_0) is a neighborhood of e_0. And p|_{V_0} is a homeomorphism on to U_0.

Since f is continuous, then f^{-1}(U_0) is open in I and we can find some small open neighborhood [0,s_1], such that f^{-1}([0,s_1])\subset V_0.

Then we define \tilde{f}:[0,s_1]\to E, by \tilde {f}(t)=(p|_{V_0})^{-1}\circ f.

Continue with compactness property... Continue on Wednesday.

Lemma for unique lifting homotopy for covering map

Let p: E\to B be a covering map, and e_0\in E and p(e_0)=b_0. Let F:I\times I\to B be continuous with F(0,0)=b_0. There is a unique lifting of F to a continuous map \tilde{F}:T\times I\to E, such that \tilde{F}(0,0)=e_0.

Further more, if F is a path homotopy, then \tilde{F} is a path homotopy.

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