Question

6. Let (X,p) be a covering space of X, let io e X and Xo = plão). (a) Show that if (X.p) is a universal covering space of X, i.e. X is simply connected, then there is a bijective function : 72(X, 10) --'(70). (b) Use (a) to give a proof of the fact (S',(1,0)) is isomorphic to Z.

Answer 1

6.

Let be a covering map.

p: (X,00) + (X, 10)

(a)

If
f
is a path in starting at
07
, then define $\tilde{f}$ to be corresponding lifting to a path in $\tilde{X}$ beginning at $\tilde{x}_0$ .

Claim:  If
f
is path homotopic to in , then $\tilde{f}$ is path homotopic to $\tilde{g}$ in $\tilde{X}$ and hence $\tilde{f}(1)=\tilde{g}(1)$

= TO Proof. Let F :IXI + X be the path homotopy between f and g, then F(0,0) Let F :IXI + be the lifting of F to 8 such that F(0,0) = čo. Since is a path homotopy, therefore F(0 x 1) = {ão} and F(1 x 1) = {ğı}. The restriction f\I x 0 is a path in X beginning at ło that is a lifting of F\I x 0(= f). Since lifting of a path is unique, we have F(8,0) = f(s). Similarly, the restriction f\I x 1 is a path in beginning at to that is a lifting of F\I x1(=9). Since lifting of a path is unique, we have F(s, 1) = g(s). Therefore, F is a path homotopy between f and ğ and both end at the same point či
Define a map $\phi:\pi_1(X,x_0) \rightarrow p^{-1}(x_0)$ by $\phi([f])=\tilde{f}(1)$ .

This map is well defined since if
f
is path homotopic to , i.e. $[f]=[g] \in \pi_1(X_0,x_0)$ , then $\tilde{f}$ is path homotopic to $\tilde{g}$ and hence $\tilde{f}(1)=\tilde{g}(1)$ .

Since $\tilde{X}$ is simply connected, we have $\tilde{X}$ is path connected.

Given ſi ep-'10), there is a path f in X from ło to ži. Then po f = f is a loop in X based at to, and ©( [f]) = čı. Therefore, o is surjective. Let (f), (g) € 71(X,Co) such that o( [f]) = $([9]). Let f and ĝ be the liftings of f and g in ï that begins at ito, since o([f]) = (([9]) we have f(1) = g(1). Since E is simply connected, there is a path homotopy F between f and ğ. Then po F is a path homotopy between f and g. Thus, [f] = [g] € 11(X, 20). Therefore, o is injective. Thus, o is bijective.

(b)

Let p : R + Sl be the covering map p() (cos(27x), sin(272)), let to 0 and 20 = pło) = (1,0). Then p-20) is the set Z of integers. Since R is simply connected, the lifting correspondence 0 : 71(S1, x0) + Z is bijective. Given [f] and [g] in 71(S1, xo), let f and ğ be their respective liftings in R beginning at 0. Let n = f(1) = o( [f]) and m = g(1) = 0([9]), then define ħ(s) = n +ğ(s) on R. The path ñ is the lifting of g beginning at n. Then the product f* ħ is defined, and it is the lifting of f*g that begins at 0. Since ħ(1) = n +m, therefore o( [f] * [g]) = n +m ( [f]) + ([g]). Therefore, o is a homomorphism and hence an isomorphism.