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\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 08, 2017}}}  %
\fancyhead[R]{{\emph{13. 4. 2017}}}                 % Nastavení pro titulní stranu
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\ex[Five lemma] 
Let
$$\begin{xy}
\xymatrix{%@L+2mu@+1em{
 A \ar[r]^*{}
     \ar[d]^*{\cong}
&
 B \ar[r]^*{}
     \ar[d]^*{\cong}
&
 C \ar[r]^*{}
     \ar[d]^*{f}
&
 D \ar[r]^*{}
     \ar[d]^*{\cong}
&
 E \ar[d]^*{\cong}
\\
 \ov{A} \ar[r]_*{}
&
 \ov{B} \ar[r]_*{}
&
 \ov{C} \ar[r]_*{}
&
\ov{D} \ar[r]_*{}
&
 \ov{E}
}\end{xy}
$$
 be a commutative diagram of modules with exact rows. Show that the middle homomorphism $f$ is an isomorphism.
\begin{proof}
We will show the surjectivity of $f$, injectivity is dual. Let $\ov{c}\in \ov{C}$ be arbitrary and suppose it maps to $\ov{d}\in \ov{D}$. This $\ov{d}$ corresponds to some $d\in D$, and since $\ov{d}$ maps to $0$ in $\ov{E}$ by exactness, $d$ also has to map to $0\in E$ by commutativity. By exactness, there exists $c\in C$ which maps to $d$. Since $\ov{c}$ and $f(c)$ both map to $\ov{d}$ by commutativity, there is (by exactness) some $\ov{b}\in \ov{B}$ which maps to $f(c)-\ov{c}$. This $\ov{b}$ corresponds to some $b\in B$, which maps to some $c'\in C$. Then by commutativity, $f(c-c')=f(c)-(f(c)-\ov{c})=\ov{c}$, as desired. (Note that instead of the four vertical maps being isomorphisms, we only needed the surjectivity of $B\to\ov{B}$, $D\to\ov{D}$ and the injectivity of $E\to\ov{E}$).
\end{proof}

\ex 
Show that for a finite CW-complex $X$ and $H^*(Y)$ being finitely generated free group in all dimensions, the cross product
$$H^*(X)\otimes H^*(Y)\xrightarrow{\mu} H^*(X\times Y)$$ is an isomorphism. (In fact, the same is true for $X$ being an infinite CW-complex.)

\begin{proof}
First let $X=pt$ be a point. Then $H^*(pt)=\Z$ with $1\in H^0(pt)$ and $pt\times Y$ is homeomorphic to $Y$, hence $$H^*(pt)\otimes H^*(Y)=\Z\otimes H^*(Y)\cong H^*(Y)\cong H^*(pt \times Y).$$
Now let $X=p_1\sqcup p_2\sqcup\dots \sqcup p_k$ be a finite disjoint union of points (i.e, a discrete set). Then $H^*(X)=\bigotimes_{i=1}^k \Z$, hence
\begin{equation*}
\begin{split}
H^*(X)\otimes H^*(Y)&=\left( \bigoplus_{i=1}^k \Z \right)\otimes H^*(Y)\cong \bigoplus_{i=1}^k\left(  \Z \otimes H^*(Y)\right)\cong \bigoplus_{i=1}^k H^*(Y)\\
&\cong H^*(\underbrace{Y\sqcup Y\sqcup\dots\sqcup Y}_{n \text{ times}})\cong H^*(X \times Y)
\end{split}
\end{equation*}
(We should also show that the isomorphism is indeed given by $\mu$, but if $e_1,\dots,e_k\in H^0(X)$ are such that $$e_i(p_j)=\begin{cases} 1 & \text{ if } i=j\\ 0 & \text{ if } i\neq j\end{cases},$$ it's not hard to see that $$\mu(e_i\otimes a)=(0,0,\dots,\underbrace{a}_{i\text{-th place}},0,\dots,0)$$ using projections and the definition of the cup product.)
\newpage
Now we can proceed inductively:

\begin{enumerate}[i)]
\item We now know that the theorem is true for $X$ of dimension $0$.
\item Since $D^n$ is homotopy equivalent to a point, the theorem is also true for $X=\bigsqcup_{\alpha=1}^k D_{\alpha}^n$.
\item Suppose that the theorem holds for finite CW-complexes of dimension $n-1$. Then it is also true for pairs $(\bigsqcup D_{\alpha}^n,\bigsqcup S_{\alpha}^{n-1}).$
\item Since $H^*(X,A)\cong H^*(X/A)$ for a subcomplex $A\subseteq X$, the theorem also holds for $\bigvee S_{\alpha}^n\cong\bigsqcup D_{\alpha}^n/\bigsqcup S_{\alpha}^{n-1}$.
\item Now let $X=X^{(n)}$ be an $n$-dimensional CW-complex and consider the diagram
\[
\xymatrix{
H^*(X,A) \otimes H^*(Y) \ar[rr] \ar[dd]^{\mu} &  & H^*(X)\otimes H^*(Y)\ar[dd]^{\mu} \ar[ld]\\
 & H^*(A)\otimes H^*(Y)\ar[dd]^<<<<{\mu} \ar[ul]_{\delta^*\otimes\id} & \\
H^*(X\times Y,A\times Y)\ar[rr] &  & H^*(X\times Y)\ar[ld]\\
 & H^*(A\times Y) \ar[lu]_{\delta^*}& 
},
\]
where the lower triangle represents a long exact cohomology sequence by definition, and the same is true for the upper triangle (since free modules are flat). We know that the theorem holds for $X^{(n-1)}$, and also for $X^{(n)}/X^{(n-1)}\cong \bigvee S_{\alpha}^{n-1}$. Therefore we can use the Five lemma after unfolding the diagram in the appropriate dimensions, from which it follows that the theorem also holds for $X^{(n)}$. This completes the induction.
\end{enumerate}
\end{proof}

\ex Compute the cohomology rings of $\C P^2 \times S^6$ and $\C P^2 \vee S^6$.
\begin{proof}
We have $H^*(\C P^2)=\Z[w]/\langle w^3 \rangle$ for $w\in H^2$ and $H^*(S^6)=\Z[a]/\langle a^2 \rangle$ for $a\in H^6$, hence
$$H^*(\C P^2 \times S^6)=\Z[w]/\langle w^3 \rangle\otimes\Z[a]/\langle a^2 \rangle \cong \Z[w,a]/\langle w^3,a^2 \rangle.$$
Next, it is true in general that $\ov{H}^*(X\wedge Y)\cong \ov{H}^*(X)\oplus \ov{H}^*(Y) $ is an isomoprhism of graded rings (this can be proven proven straight from the definitions, but it takes some time). Since $\C P^2 \vee S^6$ is connected, we have $H^*(\C P^2 \vee S^6)\cong \ov{H}^*(\C P^2)\oplus \ov{H}^*(S^6)\oplus\Z$. Now $w\cup a\in H^8=0$ (more generally, we could use that fact that $(w,0)\cup (0,a)=(0,0)$). Therefore $$H^*(\C P^2 \vee S^6)\cong \Z[w,a]/\langle w^3,a^2,wa \rangle.$$
\end{proof}

\ex Show that the  $\C P^2 \vee S^6$ is not homotopy equivalent to $\C P^3$.
\begin{proof}
It suffices to show that the cohomology rings of these spaces are not isomorphic (note that the additive group structure is not enough to distinguish them). We already know that $$H^*(\C P^2 \vee S^6)\cong \Z[w,a]/\langle w^3,a^2,wa \rangle$$ and we have $H^*(\C P^3)=\Z[b]/\langle b^4 \rangle$ for $b\in H^2$. Any isomorphism would have to map $w$ and $b$ to $\pm$ each other (these are the respective generators in dimension 2), but $w^3=0$ while $b^3\neq 0$, so this is not possible.
\end{proof}

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