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.

Universal Property of Kernel Question

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In this blog post, we will discuss a category theory question, in the framework of homomorphisms of abelian groups.

Let \phi:M'\to M be a homomorphism of abelian groups. Suppose that \alpha:L\to M' is a homomorphism of abelian groups such that \phi\circ\alpha is the zero map. (One example is the inclusion \mu:\ker\phi\to M')

Are the following true or false?

(i) There is a unique homomorphism \alpha_0:\ker\phi\to L such that \mu=\alpha\circ\alpha_0.

(ii) There is a unique homomorphism \alpha_1:L\to\ker\phi such that \alpha=\mu\circ\alpha_1.

It turns out that (i) is false. We may construct a trivial counterexample as follows. Consider L=M=0, and M'=\mathbb{Z}/2\mathbb{Z}. Let \alpha, \phi be both the zero maps. Then certainly \phi\circ\alpha=0. \ker\phi=\mathbb{Z}/2\mathbb{Z}. Then, for any \alpha_0, \alpha\circ\alpha_0(x)=0, and hence is not equals to the the inclusion map \mu.

It turns out that (ii) is true, in fact it is the famous universal property of the kernel, that any homomorphism yielding zero when composed with \phi has to factor through \ker\phi.

First we will prove uniqueness. Let \alpha=\mu\circ\alpha_1=\mu\circ\beta, where \beta is another such map with the property (ii). Then for all x\in L, \mu\alpha_1(x)=\mu\beta(x), which implies \mu(\alpha_1(x)-\beta(x))=0. Since \mu is the inclusion map, this means that \alpha_1(x)-\beta(x)=0 and thus \alpha_1(x)=\beta (x).

Next, we will prove existence. Consider \alpha_1:L\to\ker\phi, \alpha_1(l)=\alpha(l). Note that \phi(\alpha(l))=0 by definition thus \alpha(l)\in\ker\phi.

Next we prove it is a homomorphism. \alpha_1(l_1l_2)=\alpha(l_1l_2)=\alpha(l_1)\alpha(l_2)=\alpha_1(l_1)\alpha_1(l_2).

Finally by construction it is easy to see that \mu\alpha_1(l)=\mu\alpha(l)=\alpha(l) for all l\in L.