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{\Large Exercises from February 25th} \\[.3em]
{\small 18.904 Spring 2011}
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These are exercises from Noah's and Gabriels's talks. Solutions are on the following page.
\begin{exercise}[Noah, from Munkres, 54.8]
\label{noah1}
Let $p:\tilde{X}\to X$ be a covering map. Show that if $\tilde{X}$ is path connected and $X$ is simply connected, then $p$ is a homeomorphism.
\end{exercise}
\begin{exercise}[Gabriel]
\label{gabriel1}
Let $X=S^1$, $\tilde{X}=\{(e^{ix},x)|x\in\mathbb{R}\}$ for some continuous function $f:\mathbb{R}\to\mathbb{R}$. Show that the standard projection $p:(x,y)\to x$ is a covering space of $X$ only if $f$ monotonic, no-where constant.
\end{exercise}
\newpage
\par\noindent{\it Solution to Exercise~\ref{noah1}}:
As $p$ is a covering map, it suffices to prove that $p$ is injective. Suppose $x_0\in X$, and $\tilde{x_0},\tilde{x_1}\in p^{-1}(x_0)$. As $X$ is path connected, there exists some path $\tilde{f}_0$ in $\tilde{X}$ from $\tilde{x_0}$ to $\tilde{x_1}$. Then as $X$ is simply connected, there exists a path homotopy between the loop $p\tilde{f}_0$ based at $x_0$ and the trivial loop based at $x_0$. By the path homotopy lifting property, there exists a path homotopy between the loop $\tilde{f}_0$ and a lift of the trivial loop. But the only loop of the trivial loop is the trivial loop, so we must have $\tilde{x}_0=\tilde{f}_0(0)=\tilde{f}_0(1)=\tilde{x}_1$, completing the proof.
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\par\noindent{\it Solution to Exercise~\ref{gabriel1}}:
There are several possible ways of pursuing this result. We may, however, choose a fairly simple one: suppose $f$ somewhere constant, that is, $f[a,b]=\tilde{x}$. Then $[a,b]\in p^{-1}(p(\tilde{x}))$, so that $p^{-1}p\tilde{x}$ not discrete, and hence $p:\tilde{X}\to X$ not a covering space.\par
Suppose, that $f$ has an extremal point --- without loss of generality a local maximum. Then there exists a point $a\in \mathbb{R}$, neighborhood $U\ni a$ such that $f(U)$ simply connected and for $b\in U$ $f(b)$ clockwise of $f(a)$ in the standard orientation. In particular, this means we can find $b_1