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\begin{center}
Math 205B - Topology\\
\emph{ }\\
Dr. Baez\\
\emph{ }\\
January 26, 2007\\
\emph{ }\\
Christopher Walker\\
\emph{ }\\
\end{center}
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\begin{exercise}\rm{
Generalize the proof of Theorem 54.5 to show that the fundamental group of the torus is isomorphic to the group $\mathbb{Z}\times \mathbb{Z}$.}
\end{exercise}
\begin{proof}
Recall that a covering map of $S^{1}$ by $\mathbb{R}$ is the map $p:\mathbb{R}\to S^{1}$ given by $p(x)=e^{2\pi ix}$. The torus is the set $T=S^{1}\times S^{1}$. By Theorem 53.3 we have that the map $p\times p:\mathbb{R}\times\mathbb{R}\to S^{1}\times S^{1}$ is a covering map, since $p$ is a covering map. Let $e_0=(0,0)\in \mathbb{R}\times \mathbb{R}$ and let $b_0=p(e_0)$. This gives us that $p^{-1}(b_0)$ is the set $\mathbb{Z}\times \mathbb{Z}$. Since $\mathbb{R}$ is simply connected, then $\mathbb{R}\times \mathbb{R}$ is simply connected, and we have a bijective lifting correspondence
$$\phi:\pi_1(S^{1}\times S^{1},b_0)\to \mathbb{Z}\times \mathbb{Z}.$$
We now prove that $\phi$ is a homomorphism. Let $[f],[g]\in \pi_1(S^{1}\times S^{1}, b_0)$ and let $\tilde{f}$ and $\tilde{g}$ by their liftings to paths in $\mathbb{R}\times \mathbb{R}$ begininning at $(0,0)$. If we let $(a,b)=\tilde{f}(1)$ and $(c,d)=\tilde{g}(1)$, then by definition $\phi([f])=(a,b)$ and $\phi([g])=(c,d)$. Now consider the path $\tilde{\tilde{g}}$ of $\mathbb{R}\times \mathbb{R}$ given by
$$\tilde{\tilde{g}}(s)=(a,b)+\tilde{g}(s).$$
The path $\tilde{\tilde{g}}$ is a lifting of $\tilde{g}$ since for all $(x_1,x_2)\in \mathbb{R}\times\mathbb{R}$ we have:
$$\begin{array}{rl}
(p\times p)((a,b)+(x_1,x_2)) & =(p(a+x_1),p(b+x_2))\\
& =(p(x_1),p(x_2))\\
& = (p\times p)((x_1,x_2))\\
\end{array}$$
and $\tilde{\tilde{g}}$ is a path beginning at $(a,b)$. This means the product $\tilde{f}*\tilde{\tilde{g}}$ is defined, and it is a lifting of $f*g$ that begins at $(0,0)$. This path ends at $\tilde{\tilde{g}}(1)=(a+c, b+d)$. We can then show that $\phi$ preserves the group operation:
$$\begin{array}{rl}
\phi([f]*[g]) & = \phi([f*g])\\
& = (a+c, b+d)\\
& = (a,b)+(c,d)\\
& = \phi([f])+\phi([g]) \\
\end{array}$$
and so $\phi$ is a homomorphism. Since $\phi$ is also bijective, then
$$\pi_1(S^{1}\times S^{1},b_0)\im \mathbb{Z}\times \mathbb{Z}.$$
\end{proof}
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\begin{exercise}\rm{
Suppose that you are given the fact that for each $n$, there is no retraction $r:B^{n+1}\to S^n$. Prove the following:
\begin{enumerate}
\item The identity map $\iota:S^n\to S^n$ is not nulhomotopic.
\item The inclusion map $j:S^n\to \mathbb{R}^{n+1}-\textbf{0}$ is not nulhomotopic.
\item Every non-vanishing vector field on $B^{n+1}$ point directly outward at some point of $S^n$, and directly inward at some point of $S^n$.
\item Every continuous map $f:B^{n+1}\to B^{n+1}$ has a fixed point.
\end{enumerate}
}
\end{exercise}
\begin{proof}\emph{ }\\
\emph{ }\\
(a) First, the fact that there is no retraction $r:B^{n+1}\to S^n$ tells us that there is no extension of $\iota$ to a map $B^{n+1} \to S^{n}$. Now assume $\iota$ is nulhomotopic. Let $H:S^{n}\times I\to S^{n}$ be a homotopy between $\iota$ and a constant map $c$. Let $\pi: S^{n}\times I\to B^{n+1}$ be the map
$$\pi(x,t)=(1-t)x.$$
$\pi$ is a quotient map since it is continuous, surjective, and closed. Also, $\pi$ has the property that $S^{n}\times \{1\}$ goes to zero, and $\pi$ is injective on the rest of the domain. Since $H$ is also constant on $S^{n}\times \{1\}$ then it induces, through $\pi$, a continuous map $f:B^{n+1}\to S^{n}$ that is an extension of $\iota$ (a contradiction). Thus $\iota$ is not nulhomotopic.\\
\emph{ }\\
(b) Assume the inclusion map $j$ is nulhomotopic. If we consider the map $r:\mathbb{R}^{n+1}-\textbf{0}\to S^n$ given by the vector function $r(x)=\frac{x}{\left\|x\right\|}$, this is a retraction since for $y\in S^n$ we have $\left\|y\right\|=1$. We thus have that the inclusion map $\iota$ from part (a) is given by $\iota=r\circ j$. Since $j$ is nulhomotopic there exists a constant map $c:S^{n}\to \mathbb{R}^{n+1}-\mathbf{0}$ such that $j\simeq c$. This gives us by Exercise 51.1 that $\iota=r\circ j\simeq r\circ c$, but $r\circ c$ is a constant map, so $\iota$ is nulhomotopic (a contradiction to part (a)). Thus $j$ is not nulhomotopic.\\
\emph{ }\\
(c) We have a vector field on $B^{n+1}$ given by ordered pairs $(x,v(x))$ where $v$ is a continuous map $v: B^{n+1}\to \mathbb{R}^{n+1}$. To say that this vector field is non-vanishing implies $v(x)\neq 0$ for all $x\in B^{n+1}$. This tells us that we actually have $v: B^{n+1}\to \mathbb{R}^{n+1}-\textbf{0}$. Now since the identity map on $B^{n+1}$ is nulhomotopic and $R^{n+1}-\mathbf{0}$ is path connected, then by Exercise 51.3 the set of homotopy class of maps from $B^{n+1}\to R^{n+1}-\mathbf{0}$ contain a single element. This means that all of these maps are nulhomotopic, and in particular $v$ is nulhomotopic. So we get that the restriction of $v$ to $S^{n}$ (say $w$) is also nulhomotopic. \\
\par\indent We will now proceed by contradiction. Assume that $v(x)$ does not point directly inward at any point $x\in S^n$. We will show that this gives us that $w$ is homotopic to the inclusion map $j: S^n\to \mathbb{R}^{n+1}$. Consider the straight line homotopy define by:
$$F(x,t)=tx +(1-t)w(x),$$
For all $x\in S^n$. We need that $F(x,t)\neq 0$ for any value of $t$ to verify this homotopy is continuous. $F(x,0)\neq 0$ since $w(x)\neq 0$ for all $x$. also $F(x,1)=x\neq 0$ since $0\notin S^n$. Now if $F(x,t)=0$ for some $00$, a contradiction. Therefore there is at least one point $x\in B^{n+1}$ with $f(x)=x$.
\end{proof}
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