\section{Homology of CW-complexes and applications} \begin{cislo}\label{HOCW1appl}{\bf First applications of homology.} Using homology groups we can easily prove the following statements: \begin{enumerate} \item $S^n$ is not a retract of $D^{n+1}$. \item Every map $f:D^n\to D^n$ has a fixed point, i.e. there is $x\in D^n$ such that $f(x)=x$. \item If $\emptyset\ne U\subseteq \mathbb R^n$ and $\emptyset\ne V\subseteq \mathbb R^m$ are open homeomorphic sets, then $n=m$. \end{enumerate} \end{cislo} \begin{proof}[Outline of the proof] (1) Suppose that there is a retraction $r:D^{n+1}\to S^n$. Then we get the commutative diagram $$ \xymatrix{ \mathbb Z=H_n(S^n) \ar[rr]^{\id} \ar[rd]_{i_*} &{}& H_n(S^n)=\mathbb Z \\ {} & H_n(D^{n+1})=0 \ar[ru]_{r_*} &{} } $$ which is a contradiction. \noindent (2) Suppose that $f:D^n\to D^n$ has no fixed point. Then we can define the map $g:D^n\to S^{n-1}$ where $g(x)$ is the intersection of the ray from $f(x)$ to $x$ with $S^{n-1}$. However, this map would be a retraction, a contradiction with (1). \noindent (3) The proof of the last statement follows from the isomorphisms: $$H_i(U,U-\{x\})\cong H_i(\mathbb R^n,\mathbb R^n-\{x\})\cong \tilde H_{i-1}(\mathbb R^n-\{x\})\cong \tilde H_{i-1}(S^{n-1})= \begin{cases}\mathbb Z\quad&\text{for }i=n,\\ 0\quad&\text{for }i\ne n. \end{cases}$$ \end{proof} \begin{cislo}\label{HOCWdeg}{\bf Degree of a map.} Consider a map $f:S^n\to S^n$. In homology $f_*:\tilde H_n(S^n)\to H_n(S^n)$ has the form $$f_*(x)=ax,\quad a\in\mathbb Z.$$ The integer $a$ is called the \emph{degree}\index{dgree of a map} of $f$ and denoted by $\deg f$. \noindent The degree has the following properties: \begin{enumerate} \item $\deg \id=1$. \item If $f\sim g$, then $\deg f=\deg g$. \item If $f$ is not surjective, then $\deg f=0$. \item $\deg(fg)=\deg f\cdot\deg g$. \item Let $f:S^n\to S^n$, $f(x_0,x_1,\dots, x_n)=(-x_0,x_1,\dots, x_n)$. Then $\deg f=-1$. \item The antipodal map $f:S^n\to S^n$, $f(x)=-x$ has $\deg f=(-1)^{n+1}$. \item If $f:S^n\to S^n$ has no fixed point, then $\deg f=(-1)^{n+1}$. \end{enumerate} \begin{proof} We outline only the proof of (5) and (7). The rest is not difficult and left as an exercise. We show (5) by induction on $n$. The generator of $\tilde H_0(S^0)$ is $1-(-1)$ and $f_*$ maps it in $(-1)-1$. Hence the degree is $-1$. Suppose that the statement is true for $n$. To prove it for $n+1$ we use the diagram with rows coming from a suitable Mayer-Vietoris exact sequence $$ \xymatrix{ 0\ar[r] & \tilde H_{n+1}(S^{n+1}) \ar[r]^-{\cong} \ar[d]_{f_*} & \tilde H_n(S^n)\ar[r] \ar[d]^{(f/S^n)_*} & 0\\ 0\ar[r] & \tilde H_{n+1}(S^{n+1}) \ar[r]^-{\cong} & \tilde H_n(S^n)\ar[r] & 0 } $$ If $(f/S^n)_*$ is a multiplication by $-1$, so is $f_*$. To prove (7) we show that $f$ is homotopic to the antipodal map through the homotopy $$H(x,t)=\frac{tf(x)-(1-t)x}{\Vert tf(x)-(1-t)x \Vert}.$$ \end{proof} \begin{cor*} $S^n$ has a nonzero continuous vector field if and only if $n$ is odd. \end{cor*} \begin{proof} Let $S^n$ has such a field $v(x)$. We can suppose $\Vert v(x)\Vert=1$. Then the identity is homotopic to antipodal map through the homotopy $$H(x,t)=\cos t\pi\cdot x+\sin t\pi\cdot v(x).$$ Hence according to properties (2) and (6) $$(-1)^{n+1}=\deg (-\id)=\deg (\id)=1.$$ Consequently, $n$ is odd. On the contrary, if $n=2k+1$, we can define the required vector field by prescription $$v(x_0,x_1,x_2,x_3,\dots, x_{2k}, x_{2k+1})= (-x_1,x_0,-x_3,x_2,\dots,-x_{2k+1},x_{2k}).$$ \end{proof} \begin{ex*} Prove the properties (3), (4) and (6) of the degree. \end{ex*} \end{cislo} \begin{cislo}\label{HOCWld} {\bf Local degree.} Consider a map $f:S^n\to S^n$ and $y\in S^n$ such that $f^{-1}(y)=\{x_1,x_2,\dots ,x_m\}$. Let $U_i$ be open disjoint neighbourhoods of points $x_i$ and $V$ a neighbourhood of $y$ such that $f(U_i)\subseteq V$. Then \begin{multline*} (f/U_i)_*:H_n(U_i,U_i-\{x_i\})\cong H_n(S^n,S^n-\{x_i\})=\mathbb Z\\ \longrightarrow H_n(V,V-\{y\})\cong H_n(S^n,S^n-\{y\})=\mathbb Z \end{multline*} is a multiplication by an integer which is called a \emph{local degree} \index{local degree} and denoted by $\deg f\vert x_i$. \begin{thma} Let $f:S^n\to S^n$, $y\in S^n$ and $f^{-1}(y)=\{x_1,x_2,\dots ,x_m\}$. Then $$\deg f=\sum_{i=1}^m\deg f\vert x_i.$$ \end{thma} For the proof see [Hatcher], %\cite{Ha}, Proposition 2.30, page 136. The suspension $Sf$ of a map $f:X\to Y$ is given by the prescription $Sf(x,t)=(f(x),t)$. \begin{thma} $\deg Sf=\deg f$ for any map $f:S^n\to S^n$. \end{thma} \begin{proof} $f$ induces $Cf:CS^n\to CS^n$. The long exact sequence for the pair $(CS^n,S^n)$ and the fact that $SS^n=CS^n/S^n$ give rise to the diagram $$ \xymatrix{ \tilde H_{n+1}(S^{n+1}) \ar[r]_-{\cong} \ar[d]_{Sf_*} & \tilde H_{n+1}(CS^n,S^n)\ar[r]^-{\partial_*}_-{\cong} \ar[d]_{Cf_*} & \tilde H_n(S^n) \ar[d]^{f_*}\\ \tilde H_{n+1}(S^{n+1}) \ar[r]_-{\cong} & \tilde H_{n+1}(CS^n,S^n) \ar[r]^-{\partial_*}_-{\cong} & \tilde H_n(S^n) } $$ which implies the statement. \end{proof} \begin{cor*} For any $n\ge 1$ and given $k\in \mathbb Z$ there is a map $f:S^n\to S^n$ such that $\deg f=k$. \end{cor*} \begin{proof} For $n=1$ put $f(z)=z^k$ where $z\in S^1\subset \mathbb C$. Using the computation based on local degree as above, we get $\deg f=k$. The previous theorem implies that the degree of $S^{n-1}f:S^n\to S^n$ is also $k$. \end{proof} \end{cislo} \begin{cislo}\label{HOCWcomp}{\bf Computations of homology of CW-complexes.} If we know a CW-structure of a space $X$, we can compute its cohomology relatively easily. Consider the sequence of Abelian groups and its morphisms \begin{equation*} (H_n(X^n,X^{n-1}),d_n) \end{equation*} where $d_n$ is the composition $$H_n(X^n,X^{n-1})\xrightarrow{\ \partial_n\ } H_{n}(X^{n-1}) \xrightarrow{\,j_{n-1}\,}H_{n-1}(X^{n-1},X^{n-2}).$$ \begin{thm*} Let $X$ be a CW-complex. $(H_n(X^n,X^{n-1}),d_n)$ is a chain complex with homology $$H^{CW}_n(X)\cong H_n(X).$$ \end{thm*} \begin{proof} First, we show how the groups $H_k(X^n,X^{n-1})$ look like. Put $X^{-1}=\emptyset$ and $X^0/\emptyset=X^0\sqcup\{*\}$. Then $$H_k(X^n,X^{n-1})=\tilde H_k(X^n/X^{n-1})=\tilde H_k(\bigvee S^n_{\alpha})= \begin{cases}\bigoplus_{\alpha}\mathbb Z \ & n=k,\\ 0\ &n\ne k. \end{cases} $$ Now we show that $$H_k(X^n)=0\quad\text{for }k>n.$$ From the long exact sequence of the pair $(X^n,X^{n-1})$ we get $H_k(X^n)=H_k(X^{n-1})$. By induction $H^k(X^n)=H_k(X^{-1})=0$. Next we prove that $$H_k(X^n)=H_k(X)\quad\text{for }k\le n-1.$$ From the long exact sequence for the pair $(X^{n+1},X^n)$ we obtain $H_k(X^n)=H_k(X^{n+1})$. By induction $H_k(X^n)=H_k(X^{n+m})$ for every $m\ge 1$. Since the image of each singular chain lies in some $X^{n+m}$ we get $H_k(X^n)=H_k(X)$. To prove Theorem we will need the following diagram with parts of exact sequences for the pairs $(X^{n+1},X^n)$, $(X^n,X^{n-1})$ and $(X^{n-1},X^{n-2})$. $$ \xymatrix{ {} & 0 &{ }&{}\\ 0 \ar[rd] & H_n(X^{n+1})\ar[u] &{} & {}\\ {}& H_n(X^n)\ar[u]\ar[rd]^-{j_n} & {} & {} \\ {} & H_{n+1}(X^{n+1}, X^n)\ar[u]^{\partial_{n+1}} \ar[r]^-{d_{n+1}}& H_n(X^n,X^{n-1}) \ar[r]^-{d_n} \ar[rd]^{\partial_n}& H_{n-1}(X^{n-1},X^{n-2})\\ {} & {} & {} & H_{n-1}(X^{n-1})\ar[u]^{j_{n-1}}\\ {} & {} & {} & 0 \ar[u] } $$ From it we get $$d_nd_{n+1}=j_{n-1}(\partial_nj_n)\partial_{n+1}=j_{n-1}(0) \partial_{n+1}=0.$$ Further, $$\ker d_n=\ker\partial_n=\im j_n\cong H_n(X^n)$$ and $$\im d_{n+1}\cong\im \partial_{n+1},$$ since $j_{n-1}$ and $j_n$ are monomorphisms. Finally, $$H^{CW}_n(X)=\frac{\ker d_n}{\im d_{n+1}}\cong\frac{H_n(X^n)} {\im\partial_{n+1}}\cong H_n(X^{n+1})\cong H_n(X).$$ \end{proof} \begin{example*} $H_n(X)=0$ for CW-complexes without cells in dimension $n$. $$H_k(\mathbb {CP}^n)= \begin{cases}\mathbb Z\quad&\text{for }k\le 2n \text{ even},\\ 0\quad&\text{in other cases.} \end{cases}$$ \end{example*} \end{cislo} \begin{cislo}\label{HOCWdn}{\bf Computation of $d_n$.} Let $e^n_{\alpha}$ and $e^{n-1}_{\beta}$ be cells in dimension $n$ and $n-1$ of a CW-complex $X$, respectively. Since $$H_n(X^n,X^{n-1})=\bigoplus_{\alpha}\mathbb Z,\quad H_{n-1}(X^{n-1},X^{n-2})=\bigoplus_{\beta}\mathbb Z,$$ they can be considered as generators of these groups. Let $\varphi_{\alpha}:\partial D^n_{\alpha}\to X^{n-1}$ be the attaching map for the cell $e^n_{\alpha}$. Then $$d_n(e^n_{\alpha})=\sum_{\beta}d_{\alpha\beta}e^{n-1}_{\beta}$$ where $d_{\alpha\beta}$ is the degree of the following composition $$S^{n-1}=\partial D^n_{\alpha}\xrightarrow{\varphi_{\alpha}} X^{n-1}\to X^{n-1}/X^{n-2}\to X^n/(X^{n-2}\cup \bigcup_{\gamma\ne\beta} e^{n-1}_{\gamma})=S^{n-1}.$$ For the proof we refer to [Hatcher], %\cite{Ha}, pages 140 and 141. \begin{ex*} Compute homology groups of various 2-dimensional surfaces (torus, Klein bottle, projective plane) using their CW-structure with only one cell in dimension 2. \end{ex*} \end{cislo} \begin{cislo}\label{HOCWproj}{\bf Homology of real projective spaces.} The real projective space $\mathbb{RP}^n$ is formed by cell $e^0,e^1,\dots,e^n$, one in each dimension from 0 to $n$. The attaching map for the cell $e^{k+1}$ is the projection $\varphi:S^k\to \mathbb{RP}^k$. So we have to compute the degree of the composition $$f:S^k\xrightarrow{\varphi}\mathbb{RP}^k\to \mathbb{RP}^k/\mathbb{RP}^{k-1}=S^k.$$ Every point in $S^k$ has two preimages $x_1$, $x_2$. In a neihbourhood $U_i$ of $x_i$ $f$ is a homeomorphism, hence its local degree $\deg f\vert x_i=\pm 1$. Since $f/U_2$ is the composition of the antipodal map with $f/U_1$, the local degrees $\deg f\vert x_1$ and $\deg f\vert x_1$ differs by the multiple of $(-1)^{k+1}$. (See the properties (4) and (6) in \ref{HOCWdeg}.) According to \ref{HOCWld} $$\deg f=\pm 1(1+(-1)^{k+1})= \begin{cases}0\quad & \text{for }k+1\text{ odd},\\ \pm 2\quad & \text{for }k+1\text{ even}. \end{cases}$$ So we have obtained the chain complex for computation of $H_*^{CW}(\mathbb {RP}^n)$. The result is $$H_k(\mathbb{RP}^n)= \begin{cases}\mathbb Z\quad & \text{for }k=0\text{ and } k=n\text{ odd},\\ \mathbb Z_2\quad & \text{for }k\text{ odd , }0