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.