\documentclass[11pt,reqno]{amsart}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{mathrsfs}
%% set margins
\topmargin=0in
\oddsidemargin=0in
\evensidemargin=0in
\textwidth=6.5in
\textheight=9in
\newcommand{\R}{\mathbf{R}}
\theoremstyle{definition}
\newtheorem{exercise}{Exercise}
\begin{document}
\begin{center}
{\Large Exercises from February 11th} \\[.3em]
{\small 18.904 Spring 2011}
\end{center}
\vskip 1ex
These are exercises from Marcel's and Danny's talks. Solutions are on the following page.
\begin{exercise}[Marcel]
\label{marcel1}
Prove that any convex subset $X$ of $\R^n$ is contractible.
\end{exercise}
\begin{exercise}[Marcel, adapted from Ch.~2 Ex.~2 of Hatcher]
\label{marcel2}
Construct an explicit homotopy between $\R^n - \{0\}$ and the subset $S^{n-1}$ which is constant on $S^{n-1}$.
This is an example of a ``deformation retraction,'' a special case of homotopy. Use this to prove that $\R^n-\{0\}$
is not contractible.
\end{exercise}
\begin{exercise}[Danny, adapted from Hatcher Ex. 1.1.14]
\label{danny1}
Let $X$ and $Y$ be spaces containing the points $x_0$ and $y_0$ respectively.
Find (with proof) an explicit isomorphism between
$\pi_1(X \times Y, (x_0, y_0))$ and $\pi_1(X, x_0) \times \pi_1(Y, y_0)$.
\end{exercise}
\begin{exercise}[Danny, adapted from Munkres 60.1]
\label{danny2}
Find the fundamental group of the ``solid torus'' consisting of the torus plus its interior (as in a solid donut).
\end{exercise}
\newpage
\par\noindent{\it Solution to Exercise~\ref{marcel1}}:
We show that $X$ is contractible by providing a homotopy between $X$ and a point $x_0$ in $X$. We claim that
the map $f : I \times X \to X$ defined by $f_t(x) = tx_0 + (1 - t)x$ is such a homotopy. We see that
$f_0(x) = x$ and $f_1(x) = x_0$. This map is continuous because it is a linear addition of vectors. Finally,
this map is well defined because it defines a straight line between $x$ and $x_0$, which by our assumption of
$X$ being convex, is contained entirely in $X$.
\vskip\baselineskip
\par\noindent{\it Solution to Exercise~\ref{marcel2}}:
The map $f_t(x) = x/|x|^t$ works. $f_0(x) = x$ and $f_1(x) = x/|x|$, which normalizes the length of every vector
to 1 so that it is in $S^{n-1}$. For $x$ in $S^{n-1}$, $|x| = 1$ so $f_t(x) = x$. This map is continuous because
it is a composition of algebraic functions. It is well defined because $|x|$ is never 0. Because $\R^n - \{0\}$
is homotopic to the space $S^{n-1}$, which is not contractible, the space $\R^n - \{0\}$ is not contractible.
\vskip\baselineskip
\par\noindent{\it Solution to Exercise~\ref{danny1}}:
Let $f$ be a loop in $X \times Y$ based at $(x_0, y_0)$ and $p_i$ be the projection map from $X \times Y$ on the $i$th
coordinate.
We claim that $\phi: \pi_1(X \times Y, (x_0, y_0)) \rightarrow \pi_1(X, x_0) \times \pi_1(Y, y_0)$ which defined by
$[f] \mapsto ([p_1 \circ f], [p_2 \circ f])$ is the desired isomorphism. We need to show:
\begin{enumerate}
\item $\phi$ is well-defined. For this we need to show that if $\tilde{f}$ is another loop with the same base point as $f$
for which $f \simeq \tilde{f}$, then $p_i \circ f \simeq p_i \circ \tilde{f}$ for each $i$. First notice that since
$f(0) = f(1) = \tilde{f}(0) = \tilde{f}(1) = (x_0, y_0)$, then certainly $p_i \circ f(0) = p_i \circ \tilde{f}(0) = x_0$ or $y_0$
(depending on $i$), and the same holds for $f(1)$. Moreover, projections are continuous and compositions of continuous maps
are continuous, so the projections are continuous. This tells us that $p_i \circ f$ and $p_i \circ \tilde{f}$ are loops in
$X$ and $Y$ based at $x_0$ or $y_0$ (depending on $i$).
Now let $F(t,s)$ be a homotopy with $F(0,s) = f(s)$ and $F(1, s) = \tilde{f}(s)$. $\pi_i \circ F$ is continuous since $F$
is, and we have both $\pi_i \circ F(0,s) = \pi_i \circ f(s)$ and $\pi_i \circ F(1, s) = \pi_i \circ(\tilde{f}(s))$ by
construction. Thus $\pi_i \circ F$ is a homotopy from $\pi_i \circ f$ to $\pi_i \circ \tilde{f}$.
\item $\phi^{-1}$ is well-defined. For this we need to show that if $g \simeq \tilde{g}$ are loops in $X$ and $h \simeq
\tilde{h}$ are loops in $Y$, then $(g, h) \simeq (\tilde{g}, \tilde{h})$. It is an elementary fact of point-set topology that if $g: Z \rightarrow X$ and
$h: Z \rightarrow Y$ are continuous, then the map $f:Z \rightarrow X \times Y$ defined by $f(z) = (g(z), h(z))$ is continuous.
This allows us to reverse the logic in our previous proof to get the desired homotopy (treating the homotopies from
$g$ to $\tilde{g}$ and $h$ to $\tilde{h}$ as component functions of a new homotopy).
\item $\phi$ is a homomorphism. For this we have:
\begin{eqnarray*}
\phi([f] \cdot [g]) &=& \phi([fg])\\
&=& ([p_1 \circ (fg)], [p_2 \circ (fg)]\\
&=& ([p_1 \circ f][p_1 \circ g],[p_2 \circ f][p_2 \circ g]) \\
&=& ([p_1 \circ f], [p_2 \circ f])([p_1 \circ g], [p_2 \circ g]) = \phi([f]) \phi([g])
\end{eqnarray*}
\end{enumerate}
\vskip\baselineskip
\par\noindent{\it Solution to Exercise~\ref{danny2}}:
We can write the solid torus as $S^1 \times D^2$. Hence the solid torus has fundamental
group isomorphic to $\pi_1(\mathbb{Z}) \times \pi_1(1) \approx \mathbb{Z}$.
\end{document}