Covering space projection induces isomorphisms

Proposition 4.1 (from Hatcher): A covering space projection p:(\tilde{X},\tilde{x}_0)\to (X,x_0) induces isomorphisms p_*:\pi_n(\tilde{X},\tilde{x}_0)\to\pi_n(X,x_0) for all n\geq 2.

We will elaborate more on this proposition in this blog post. Basically, we will need to show that p_* is a homomorphism and also bijective (surjective and injective).

Homomorphism

p_*([f]):=[pf]

p_*([f]+[g])=[p(f+g)]

p(f+g)(s_1,s_2,\dots,s_n)=\begin{cases}pf(2s_1,s_2,\dots,s_n)&s_1\in[0,\frac 12]\\    pg(2s_1-1,s_2,\dots,s_n)&s_1\in[\frac 12,1]    \end{cases}

p_*[f]+p_*[g]=[pf]+[pg]

(pf+pg)(s_1,s_2,\dots,s_n)=    \begin{cases}pf(2s_1,s_2,\dots,s_n)&s_1\in[0,\frac 12]\\    pg(2s_1-1,s_2,\dots,s_n)&s_1\in[\frac 12,1]    \end{cases}, which we can see is the same.

Thus, p_* is a homomorphism.

Surjective

For surjectivity, we need to use a certain Proposition 1.33: Suppose given a covering space p:(\tilde{X},\tilde{x}_0)\to (X,x_0) and a map f:(Y,y_0)\to (X,x_0) with Y path-connected and locally path-connected. Then a lift \tilde{f}:(Y,y_0)\to (\tilde{X},\tilde{x}_0) of f exists iff f_*(\pi_1(Y,y_0))\subset p_*(\pi_1(\tilde{X},\tilde{x}_0)).

Let [f]\in\pi_n(X,x_0), where f:(S_n,s_0)\to(X,x_0), n\geq 2. Since S^n is simply connected for n\geq 2, \pi_1(S_n,s_0)=0. Thus f_*(\pi_1(S_n,s_0))=0\subset p_*(\pi_1(\tilde{X},\tilde{x}_0)). By Proposition 1.33, a lift \tilde{f}:(S_n,s_0)\to (\tilde{X},\tilde{x}_0) of f exists, where p\tilde{f}=f.

i.e. we have \boxed{p_*[\tilde{f}]=[p\tilde{f}]=[f]}. Hence p_* is surjective.

Injective

Let [\tilde{f}_0]\in\ker p_*, where \tilde{f}_0:I^n\to \tilde{X} with a homotopy f_t:I^n\to X of f_0=p\tilde{f}_0 to the trivial loop f_1.

By the covering homotopy property (homotopy lifting property), there exists a unique homotopy \tilde{f}_t:I^n\to \tilde{X} of \tilde{f}_0 that lifts f_t, i.e. p\tilde{f}_t=f_t. There is a lifted homotopy of loops \tilde{f}_t starting with \tilde{f}_0 and ending with a constant loop. Hence [\tilde{f}_0]=0 in \pi_n(\tilde{X},\tilde{x}_0) and thus p_* is injective.

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About mathtuition88

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