\documentclass[10pt]{article}
\usepackage{latexsym,amsfonts,amsmath}
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}

\newcommand{\T}{\mathbb{\tau}}

\newcommand{\x}{\underline{x}}
\newcommand{\y}{\underline{y}}
\begin{document}
\thispagestyle{empty}

\begin{center}
{\Large Intro to Topology -- HW 3.}
\end{center}

\bigskip
\smallskip


\noindent
\begin{enumerate}

\item Give an argument that shows: for a top space $(X, \T_X)$ and a subset $A\subset X$ with the subspace topology $\T_A$ we have $V\subset A$ is closed
wrto $\T_A$  if and only if there is a $W\subset X$ that is
closed wrto $\T_X$ and $V=W\cap A$.

\item  Read the three excerpts from "Messer and Straffin", "Munkres" as well as "Weeks"about quotient topologies and the torus/Mobius strip/projective plane case. These are posted separately.
\item  Verify that the following limerick is true 

\begin{quote}
A matematician named Klein\newline
Thought the M\"obius band was divine\newline
Said he, "If you glue\newline
the edges of two,\newline
you get a weird bottle like mine."
\end{quote}

(i.e.
the Klein bottle can be obtained by gluing two Mobius strips along their bounding circles.)


\item Consider the unit square and all combinatorially possible {\it pairwise} identifications
of its edges (up to rotation and mirror images). 

Identify what you got in each case, topologically. 

{\it (No formal proofs are necessary,
however, you may have to use a cut-and-paste argument on a diagram,  to recognize what
you got.)}

\item a.) Consider the  hexagon $X$ whose edges are identified according to the word $abca^{-1}b^{-1}c^{-1}$. 

The identification
provides a quotient space $(X/\sim, \tau_{x/\sim})$.

i.) How are the vertices of the original hexagon identified in this space?  Label the vertices and write it clearly which vertices of the original hexagon are in the same equivalence class.

{\it Note: if two edges are identified (glued), that means their beginning and endpoints -which are vertices - have to be the same.
Thus identifying edges determines how the vertices are identified.   }

ii.) The quotient is a surface, as every point in the quotient is contained in an open set that is homeomorphic to an open disc of $\R^2$.

While if you consider images (under the quotient map) of points that are originally inside the hexagon or on an edge this is relatively straightforward to "see" intuitively, it is not so apparent for points "coming from" vertices.

Label the corners of the hexagon and "show" that at each point in the quotient, that "comes from" a vertex of the hexagon, the corners glue in such a way that you get a disc around them. 

(No formal proof is necessary just draw appropriate figures. However, make sure to label correctly, so that it is possible to follow what you are doing.)

iii.) Show that $abca^{-1}b^{-1}c^{-1}$ is actually a torus by a cut-and-paste argument.

{\it I posted an illustration of how the doughnut surface (which is homeomorphic to the torus) can be glued from the hexagon.}

\item  [Extra credit]
{\it We will discuss background needed for this problem on Tuesday.}

Start with the solid cube $X=\{(x,y,z) \, |\, x,z,y\in [-1,1]\}$. Glue its faces
as follows: $(1,y,z)\sim (-1,y,z)$, $(x,1,z)\sim (-x,-1,z)$ and $(x,y,1)\sim (x,y,-1)$.
The resulting quotient space is also
a product space. What product is it? No formal proof is necessary, however, explain your reasoning.

Find in this space a one sided torus.

\end{enumerate}

\end{document}



