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\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 05, 2017}}}  %
\fancyhead[R]{{\emph{23. 3. 2017}}}                 % Nastavení pro titulní stranu
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%13:45



\ex Prove that $S^n$ has a nonzero vector field if and only if $n$ is odd.
\begin{proof}
First note, that we have $v\colon S^n\to \R^{n+1}$ such that $v(x)\perp x$. Consider the case of $S^1$ and $(x_0,x_1)\mapsto (x_1,-x_0)$. Take $(x_0,x_1,x_2,x_3,\dots,x_{2n+1})\in S^{2n+1}\subseteq \R^{2n}$. Then we get $(x_1,-x_0,x_3,-x_2,\dots, x_{2n+1},-x_{2n})$ as image and there is nothing more obvious than that the product is zero, i.e. it's perpendicular.\par 
Note, for $S^1\subseteq \C$ it is $z\mapsto ez$, where $e$ is the complex unit, usually denoted as $i$.\par 
Now we want to prove that if $S^n$ has a nonzero vector field, then $n$ is odd. We use the fact that $\de(\id)=1$ and $\de(-\id)=(-1)^{n+1}$. Take $v\colon S^n\to S^n$. If we show $\id\sim -\id $, then $1=(-1)^{n+1}\Rightarrow n$ is odd. The homotopy is $h(x,t)$ we are looking for is $h(x,t)=cos(t)x+sin(t)v(x)$, where $t\in[0,\pi]$. Also note $\parallel\! h(x,t)\!\parallel=cos^2t+sin^2t=1$. We are done.
\end{proof}

\ex Prove $\de(f)=\sum_{i=1}^k \de(f/x_i)$.

\begin{proof}
We have $f\colon S^n\to S^n$ and $y\in S^n$ has a neighborhood (nbhd) $V$. Denote $f^{-1}(y)=\{x_1,\dots,x_k\}$ and $U_i$ nbhd's of $x_i$, $U_i\cap U_j=\emptyset$. Let degree of $f/x_i$ be $n_i$. We have $f\colon (U_i,U_i-x_i)\to (V,V-y)$ and $(f/U_i)_*\colon H_n(U_i,U_i-x_i)\cong \Z \to H_n(V,V-y)\cong \Z$, so $z\mapsto n_i z$. Consider the following diagram, its evolution (i.e. drawing order) was revealed in the tutorial class:
\[
\xymatrix{
 &\bigoplus H_n(U_i,U_i-x_i) \ar[r]^{(f/U_i)_*} \ar[ld]_{\cong} \ar[d]^k & H_n(V,V-y) \ar[d]^h\\
H_n(S^n,S^n-x_i) & H_n(S^n,S^n-f^{-1}(y)) \ar[r] &H_n(S^n,S^n-y)\\
 &H_n(S^n) \ar[lu]^{\cong} \ar[u]^{j_*} \ar[r]^{f_*} & H_n(S^n) \ar[u]^{j_{y*}}
}
\] 

We have $j_{y*}$ iso and $h$ is iso by excision theorem.\par 
So, we have $H_n(S^n,S^n-f(y))\stackrel{\cong}{\leftarrow} \bigoplus H_n(U_i,U_i-x_i)\cong H_n(\bigsqcup_1^k(U_i,U_i-x_i))$. Iso between $H_n(S^n,S^n-f(y))$ and $H_n(\bigsqcup_1^k(U_i,U_i-x_i))$ by excision. So $k$ is also iso.\par 
We can take $j_{y*}=h\circ \sum(f/U_i)_* \circ k^{-1}\circ j_*$ and for generator $1\in H_n(S^n)$ we have $((j_y)_*\circ f_*)(1)=\de(f)\cdot 1$. We are done.
\[
\xymatrix{
(1,\dots,1) \ar@{->}[r]  & \sum deg(f/x_i) \ar@{->}[d]\\
1 \ar@{|->}[u] &\sum \de(f/x_i)}
\]

\end{proof}

\newpage

\ex Compute homology groups of $\mathbb{RP}^n$ using CW-structure.

\begin{proof}
Let $X$ be the $CW$-complex, we have $C^{CW}_n=\bigoplus \Z[e^n_{\alpha}]$, $H_n(C^{CW}_*,d)=H_n(X)$. Also, $d[e^n_{\alpha}]=\sum_{\beta}d^{\beta}_{\alpha}[e^{n-1}_{\beta}]$, where $\alpha^{\beta}_{\alpha}$ is the degree of the map 
$$S^{n-1}\stackrel{f_{\alpha}}{\longrightarrow}X^{n-1}\to X^{n-1}/(X^{n-1}-e^{n-1}_{\beta})\cong S^{n-1}.$$
We know that $\mathbb{RP}^n=e^0\cup e^1 \cup \dots \cup e^n$, cell in every dimension. Attaching map is
$$f\colon S^i\to (\mathbb{RP})^{(i)}=\mathbb{RP}^i=S^i/(x\sim -x).$$
Then have 
$$0\to \Z \to \Z \to \cdots \to\Z\to\Z\to\Z\to 0,$$
where $\Z$ appears $n+1$ times. Now, $S^i\stackrel{f}{\to}\mathbb{RP}^i\to \mathbb{RP}^i/\mathbb{RP}^{i-1}\cong S^i$, hence we can apply the attaching map. Class $[y]$ has two preimages $y,-y$, there are nbhd's, $deg(f/y)=\pm 1$. Note that $f/-y=f/y\circ (-\id)$. Now we can compute the degree: $\de(f)=\de(f/y)+\de(f/-y)=\pm(1+(-1)^{i+1})$. It is $\pm 2$ if $i$ is odd and $0$ otherwise. So, we can add to the sequence above:
$$0\to \Z \to \Z \to \cdots \to\Z\stackrel{2\cdot}{\to}\Z \stackrel{0}{\to}\Z\to 0,$$
and compute for $X=\mathbb{RP}^n$, $n$ even: $H_0(X)=\Z,H_1(X)=\Z_2,H_2(X)=0,H_i(X)=0$ for $i$ even, $H_i(X)=\Z_2$ for $i$ odd. The homology is zero for $i>0$. For $n$ even the only difference is $H_n(X)=\Z$.\par 
(Note: this is because for even they are not orientable, for odd it is oriented space.)
\end{proof}

\ex Compute homology groups of oriented two dimensional surfaces. What about nonorientable?

\begin{proof}
Denote $M_g$ surface of genus $g$ (it is the same as sphere with $g$ handles, i.e. $M_1$ is torus and $M_2$ is double torus (homework 4). The $CW$-model is $e^0\cup e^1_1 \cup\dots\cup e^1_{2g}\cup e^2$ and we have 
$$0\to\Z\to \bigoplus_1^{2g}\Z \to \Z\to 0.$$
Second differential is zero, $e^0-e^0=0$. The first one is zero as well (glue the model, the arrows go with $+$ and then $-$). Then we get $H_0=\Z,H_1=\bigoplus_1^{2g}\Z,H_2=\Z$.\par 
For nonorientable surfaces, $N_g$ is modelled by one $2$-dimensional disc which has boundary composed with $g$ segments every of which repeats twice with the same orientation. So we have one cell in dimensions $2$ and $0$ and $g$ cells in dimension one. We get (quite similarly)
$$0\to\Z\to \bigoplus_1^{g}\Z \to \Z\to 0.$$
This equality holds: $d[e^2]=2[e^1_1]+2[e^1_2]+\dots,$ so $H_2=0,H_0=\Z$ and the only interesting case is 
$$H_1=\dfrac{\Z[e^1_1,\dots,e^1_g]}{\Z[2e^1_1+\cdots +2e^1_g]}=\Z_2+\bigoplus_1^{g-1}\Z.$$
\end{proof}

\ex Have $f\colon S^n\to S^n$ map of degree $k$. (such map always exists). Let $X=D^{n+1}\cup_{f}S^n$ and compute homology of $X$ and the projection $p\colon X\to X/S^n$ in homology.

\begin{proof}
Easy, $X=e^0\cup e^n\cup e^{n+1}$ and $0\to\Z\stackrel{k\cdot}{\to}\Z\to0\to\dots\to0\to\Z\to0$, also
$$S^n\to X^{(n)}\to X^{(n)}/(X^{(n)}-e^n)=S^n.$$
We get $H_{n+1}(X)=0,H_n=\Z_k,H_0(X)=\Z$. Note, $X/S^n\cong S^{n+1}$. So, for the $p_*$ we have 
$$p_*\colon H_{n+1}(X)=0\stackrel{0}{\to}H_{n+1}(S^{n+1})=\Z$$
and 
$$p_*\colon H_n(X)=\Z_k\stackrel{0}{\to}H_n(S^{n+1})=0$$
and at $H_0$ it is identity $\Z\to\Z$.\par 
\end{proof}


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