\chapter{Applications of $\Z_2$-Homology: A Topological Superhero}

\begin{thm}[No Retraction Theorem]
\index{No Retraction Theorem!$n$-dimensional}
Let $M^n$ be a
connected triangulated $n$-manifold with $\partial
M^n\neq\emptyset$. Then there is no retraction $r:M^n \to \partial M^n$, i.e., no continuous function $r:M^n \to \partial M^n$ such that for each $x \in \partial M^n$, $r(x)=x$.
\end{thm}

\oneperpage

\begin{thm}[$n$-dimensional Brouwer Fixed Point Theorem]
\index{Brouwer Fixed Point Theorem!$n$-dimensional} 
Let $\B^n$ be the $n$-dimensional ball. For every
continuous function $f:\B^n\to\B^n$ there exists a point $x\in\B^n$
such that $f(x)=x$.
\end{thm}

\oneperpage

\begin{lem}
Let $M^n$ be a triangulated, connected $n$-manifold. Let $f:M^n\to
M^n$ be a simplicial map.  Then $f_*:\HH_n(M^n)\to \HH_n(M^n)$
is surjective if and only if $f_\#(M^n)=M^n$.
\end{lem}

\oneperpage

\begin{thm}
Let $f:\Sph^1\to\Sph^1$ be an antipode preserving continuous map (that is, for
every $x\in\Sph^1$, $f(-x)=-f(x)$. Then $f_*:\HH_1(\Sph^1)\to
\HH_1(\Sph^1)$ is surjective.
\end{thm}

\oneperpage

\begin{thm}[Borsuk-Ulam Theorem for $\Sph^2$]
Let $f:\Sph^2\to\R^2$ be a continuous map. Then there exists
an $x\in\Sph^2$ such that $f(-x)=f(x)$.
\end{thm}

\oneperpage

\begin{thm}
Let $f:\Sph^n\to\Sph^n$ be an antipode preserving map (that is, for
every $x\in\Sph^n$, $f(-x)=-f(x)$). Then $f_*:\HH_n(\Sph^n)\to
\HH_n(\Sph^n)$ is surjective.
\end{thm}

\oneperpage

\begin{thm}[Borsuk-Ulam]
\index{Borsuk-Ulam Theorem}
Let $f:\Sph^n\to\R^n$ be a continuous function. Then there is an
$x\in\Sph^n$ such that $f(-x)=f(x)$.
\end{thm}

\oneperpage

\begin{thm}[Ham Sandwich Theorem]
\index{Ham Sandwich Theorem}
Let $A_1, A_2,..., A_n$ be measurable sets of finite measure in $\R^n$. Then there exists an $(n-1)$-dimensional hyperplane $H$ in $\R^n$ that simultaneously cuts each $A_i$ in half. 
\end{thm}

\oneperpage

\begin{exercise}
Draw a sequence of pictures to demonstrate a sequence of elastic moves with no tricks, no cutting, and no gluing that takes the left hand picture of Figure \ref{surprising-unlinking} and turns it into the right hand picture.
\end{exercise}

\oneperpage

\begin{thm}\label{invariance}If $m \neq n$ then $\R^m$ is not homeomorphic to
$\R^n$.
\end{thm}

\oneperpage

\begin{thm}[Invariance of Domain or Invariance of Dimension]
\index{Invariance of Domain Theorem} 
A space cannot be both an $n$-manifold and an $m$-manifold if $n\neq m$.
\end{thm}

\oneperpage

\begin{exercise}
Describe a continuous function from $[0,1]$ to $[0,1]$ that is nowhere differentiable.  
\end{exercise}

\oneperpage

\begin{thm}
Let $B$ be the set of all nowhere differentiable continuous functions from $[0,1]$ to $[0,1]$. Then $B$ is the intersection of countably many dense open sets in $\mathcal{C}$. 
\end{thm}

\oneperpage

\begin{exercise}
Describe an embedding of $[0,1]$ into the plane that has infinite length. In fact, you might choose the graph of a differentiable function.  
\end{exercise}

\oneperpage

\begin{exercise}
Describe an embedding of $[0,1]$ into the unit square and two points $x$ and $y$ in the unit square not on the embedded arc such that to connect $x$ to $y$ by a polygonal path missing the embedded arc requires a polygonal path of length at least a mile.  
\end{exercise}

\oneperpage

\begin{lemma}
Let $h: [0,1] \to \R^2$ be an embedding and let $p$ and $q$ be points in $\R^2 - h([0,1])$. If $p$ and $q$ are connected in $\R^2 - h([0,\frac{1}{2}])$ and $p$ and $q$ are connected in $\R^2 - h([\frac{1}{2},1])$, then $p$ and $q$ are connected in $\R^2 - h([0,1])$. 
\end{lemma}

\oneperpage

\begin{thm}
Let $h: [0,1] \to \R^2$ be an embedding. Then $h([0,1])$ does not separate $\R^2$.
\end{thm}

\oneperpage

\begin{lemma}
For any natural number $n$, let $h: [0,1] \to \R^n$ be an embedding and let $p$ and $q$ be points in $\R^n - h([0,1])$. If $p$ and $q$ are connected in $\R^n - h([0,\frac{1}{2}])$ and $p$ and $q$ are connected in $\R^n - h([\frac{1}{2},1])$, then $p$ and $q$ are connected in $\R^n - h([0,1])$. 
\end{lemma}

\oneperpage

\begin{thm}
For any natural number $n$, let $h: [0,1] \to \R^n$ be an embedding. Then $h([0,1])$ does not separate $\R^n$.
\end{thm}

\oneperpage

\begin{lemma}
For any natural number $n$, let $h: [0,1] \to \R^n$ be an embedding and let $Z$ be a $\Z_2$ $1$-cycle in $\R^n - h([0,1])$. If $Z$ bounds a $2$-chain in $\R^n - h([0,\frac{1}{2}])$ and $Z$ bounds a $2$-chain in $\R^n - h([\frac{1}{2},1])$, then $Z$ bounds a $2$-chain in $\R^n - h([0,1])$.
\end{lemma}

\oneperpage

\begin{thm}
For any natural number $n$, let $h: [0,1] \to \R^n$ be an embedding and let $Z$ be a $\Z_2$ $1$-cycle in $\R^n - h([0,1])$. Then $Z$ bounds a $2$-chain in $\R^n - h([0,1])$.
\end{thm}

\oneperpage

\begin{thm}
For any natural numbers $n$ and $k$ with $k<n$, let $h: [0,1] \to \Sph^n$ be an embedding and let $Z$ be a $\Z_2$ $k$-cycle in $\Sph^n - h([0,1])$. Then $Z$ bounds a $(k+1)$-chain in $\Sph^n - h([0,1])$.
\end{thm}

\oneperpage

\begin{thm}[Jordan-Brouwer Separation Theorem]
\index{Jordan-Brouwer Separation Theorem} 
Let $h:\Sph^{n-1}\to \Sph^n$ be a topological embedding.  Then $h(\Sph^{n-1})$ separates
$\Sph^n$ into precisely two components and is the boundary of each.
\end{thm}

\oneperpage

\begin{thm}[Two Chains Theorem]
Let $f:\Sph^{n-1}\to\Sph^n$ be a simplicial map. Then there
exist $\Z_2$ $n$-chains $A^n$ and $C^n$ such that
$\partial(A^n)=\partial(C^n)=f_\#(\Sph^{n-1})$ and $A^n\cup
C^n=\Sph^n$.
\end{thm}

\oneperpage

\begin{thm}[Absolute Neighborhood Retract Theorem]
For every $k$, $\Sph^k$ is an absolute neighborhood retract. 
\end{thm}

\oneperpage

\begin{thm}
Let $h:\Sph^{n-1}\to \Sph^n$ be a topological embedding and let $U$ be a neighborhood of $h:\Sph^{n-1}$ that retracts to it. Then $U \neq \Sph^n$.
\end{thm}

\oneperpage

\begin{cor}
Let $h:\Sph^{n-1}\to \Sph^n$ be a topological embedding and let $U$ be a neighborhood of $h:\Sph^{n-1}$ with retraction $r:U \to h(\Sph^{n-1})$. Then there exists an open set $V \subset U$ such that for every point $v \in V$, the straight line between $v$ and $r(v)$ is contained in $U$.
\end{cor}

\oneperpage

\begin{lem} \label{J2}
Let $h:\Sph^{n-1}\to\Sph^n$ be a topological embedding.
Then there exists an $\epsilon>0$ such that if
$f:\Sph^{n-1}\to\Sph^n$ is a simplicial map such that
$d(f(x),h(x))<\epsilon$ for all $x\in\Sph^{n-1}$, then
$f_\#(\Sph^{n-1})$ does not bound an $n$-chain in the
$\epsilon$-neighborhood of $h(\Sph^{n-1})$.
\end{lem}

\oneperpage

\begin{lem}
Let $h:\Sph^{n-1}\to\Sph^n$ be a topological embedding and let $\epsilon>0$. Then there exists a $\delta >0$ such that if
$f,g:\Sph^{n-1}\to\Sph^n$ are simplicial maps such that
$d(f(x),h(x))<\delta$ and $d(g(x),h(x))<\delta$ for all $x\in\Sph^{n-1}$, then
$f_\#(\Sph^{n-1})$ and $g_\#(\Sph^{n-1})$ bound an $n$-chain in the
$\epsilon$-neighborhood of $h(\Sph^{n-1})$.
\end{lem}

\oneperpage

\begin{lem}
Let $h:\Sph^{n-1}\to\Sph^n$ be a topological embedding, $U$ be a neighborhood that retracts to $h(\Sph^{n-1})$, let $f:\Sph^{n-1}\to\Sph^n$ be a simplicial map such that for each point $x \in \Sph^{n-1}$, the straight line segment between $h(x)$ and $f(x)$ lies in $U$. Let $A^n$ and $C^n$ be $\Z_2$ $n$-chains  such that $\partial(A^n)=\partial(C^n)=f_\#(\Sph^{n-1})$ and $A^n\cup C^n=\Sph^n$. Then there exists a point $a \in (A - U)$ and a point $c \in (C - U)$. 
\end{lem}

\oneperpage

\begin{lem}
Let $h:\Sph^{n-1}\to\Sph^n$ be a topological embedding, $U$ be a neighborhood that retracts to $h(\Sph^{n-1})$, let $f, g:\Sph^{n-1}\to\Sph^n$ be simplicial maps such that for each point $x \in \Sph^{n-1}$, the straight line segments between $h(x)$ and $f(x)$ and $h(x)$ and $g(x)$ lie in $U$. Let $A_f^n$ and $C_f^n$ be $\Z_2$ $n$-chains (from the Two Chains Theorem) such that $\partial(A_f^n)=\partial(C_f^n)=f_\#(\Sph^{n-1})$ and $A_f^n\cup C_f^n=\Sph^n$ and let $a \in (A - U)$ and $c \in (C - U)$. Let $A_g^n$ and $C_g^n$ be $\Z_2$ $n$-chains such that $\partial(A_g^n)=\partial(C_g^n)=g_\#(\Sph^{n-1})$ and $A_g^n\cup C_g^n=\Sph^n$ where $a \in A_g$. Then $c \notin A_g$. 
\end{lem}

\oneperpage

\begin{lem}
Let $h:\Sph^{n-1}\to\Sph^n$ be a topological embedding, $U$ be a neighborhood such that there is a retract $r:U \to h(\Sph^{n-1}$, and $V$ be an open set in $U$ such that for each $x \in V$, the straight line segment from $x$ to $r(x)$ is in $U$. Let $a \notin U$ and let $T$ be a standard triangulation of $\Sph^n$ with simplexes so small that for any simplex $\sigma \in T$, if $\sigma \cap (\Sph^n - V) \neq \emptyset$, then $\sigma \cap h(\Sph^{n-1} = \emptyset$.  Let $\tau_0$ be an $n$-simplex in $T$ such that $a \in \tau_0$. Let $A$ be the union of all $n$-simplexes $\tau_k$ in $T$ such that there are $n$-simplexes $\{\tau_i\}_{i = 0,\cdots, k}$ such that (1) each $\tau_i$ contains a point in $\Sph^n - V$, and (2) for each $i$, $\tau_i$ and $\tau_{i+1}$ share an $(n-1)$-face. Then $\partial A \subset V$ and $h^{-1}(r(\partial A))$ is the non-trivial element of $H_{n-1}(\Sph^{n-1})$. Also, $r(\partial A) = h(\Sph^{n-1})$.
\end{lem}

\oneperpage

\begin{lem}
Let $h:\Sph^{n-1}\to \Sph^n$ be a topological embedding, let $\{U_i\}_{i \in \N}$ be open sets each containing $h(\Sph^{n-1})$ and each contained in the $\frac{1}{n}$-neighborhood of $h(\Sph^{n-1})$ with retract $r:U_1 \to h(\Sph^{n-1})$ and such that for every $i \in \N$ and every point $x \in U_{i+1}$, the straight line homotopy between $x$ and $r(x)$ lies in $U_i$. Let $\{T_i\}_{i \in \N}$ be a sequence of triangulations of $\Sph^n$ where each triangulation $T_{i+1}$ is a subdivision of $T_i$ with simplices so small that any simplex of $T_i$ that intersects $h(\Sph^{n-1})$ lies entirely in $U_i$. Let $a \in (\Sph^n - U_1)$.  Let $A_i$ be the component containing $a$ of the union of all $n$-simplices of $T_i$ that miss $h(\Sph^{n-1})$. Then $\partial(A_{i+1})$ and $\partial(A_{i+2})$ co-bound an $n$-chain in $U_i$, $\cup_{i \in \N}A_i \cap h(\Sph^{n-1}) = \empty$, each point $x \in h(\Sph^{n-1})$ is a limit point of $\cup_{i \in \N}A_i$, there exists a point $c$ in $(\Sph_n - \cup_{i \in \N}A_i - U_1)$, and if we do the same process that we did for $a$ for $c$ creating $C_i$'s, then $\Sph_n = (\cup_{i \in \N}A_i) \cup h(\Sph^{n-1}) \cup (\cup_{i \in \N}C_i )$.
\end{lem}

\oneperpage

\begin{thm}[Jordan-Brouwer Separation Theorem]
\index{Jordan-Brouwer Separation Theorem} Let $h:\Sph^{n-1}\to
\Sph^n$ be a topological embedding.  Then $h(\Sph^{n-1})$ separates
$\Sph^n$ into precisely two components and is the boundary of each.
\end{thm}

\oneperpage

\begin{cor}[Jordan-Brouwer Separation Theorem]\label{JBST}
Let $h:\Sph^{n-1}\to \R^n$ be a topological embedding.  Then
$h(\Sph^{n-1})$ separates $\R^n$ into precisely two components and
is the boundary of each.  The two components are distinguished
topologically by the fact that one has a compact closure and the other does
not.\end{cor}

\oneperpage

\begin{thm}
Every connected, compact topologically embedded $(n-1)$-manifold in $\R^n$ separates $\R^n$ into two components and is the topological boundary of each. 
\end{thm}

\oneperpage

\begin{thm} 
The Klein Bottle cannot be embedded in $\R^3$.
\end{thm}

\oneperpage

