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\Large \noindent \underline{\textbf{SAMPLE}} \hfill\underline{\textbf{SAMPLE}}

\scriptsize \noindent 31 May, 2011  \hfill  \tiny [AUTHOR: Maxwell/Rhodes]
\normalsize\bigskip

\centerline{\large\textbf{Topology Comprehensive Exam}} 
\bigskip

Complete {\bf SIX} of the following eight problems.
\bigskip


\begin{enumerate}




\item 
\begin{enumerate}
\item State the Urysohn Lemma.
\item Show that a connected normal space with at least two points is uncountable.
\end{enumerate}

\item Let $A$ be a compact subset of a Hausdorff space $X$.
Show that $A$ is closed. (Your proof must be elementary and may not use
nets.)

\item Suppose $A$ and $B$ are connected subsets of $X$. Suppose $A\cap B\neq \emptyset$.  Show that $A\cup B$ is connected.

\item Let $A$ be a subset of a topological space $X$.  Suppose there exists a continuous function $r:X\ra A$
such that $r(a)=a$ for all $a\in A$ (such a map is called a {\it retraction}
onto $A$).
\begin{enumerate}
\item Show that $r$ is a quotient map.
\item Show that $A$ is closed.
\end{enumerate}

\item 
\begin{enumerate}
\item State the characteristic property of the product topology.
\item Let $\{X_\alpha\}_{\alpha\in A}$ be a family of
topological spaces, and for each $\alpha$ let $W_\alpha\subseteq X_\alpha$. 
Let $X=\prod X_\alpha$ and let $W=\prod W_\alpha$.
Without ever mentioning open or closed sets, prove that 
the product topology on $W$  and the
subspace topology on $W$ (as a subspace of $X$) are the same.
\end{enumerate}	

\item 
\begin{enumerate}
\item Define an $n$-manifold.
\item Show that $\{(x,y)\in\Reals^2:x\neq 0,y=1/x$ is a $1$-manifold\}.
\end{enumerate}

\item Let $B=\{x\in\Reals^2:|x|\le 1\}$, where $|\cdot|$ denotes the
Euclidean norm.  Define an
equivalence relation on $B$ by $x\sim y$ if $|x|=|y|=1$. Prove
that $B/\sim$ is homeomorphic to a familiar space.

\item Exhibit counterexamples (and brief justifications) 
to the following false claims.
\begin{enumerate}
\item Every open map is closed.
\item Connected components are open.
\item Every surjective continuous map is a quotient map.
\item If $A\subseteq X$ is path connected, so is $\overline A$.
\end{enumerate}

\end{enumerate}
\end{document}








\hrule


\item
\begin{enumerate}
\item Define a topological $n$-manifold.
\item Prove that $S^1=\{(x,y)\in\Reals^2:x^2+y^2=1\}$ is a $1$-manifold.  You should be
precise when working with the subspace topology on $S^1$.
\end{enumerate}

\item 
\begin{enumerate} 
\item State the Closed Map Lemma.
\item  Let $X=\{(xy,yz,zx,x^2,y^2,z^2\}\in \Reals^6: x^2+y^2+z^2=1\}$.  
Show that $X$ is homeomorphic to $\RP^2$.  You are free to use well-known facts about $\RP^2$.
\end{enumerate}

\item Suppose that $\pi : X\ra Y$ is a quotient map, that $Y$ is connected, and that each fiber
$\pi^{-1}(y)$ is connected.  Prove that $X$ is connected.

\item Recall that the finite complement topology on a set consists of the empty set together with
the subsets having finite complements.  Consider $\Cplx$ with the finite complement topology $\tau$. 
\begin{enumerate}
	\item Show that $(\Cplx,\tau)$ is not Hausdorff, but is separable.
	\item Show that every polynomial is a continuous map from $(\Cplx,\tau)$ to itself.  Note: The constant polynomials require special treatment.
\end{enumerate}
	
\item Let $G$ be a topological group.  Give a proof using nets that if $A\subseteq G$ is closed and
$B\subseteq G$ is compact then $A\cdot B$ is closed.

\item Let $\{X_\alpha\}_{\alpha \in I}$ be a family of topological spaces, infinitely many
of which are non-compact.  Show that every compact subset of $\prod X_\alpha$ has empty interior.

\item 
\begin{enumerate}
\item State the Urysohn Lemma.
\item Show that a connected normal space with at least two points is uncountable.
\end{enumerate}

\item Suppose $X$ and $Y$ are topological spaces, $Y$ is Hausdorff, and $A$ is a subset of $X$.
Suppose $f:A\ra Y$ is continuous.  Show that $f$ need not have an extension as a continuous
function to the closure of $A$, but that if such an extension exists, it is necessarily unique.

\end{enumerate}
\end{document}