\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}}%
}

\def\cy#1{\Z/{#1}\Z}
\def\RA#1{\limitsarrow{#1}{}{\rightarrowfill}}
\centerline{\bf Math 608 Final}
\centerline{\bf May 7, 1998}
\centerline{Professor Taylor}

\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
When proving things during this test, you may assume earlier parts of\
a problem even if you can not do them.
You may also assume known the homology of $S^n$,
$RP^n$ and $CP^n$ as well as their cohomology rings.

\np
\itemitem{(a)} Show $CP^3$ and $S^2\times S^4$ have isomorphic
homology.
(You may just write down $H_\ast(CP^3)$, but you need to compute
$H_\ast(S^2\times S^4)$ using the K\"unneth formula.)
\itemitem{(b)} Part (a) notwithstanding, show $CP^3$ and $S^2\times S^4$
are not homotopy equivalent by showing that any map
$f\colon S^2\times S^4\to CP^3$ has degree $0$.
Hint: Compute in cohomology.
\itemitem{(c)}  For good measure, show that any map
$g\colon CP^3\to S^2\times S^4$ also has degree $0$.
\def\hom(#1,#2){{\rm Hom}(#1,#2)}
\np
Suppose given an exact sequence $0\to G_0\to G_1\to G_2\to0$.
\itemitem{(a)}
If $A$ is an abelian group show
$$0\to \hom(A, G_0) \to \hom(A,G_1)\to \hom(A, G_2)$$
is exact.
\itemitem{(b)}
If $A$ is a free abelian group show
$$0\to \hom(A, G_0) \to \hom(A,G_1)\to \hom(A, G_2)\to 0$$
is exact.

\itemitem{$\bullet$} Apply (b) to the groups $S_\ast(X)$.
The sequence in (b) is natural so there results an exact
sequence of chain complexes.
When one applies homology, one gets a long exact sequence
$$\cdots\to H^r(X;G_0)\to H^r(X;G_1)\to H^r(X;G_2)\
\RA{\beta}\ H^{r+1}(X;G_0)\to\cdots\ .$$
The map $\beta$ is again called the {\sl Bockstein\/} associated to the exact
sequence of coefficient groups.

\itemitem{(c)}
Compute the cohomology Bockstein sequences for
$X=RP^2$ and coefficient sequence $0\to\Z\ \RA{\times2}\ \Z\to\cy2\to0$.

\np
Let $Y$ be a compact orientable manifold of dimension $7$.
Suppose we know the following groups:
$H_7(Y;\Z)\cong\Z$, $H_6(Y;\Z)\cong\Z$, $H_5(Y;\Z)\cong\cy2$ and
$H_4(Y;\Z)=\Z\oplus\cy3$.
\itemitem{(a)}Use the universal coefficients theorem to compute
$H^r(Y;\Z)$ for $r=7$, $6$, and $5$.
\itemitem{(b)} Recall the fact that all the homology groups of $Y$ are
finitely{--}generated plus the fact that $Ext(A,\Z)\cong T$
where $T\subset A$ denotes the torsion subgroup.
Using these facts, what can you say about $H^4(Y;\Z)$?
\itemitem{(c)} Use Poincar\'e duality to compute
$H_r(Y;\Z)$ for $0\leq r<3$.
\np
Let $M^m$ be a compact connected manifold of dimension $m>2$ with
boundary $\partial M=S^{m-1}$.
If the map $\partial M\to M$ is null homotopic, show that
$\widetilde H_\ast(M;\Z)=0$.

\end
