The punctured plane has fundamental group $\mathbb{Z}$, while punctured ${\mathbb{R}}^n$ is simply connected for $n\ge 3$

For $n=2$, [L1] and [L2] identify ${\pi }_1(P_2,e_0)$ with ${\pi }_1(S^1,e_0)$, and [L3] identifies the latter with $\mathbb{Z}$.

Let $n\ge 3$ and $y\in S^{n-1}$. Since $n-1\ge 2$, [L4] says that $S^{n-1}$ is simply connected, so ${\pi }_1(S^{n-1},y)$ is trivial; [L1] and [L2] therefore make ${\pi }_1(P_n,y)$ trivial.

For an arbitrary $x\in P_n$, the path ${\gamma }_x(t)=((1-t)+t\mathrm{/}\parallel x{\parallel }_2)x$ runs in $P_n$ from $x$ to $r(x)$. Concatenating an endpoint-fixed homotopy with the fixed paths ${\bar{\gamma }}_x$ and ${\gamma }_x$ preserves it, so ${\mathrm{\Phi }}_x([\alpha ])=[{\bar{\gamma }}_x\ast \alpha \ast {\gamma }_x]$ is well defined. The piecewise formula $K(s,t)={\gamma }_x(2s(1-t))$ for $s\le 1\mathrm{/}2$ and $K(s,t)={\gamma }_x(2(1-s)(1-t))$ for $s\ge 1\mathrm{/}2$ contracts ${\gamma }_x\ast {\bar{\gamma }}_x$ to the constant path at $x$; applying the same formula to ${\bar{\gamma }}_x$ contracts ${\bar{\gamma }}_x\ast {\gamma }_x$ at $r(x)$. Hence the product of ${\mathrm{\Phi }}_x([\alpha ])$ and ${\mathrm{\Phi }}_x([\beta ])$ cancels its middle ${\gamma }_x\ast {\bar{\gamma }}_x$ and equals ${\mathrm{\Phi }}_x([\alpha \ast \beta ])$, while $[\delta ]\mapsto [{\gamma }_x\ast \delta \ast {\bar{\gamma }}_x]$ is a two-sided inverse. Thus ${\mathrm{\Phi }}_x:{\pi }_1(P_n,x)\to {\pi }_1(P_n,r(x))$ is an isomorphism, and step 1.2 makes ${\pi }_1(P_n,x)$ trivial.

Given $x,z\in P_n$, follow ${\gamma }_x$ to $r(x)$, a sphere path from $r(x)$ to $r(z)$ supplied by the path-connectedness in [L4], and the reverse of ${\gamma }_z$. This gives a path from $x$ to $z$, so $P_n$ is path-connected. Together with step 2.1, this proves simple connectedness and completes both clauses.