## 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 $l\in L$.