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\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 11, 2020}}}  %
\fancyhead[R]{{\emph{6. 1. 2020}}}                 % Nastavení pro titulní stranu
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%\addtocounter{ex}{62}






\ex  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 simply 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 Show that the Hurewicz homomorphism is natural.

\ex Show that the Hurewicz homomorphism commutes with connecting homomorphisms. It means: Let $(X,A)$ be a pair. Show that the following diagram commutes:
\[
\xymatrix{
\pi_n(X,A,x_0) \ar[r]^{\dd} \ar[d]_h & \pi_{n-1}(A,x_0)\ar[d]^h \\
H_n(X,A) \ar[r]_{\dd_*} & H_{n-1}(A)
}
\]
where $\dd$ is the boundary homomorphism, $h$ is the Hurewicz homomorphism and $\dd_*$ is the connecting homomorphism. 

\begin{proof}
Take $[f]\in \pi_n(X,A,x_0)$, that is $f\colon (D^n,D^{n-1},s_0)\to (X,A,x_0)$. Then $\dd[f]=[f/S^{n-1}]$ and $h\dd[f]=h[f/S^{n-1}]=(f/S^{n-1})_*(b)$, where $b$ is  a generator in $H_{n-1}(S^{n-1})$. (We recall the definition of the Hurewicz 
homomorphism: if $g\colon S^{n-1}\to A$, $g_*\colon H_{n-1}(S^{n-1})\to H_{n-1}(A)$ and $h[g]=g_*(b)\in H_{n-1}(A)$.) 

Let $a\in H_n(D^n,S^{n-1})$ be a generator such that $\dd_*a=b$. We proceed using commutativity of the following diagram:

\[
\xymatrix{
H_n(D^n,S^{n-1}) \ar[r]^{\dd_*}_{\cong} \ar[d]_{f_*} & H_{n-1}(S^{n-1}) \ar[d]^{(f/S^{n-1})_*} \\
H_n(X,A) \ar[r]_{\dd_*} & H_{n-1}(A)
}
\]
Now $\dd_*h[f]=\dd_* (f_*a)=(f/S^{n-1})_*(\dd_*a)=h\dd [f]$ which concludes the proof. 
\end{proof}



\ex Show that the Hurewicz homomorphism $h: \pi_n(S^n)\to H_n(S^n)$ is an isomorphism.

\ex Use Hopf fibration $S^1\to S^3\to S^2$ to compute $\pi_3(S^2)$.
 
\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 What can we say in this case about Hopf inviariant for $n$ odd 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}

\ex Show that $H(f)=1$ for the Hopf fibration $f: S^3\to S^2$. 

\ex Find a map $f$ with Hopf invariant $H(f)=2$.

\begin{proof}
We study a space $X$ with a basepoint $e$. Denote construction $J_2(X)=X\times X /\sim $, where $(x,e)\sim(e,x)$. Apply this idea to $S^n$. We get a projection $p\colon S^n\times S^n \to J_2(S^n)$. On the left we have one 0-cell, two n-cells and one 2n-cell, while on the right we have one of each. We get that $J_2(S^n)$ has to be a space of the form $C_f$, so $H^n(J_2)=\Z$ given by $a$ and $H^{2n}(J_2)=\Z$ given by $b$ and $H^n(S^n\times S^n)=\Z \oplus \Z$ (generators $a_1,a_2$) and $H^{2n}(S^n\times S^n)=\Z$ (with $b_0$). Now, $p^*\colon H^i(J_2)\to H^i(S^n\times S^n)$ and $p^*(a)=a_1+a_2, p^*(b)=b_0$. 
\begin{align*}
    a^2 &= H(f)b \\
    p^*(a^2) &= H(f) p^*(b) \\
    (a_1+a_2)^2&=H(f) b_0\\
    (a_1^2+a_1a_2+a_2a_1+a_2^2) &= H(f)a_1a_2\\
    2a_1a_2&=H(f)a_1a_2\\
    H(f)&=2
\end{align*}
because $b_0=a_1a_2$ and by evenness of the dimension $a_1a_2=a_2a_1$.
\end{proof}


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