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\newcommand{\myTitle}{HW 8, due Friday, April 5}
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\begin{document}

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\begin{enumerate}
  \item Manifolds $M$ and $N$ are {\em homotopy equivalent} if there are maps
    $f\colon M\to N$ and $g\colon N\to M$ such that $g\circ f\sim
    \mbox{id}_M$ and  $f\circ g\sim \mbox{id}_N$ (where $\sim$ denote homotopy
    equivalence of maps).  Show that if $M$ and $N$ are homotopy equivalent
    then $H^kM\approx H^kN$ for $k\geq0$.
    \medskip

    {\sc solution:}

  \item Let $X$ be a topological space, and let $A\subseteq X$ be a subspace.  A
    {\em retraction} is a continuous mapping
    \[
r\colon X\to A
    \]
    such that $r(a)=a$ for all $a\in A$.  A {\em deformation retraction} is a
    homotopy between the identity and a retraction,  that is, a continuous
    mapping
    \[
h\colon [0,1]\times X\to X
    \]
    such that $h_0(x):=h(0,x)=x$ for all $x\in X$, while $h_1(x):=h(1,x)\in A$,
    and $h_1(a)=a$ for all $x\in X$ and for all $a\in A$.  (So, more accurately,
    $h$ is a homotopy between (1) the identity, and (2) a retraction composed
    with the inclusion of $A$ into $X$.) In this case, $A$ is called a {\em
    deformation retract} of $X$.

    In the category of manifolds, we have the same definitions, but all mappings
    are required to be smooth. 
    \begin{enumerate}
  \item Show that if $N$ is a deformation retract of the
    manifold $M$, then $H^k(M)\approx H^k(N)$ for $k\geq0$.
    \medskip

    {\sc solution:}

  \item Show that $S^n$ is a deformation retract of the punctured plane,
    $\R^{n+1}\setminus\{\vec{0}\}$.  (Hence, these two manifolds have the same
    cohomology.)
    \medskip

    {\sc solution:}

\end{enumerate}

\item In class, we showed that if $M$ is a closed orientable
  $n$-manifold without boundary, then $H^nM\neq0$.  Show that the result does
  not hold without the compactness assumption.  Where is compactness used in the
  proof we gave in class?
    \medskip

    {\sc solution:}


\item Use Mayer-Vietoris to compute the cohomology of the following manifolds:
  \begin{enumerate}
    \item The punctured
      plane, $\R^2\setminus\{(0,0)\}$. Here, take
      $U=\R^2\setminus (\mbox{$x$-axis}$) and take
      $V=\R^2\setminus(\mbox{$y$-axis})$.
    \medskip

    {\sc solution:}

    \item The twice punctured plane, $\R^2\setminus\{(-1,0),(1,0)\}$.
    \medskip

    {\sc solution:}

    \item The $2$-torus, $T=S^1\times S^1$.  The only way I could think of doing
      this was as follows:  Let $p\in T$, let $U$ be a small open disk on $T$
      containing $p$, and let $V=T\setminus\{p\}$.  It turns out that $V$ is
      diffeomorphic to a sort of figure-eight band (most easily seen by drawing
      the torus as a square with sides identified in the usual way, and letting
      $p$ be the center of the square).  I then used Mayer-Vietoris to compute
      the cohomology of $V$, and plugged in the result for the Mayer-Vietoris
      sequence for~$T$.  The sequence is still ambiguous, but
      problem~3, above, fixes that.
    \medskip

    {\sc solution:}

  \end{enumerate}
 
\item Two technical results used in class:
  \begin{enumerate}
    \item Let $e_1,\dots,e_n$ and $v_1,\dots,v_n$ be two ordered bases for a
      vector space $V$, and let $e_1^*,\dots,e_n^*$ and $v_1^*,\dots,v_n^*$.
      Say $v_j=\sum_i a_{ij}e_i$ and $v_j^*=\sum_i b_{ij}e_i^*$.  Define matrices
      $A=(a_{ij})$ and $B=(b_{ij})$.  Prove that $B=(A^t)^{-1}$.
    \medskip

    {\sc solution:}

    \item Let $V$ be as above, and suppose $\langle\ ,\ \rangle$ is a bilinear
      form on $V$.  Define the matrices $G=(\langle e_i,e_j\rangle)$ and $H=(\langle
      v_i,v_j\rangle)$, and let $A$ be the matrix defined above.  Show that
      \[
       H=A^tGA.
      \]
    \medskip

    {\sc solution:}

  \end{enumerate}
\end{enumerate}
\end{document}
