\magnification1200
\advance\vsize by 1in
\footline{}
\newbox\arrowtmp
\newbox\arrowtmpA
\newdimen\arrowwd
\def\limitsarrow#1#2#3{%
\setbox\arrowtmp=\hbox{$\scriptstyle11#1$}%
\setbox\arrowtmpA=\hbox{$\scriptstyle11#2$}%
\ifdim\wd\arrowtmp<\wd\arrowtmpA\relax
\arrowwd=\wd\arrowtmpA\relax\else%
\arrowwd=\wd\arrowtmp\relax\fi%
\smash{\mathop{\vbox to 3pt{\vss\hbox to \arrowwd{#3}}}\limits^{#1}\limits_{#2}}%
}
\centerline{\bf Math 608 Midterm}
\centerline{\bf April 3, 1998}
\centerline{Due 8am April 6}
\newcount\pc\pc=1
\def\nl{\medskip\noindent}
\def\np{\nl{\bf Problem \number\pc.\ }\global\advance\pc by 1}
\def\Z{{\bf Z}}
\def\C{{\bf C}}
\bigskip
\np
If $X$ and $Y$ are CW complexes, show that $X\times_C Y$ is also.
Here $X\times_C Y=(X\times Y)_C$ as in problem 3, p.~213.
\np
Let $f\colon [-1,1]\to[-1,1]$ be continuous and suppose
$f(1)=\pm1$, $f(-1)=\pm1$.
Show that
$$f_\ast\colon H_1([-1,1],\{-1, 1\})\to H_1([-1,1],\{-1, 1\})$$
is given as follows:
\item{a.} multiplication by $1$ if $f(-1)=-1$ and $f(1)=1$;
\item{b.} multiplication by $-1$ if $f(-1)=1$ and $f(1)=-1$;
\item{c.} multiplication by $0$ if $f(1)=f(-1)$.
\np
Let $z^n\colon S^1\to S^1$ for $n\in\Z$ be the map obtained
by considering $S^1$ as the unit circle in the complex plane and
raising $z$ to the $n$-th power.
Choose cell structures on the range and domain $S^1$ so
that $z^n$ is cellular and use problem 2 to prove that the
map $z^n_\ast\colon H_1(S^1)\to H_1(S^1)$ is multiplication
by $n$.
\np
Let $f\colon S^m\to S^m$ be a map.
If $f$ is not onto, prove that
$f_\ast\colon H_m(S^m)\to H_m(S^m)$ is the zero map.
\np
Consider $S^2$ as the Riemann sphere.
Equivalently, $S^2$ is the complex plane, $\C$ with the point at
infinity added and topologized as the one point compactifiaction.
Let $p(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0$
be a degree $n$ polynomial with all $a_i\in\C$.
The polynomial $p$ induces a map $p\colon \C\to\C$.
Show that $p$ extends to a map
$p\colon S^2\to S^2$ and further show that any two
such maps of the same degree are homotopic.
\np
Use the Mayer{--}Vietoris Theorem and problem 3 to show
that $z^n\colon S^2\to S^2$ induces the map
$z^n_\ast\colon S^2\to S^2$ which is multiplication by $n$.
\nl{\bf Remark:} Problems 4, 5 and 6 prove the Fundamental
Theorem of Algebra: any degree $n$ polynomial with coefficients
in $\C$ has a root in $\C$.
\np Let $K^2$ be the Klein bottle.
This is a CW complex with one $0$-cell, two $1$-cells and one $2$-cell.
Let the rectangle, $R$, below be the evident CW complex with
four $0$-cells, four $1$-cells and one $2$-cell.
There is a cellular map $f\colon R\to K$ which takes all the vertices
of $R$ to the vertex of $K$; the four $1$-cells of $R$ to the two $1$-cells
of $K$ as indicated by the arrows and labels; the $2$-cell of $R$ to
the $2$-cell of $K$ by a homeomorphism on the interiors.
Use this information to compute the boundary maps
in the cellular chain complex for $K$.
Problem 2 is useful here.
%\hfuzz=10in
\def\LL#1#2{%
\hbox to 1in{\hss$\limitsarrow{\hskip .5in #1\hskip .5in}%
{\hskip .5in #2\hskip .5in}%
{%
\hss\hbox to .6in{\rightarrowfill}\hskip-3pt%
\hbox to.8in{\cleaders\hbox{$\mkern-3mu \mathord- \mkern-2mu$}\hfil}\hss}$\hss}%
}
\catcode`\@=11
\def\uparrowfill{\vbox{\hskip-1pt$\uparrow$}\vskip-4pt%
\cleaders\vbox{$\vert $}\vfill
}
\def\downarrowfill{%
\cleaders\vbox{\hsize=0pt\hbox to0pt{\hss$\vert$\hss}}\vfill
\vskip-4pt\vbox{\hsize=0pt\hbox to0pt{\hss$\downarrow$\hss}}%
}
\def\mybig#1#2{{\setbox0=\hbox{$#1$}\dimen0=#2\advance\dimen0 by-\ht0 %
\advance\dimen0 by -\dp0 %
\hbox to 0pt{\hss$\left#1\vbox to\dimen0{}\right.\n@space$\hss}}}
\catcode`\@=12
\def\VV#1#2{\lower 4pt\hbox to0pt{\hss$\scriptstyle#1$\ }%
\vtop to40pt{\mybig{\downarrow}{20pt}\vskip-1pt\mybig{\vert}{20pt}\vss}%
\lower 16pt\hbox to0pt{\ $#2$\hss}}
\def\VX#1#2{\lower 4pt\hbox to0pt{\hss$#1$\ }%
\vtop to40pt{\mybig{\vert}{20pt}\vskip-1pt\mybig{\uparrow}{20pt}\vss}%
\lower 4pt\hbox to0pt{\ $\scriptstyle#2$\hss}}
\vtop{$$\matrix{\bullet&
\LL{a}{}&\bullet\cr\noalign{\vskip-2pt}
\VV{b}{}&&\VX{}{b}\cr\noalign{\vskip-6pt}
\bullet&\LL{}{a}&\bullet\cr}$$
\centerline{$R$}}
\end