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  \begin{document}


  
  \vskip 1cm

  \centerline{\Large {de Rham cohomology, part 2}}
  \vskip 1cm

  The problems this week are harder, so there are fewer of
  them. Finding Poincar\'e duals is optional since we didn't cover
  them in detail.

\subsection*{More de Rham cohomology}



\begin{enumerate}[1.]

\item Let $Y_{g,n}$ be the orientable surface of genus $g$ with $n$
  points removed. Compute $H^*(Y_{g,n})$. Can you find submanifolds
  whose Poincar\'e duals generate?


\item \label{suspension}
  Let $X\subset\R^n$ be a closed set. The goal of this exercise is
  to relate $H^*(\R^n\smallsetminus X)$ and
  $H^*(\R^{n+1}\smallsetminus X)$.

  Fix a continuous function $f:\R^n\to [0,1]$ such that
  $f^{-1}(0)=X$. For example $f(y)=\min\{1,d(y,X)\}$ will do. Let
  $$U=\{(y,t)\in\R^n\times\R\mid t<f(y)\}$$ and
  $$V=\{(y,t)\in\R^n\times\R\mid t>-f(y)\}$$
  Show that $U$ and $V$ are contractible by finding a deformation
  retraction to hyperplanes $t=-1$ and $t=1$ respectively. Also show
  that $U\cap V$ deformation retracts to $\R^n\smallsetminus X$. From
  the Mayer-Vietoris sequence deduce that for $p\geq 2$ we have
  $$H^p(\R^{n+1}\smallsetminus X)\cong H^{p-1}(\R^n\smallsetminus X)$$
  The formula holds even when $p=1$ or 0, but one must modify the
  definition of $H^0$ that effectively removes one dimension (so it is
  0 for connected manifolds). That is, one uses {\it reduced
    cohomology} $\tilde H^0$. If you are curious, $\tilde H^0(X)$ is
  the cokernel of the map $H^0(p)\to H^0(X)$ induced by the constant
  map $X\to \{p\}$.

\item Using only the Stokes theorem show that an orientable compact
  manifold doesn't admit a smooth retraction to its boundary.

\end{enumerate}

\subsection*{Compactly supported de Rham cohomology}
\begin{enumerate}[resume]
  \item Let $U,V$ be open subsets of a manifold $M$ with $M=U\cup
    V$. Show that for any $p$ the sequence
    $$0\to\Omega_c^p(U\cap V)\overset{i}{\to} \Omega_c^p(U)\oplus
    \Omega_c^p(V) \overset{j}{\to}
    \Omega_c^p(M)\to 0$$ is exact, where
    $i(\omega)=(\omega_U,\omega_V)$ and
    $j(\eta,\theta)=(\tilde\eta-\tilde\theta)$, with
    $\omega_U,\omega_V$ denoting the extension of $\omega$ to $U,V$ by
    0, and $\tilde\eta,\tilde\theta$ are the extensions of
    $\eta,\theta$ by 0 to $M$. This implies, by homological algebra,
    that there is a long exact sequence
    $$\cdots \to H_c^p(U\cap V)\to H_c^p(U)\oplus H_c^p(V)\to H_c^p(M)\to
    H_c^{p+1}(U\cap V)\to\cdots$$

  \item Show that $H^p_c(\R^2\smallsetminus\{0\})=\R$ for $p=1,2$ and 0
    otherwise. Can you find submanifolds whose Poincar\'e duals
    generate? Hint: Mayer-Vietoris. There are several ways of doing
    it, one possibility is to cover $\R^2\smallsetminus\{0\}$ with two
    open sets as in Problem \ref{suspension}. The other is to take
    $U,V\subset S^2$ to be the complements of the north and south poles
    respectively and then $U\cap V$ is diffeomorphic to
    $\R^2\smallsetminus\{0\}$. 

  \item Compute $H_c^p(X_n)$ where $X_n$ is $\R^2$ with $n$ points
    removed. Can you find submanifolds whose Poincar\'e duals generate
    compactly supported cohomology? The same question for the regular
    cohomology.

  \item Show that $H^p_c(\R^n\smallsetminus\{0\})=\R$ for $p=1,n$ and 0
    otherwise. Can you find submanifolds whose Poincar\'e duals
    generate? 

\end{enumerate}





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