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\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 12, 2017}}}  %
\fancyhead[R]{{\emph{11. 5. 2017}}}                 % Nastavení pro titulní stranu
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\addtocounter{ex}{62}



\ex Using homotopy groups show that $\R P^k$ is not a retract of $\R P^n$, $n>k\geq 1$.

\begin{proof}
Retract: The composition $\R P^k\stackrel{i}{\hookrightarrow} \R P^n \stackrel{r}{\to}\R P^k$ is the identity, 
so we have the following diagram and want to use it to obtain a contradiction.
\[%\xymatrixrowsep{0.25pc}
\xymatrix{
\pi_*(\R P^k) \ar[r]^{\pi_*(i)} \ar[rd]_{\operatorname{id}} & \pi_*(\R P^n) \ar[d]^{\pi_*(r)}\\
& \pi_*(\R P^k)
}
\]
From the long exact sequence of the fibration, we know that $0=\pi_1(S^k)\to \pi_1(\R P^k)\stackrel{\cong}{\to}\pi_0(S^0)\to 0$ and $\pi_i(S^k)=\pi_i(\R P^k)$ for $k\geq 2$. Then considering diagram 
\[
\xymatrix{
\pi_k(S^k)=\Z \ar[r] \ar[rd] & \pi_k(S^n)=0 \ar[d]\\
& \pi_k(S^k)=\Z
}
\]
we see that we are factoring identity through zero group and that is Mission Impossible: Factor Zero (in theaters never). Continue with $k=1$. We use the knowledge of $\R P^1\cong S^1$. Then our diagram is a triangle with $\Z,\Z/2,\Z$ with identity $\Z\to\Z$, factoring identity through finite group is Mission Impossible: Group protocol (in theaters maybe one day, one can only hope) ...and we are done.
\end{proof}

\ex Consider the map $q\colon S^1\times S^1\times S^1\to S^3$ defined as a map $S^1\times S^1\times S^1\to D^3/S^2$ where $D^3$ is a small disk in the triple torus which is the identity in the interior of $D^3$ and constant on its complement. Further,  consider the Hopf map $p\colon S^3\to S^2=\C P^1$ (described in Hopf fibration $S^1\to S^3\to S^2$). Compute $q_*$ and $(pq)_*$ in homotopy groups. Show that $pq$ is not homotopic to a constant map.

\begin{proof}
Take following diagram and treat it like your own:

\[
\xymatrix{\xymatrixcolsep{2pc}
& S^1 \ar[d] & \text{(fiber)}\\
S^1\times S^1\times S^1 \ar[r]^<<<q & S^3 \ar[d]^p& \subseteq \C^2\\
& S^2 & =\C P^1
}
\]
where $p(z_1,z_2)=\dfrac{z_1}{z_2}$. We know that only nontrivial homotopy group of the triple torus is $\pi_1(S^1\times S^1\times S^1)$, but  $\pi_1 S^3$ is is trivial, so $q_*$  is zero in homotopy groups and the composition as well. By the following diagram we have $H$, thanks to homotopy lifting property, of course. (denote $(S^1)^3$ the triple torus $S^1\times S^1\times S^1$)

\[
\xymatrix{
& S^1=p^{-1}(s_0) \ar[d]\\
(S^1)^3 \ar[r]^q \ar[ur] \ar[d] & S^3 \ar[d] \\
(S^1)^3\times I \ar[ur]^H \ar[r]_h & (S^2,s_0)
}
\]
where $h(0,-)$ is constant map, $h(1,-)=pq$, $p(H(0))=s_0$,$\operatorname{Im} H(0)\subseteq p^{-1}(s_0)=S^1$ and $H(1)=q\sim H(0)$.
However, $H(0)_*$ and $q_*$ differ in the  third homology groups:
\[
\xymatrix{
& H_3(S^1)=0\ar[d]\\
H_3((S^1)^3)\ar[r]_{q_*=id} \ar[ru]^{H(0)_*}& H_3(S^3)=\Z
}
\]
We are trying to factor through zero, a contradiction.
\end{proof}

\ex (detail for the lecture 9. 5. 2017) If $(X,A)$ is relative $CW$-complex such that there are no cells in dimension $\leq n$ in $X\setminus A$, then $(X,A)$ is $n$-connected.

\begin{proof}
Recall the definition of $n$-connectness of a pair. For $[f]\in\pi_i(X,A,x_0)$, $i\le n$,  use cell approximation of $f$: There is a cell map $q\colon (D^i,S^{i-1},s_0)\to (X,A,x_0)$, such that $q\sim f$ relatively $S^{i-1}$ and $q(D^i)\subseteq X^{(i)}=A$ since $X^{(-1)}=X^{(i)}=\cdots=X^{(n)}=A$. Note the following very useful criterion:\par 
$[f]=0$ in $\pi_i(X,A,x_0)$ $\Longleftrightarrow f\sim q$ relatively $S^{i-1}$, $g(D^i)=A$.\par 
Thus $[f]=0$ in our case, and we are done.
\end{proof}

\ex Let $[X,Y]$ denote a set of homotopy classes of maps from $X$ to $Y$. If $(X,x_0)$ is a CW-complex and $Y$ is path connected, then $[X,Y]\cong [(X,x_0),(Y,y_0)]$.

\begin{proof}
Surely, $[(X,x_0),(Y,y_0)]\subseteq [X,Y]$ and denote the class in the left set as $(g)$ and $[g]$ the class in the $[X,Y]$. The map $(g)\mapsto [g]$ is well defined and injective. To prove that it is also surjective take $[f]\in [X,Y]$. Using HEP for $f:X\times 0\to Y$ and a curve $\omega: x_0\times I\to Y$ which connects $f(x_0)$ with $y_0$ we get $g:X\times 1\to Y$ such that $f\sim g$ and  $g(x_0)=y_0$, then $[f]=[g]$ and $(g)\in[(X,x_0),(Y,y_0)]$. We are done.
\end{proof}

\ex (application)  We know that $deg(f)$ is an invariant of $[S^n,S^n]=\pi_n(S^n)$. \\ 
Study $[S^{2n-1},S^n]\cong \pi_{2n-1}(S^n)$ and describe its co called Hopf invariant $H(f)$.

\begin{proof}
Have $f\colon \partial D^{2n}=S^{2n-1}\to S^n$ and $S^n\cup_f D^{2n}$. For $f\sim g$ we have $S^n\cup_f D^{2n}\simeq S^n\cup_g D^{2n}$, moreover $S^n\cup_f D^{2n}=C_f$ (the cylinder of $f$). For $n\geq 2$ we have $C_f=e^0\cup e^n\cup e^{2n}$. Using cohomology: $H^*(C_f)=\Z$ for $\ast\in\{0,n,2n\}$ and $0$ elsewhere. Take $\alpha\in H^{n}(C_f)$ generator, we have cup product. Then $\alpha\cup\alpha\in H^{2n}(C_f)$ and for $\beta\in H^{2n}(C_f)$ we have $\alpha\cup\alpha=H(f)\beta$, where $H(f)$ is the Hopf invariant.
\end{proof}

\ex (continuation of previous exercise) For $n$ odd, what can we say in this case about Hopf inviariant? And for $n$ even? Thanks.

\begin{proof}
Knowing $\alpha\cup \beta=(-1)^{|\alpha||\beta|} \beta\cup\alpha$ we see that $\alpha\cup\alpha = 0$. So for $n$ odd Hopf invariant is zero.\par 
For $n$ even consider the Hopf fibration $S^1\to S^3\to S^2=\C P^1$. For $\C P^2=D^4\cup_f \C P^1$ (recall how $\C P^n$ is built up from $\C P^{n-1}$) we have $C_f=\C P^2$ and $H^*(\C P^2)=\Z[\alpha]/\langle \alpha^3\rangle$, with $\alpha\in H^2$. The generator of $H^4$ is $\alpha^2$. We get that  $H(f)=1$.
\end{proof}




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