\section{Basic notions and constructions} \begin{cislo}{\bf Notation.} The closure, the interior and the boundary of a topological space $X$ will be denoted by $\overline X$, $\inte X$ and $\partial X$, respectively. The letter $I$ will stand for the interval $[0,1]$. $\mathbb R^n$ and $\mathbb C^n$ will denote the vector spaces of $n$-tuples of real and complex numbers, respectively, with the standard norm $\Vert x\Vert=\sum_{i=1}^{n}\vert x_i\vert^2$. The sets \begin{align*} D^n&=\{x\in\mathbb R^n;\ \Vert x\Vert\le 1\},\\ S^n&=\{x\in\mathbb R^{n+1};\ \Vert x\Vert =1\} \end{align*} are the $n$-dimensional disc and the $n$-dimensional sphere, respectively. \end{cislo} \begin{cislo} {\bf Categories of topological spaces.} Every category consists of objects and morphisms between them. Morphisms $f:A\to B$ and $g:B\to C$ can be composed in a morphism $g\circ f:A\to C$ and for every object $B$ there is a morphism $\id_B:B\to B$ such that $\id_B\circ f=f$ and $g\circ \id_B=g$. The category with topological spaces as objects and continuous maps as morphisms will be denoted $\top$. Topological spaces with distinquished points (usually denoted by $\ast$) and continuous maps $f:(X,\ast)\to (Y,\ast)$ such that $f(\ast)=\ast$ form the category $\top_*$. Topological spaces $X$, $A$ will be called \emph{a pair of topological spaces}\index{pair of topological spaces} if $A$ is a subspace of $X$ (notation $(X,A)$). The notation $f:(X,A)\to (Y,B)$ means that $f:X\to Y$ is a continuous map which preserves subspaces, i.~e. $f(A)\subseteq B$. The category $\top^2$ consists of pairs of topological spaces as objects and continuous maps $f:(X,A)\to (Y,B)$ as morphisms. Finally, $\top^2_*$ will denote the category of pairs of topological spaces with distinquished points in subspaces and continuous maps preserving both subspaces and distinquished points. The right category for doing algebraic topology is the category of compactly generated spaces. We will not go into details and refer to Chapter 5 of [May]. %\cite{May} In fact, the majority of spaces we deal with in this text are compactly generated. From now on, a space will mean a topological space and a map will mean a continuous map. \end{cislo} \begin{cislo}{\bf Homotopy.} Maps $f,g:X\to Y$ are called \emph{homotopic}\index{homotopic}, notation $f\sim g$, if there is a map $h:X\times I\to Y$ such that $h(x,0)=f(x)$ and $h(x,1)=g(x)$. This map is called \emph{homotopy}\index{homotopy} between $f$ and $g$. The relation $\sim$ is an equivalence. Homotopies in categories $\top_*$, $\top^2$ or $\top^2_*$ have to preserve distinquished points, i.~e. $h(\ast,t)=\ast$, subsets or both subsets and distinquished points, respectively. Spaces $X$ and $Y$ are called \emph{homotopy equivalent}\index{homotopy equivalent} if there are maps $f:X\to Y$ and $g:Y\to X$ such that $f\circ g\sim\id_Y$ and $g\circ f\sim\id_X$. We also say that the spaces $X$ and $Y$ have the same \emph{homotopy type}\index{homotopy type}. The maps $f$ and $g$ are called \emph{homotopy equivalences}. A space is called \emph{contractible}\index{contractible} if it is homotopy equivalent to a point. \begin{example*} $S^n$ and $\mathbb R^{n+1}-\{0\}$ are homotopy equivalent. As homotopy equivalences take the inclusion $f:S^n\to \mathbb R^{n+1}-\{0\}$ and $g:\mathbb R^{n+1}-\{0\}\to S^n$, $g(x)=x/\Vert x\Vert$. \end{example*} \end{cislo} \begin{cislo}{\bf Retracts and deformation retracts.} Let $i:A\hookrightarrow X$ be an inclusion. We say that $A$ is a \emph{retract}\index{retract} of $X$ if there is a map $r:X\to A$ such that $r\circ i=\id_A$. The map $r$ is called a \emph{retraction}\index{retraction}. We say that $A$ is a \emph{deformation retract}\index{deformation retract} of $X$ (sometimes also strong deformation retract) if $i\circ r:X\to A\to X$ is homotopic to the identity on $X$ relative to $A$, i.e. there is a homotopy $h:X\times I\to X$ such that $h(-,0)=\id_X$, $h(-,1)=i\circ r$ and $h(i(-),t)=\id_A$ for all $t\in I$. The map $h$ is called a \emph{deformation retraction} \index{deformation retraction}. \begin{ex} Show that deformation retract of $X$ is homotopy equivalent to $X$. \end{ex} \end{cislo} \begin{cislo}{\bf Basic constructions in $\top$.} Consider a topological space $X$ with an equivalence $\simeq$. Then $X/{\simeq}$ is the set of equivalence classes with the topology determined by the projection $p:X\to X/{\simeq}$ in the following way: $U\subseteq X/{\simeq}$ is open iff $p^{-1}(U)$ is open in $X$. \begin{ex} The map $f:(X/{\simeq}) \to Y$ is continuous iff the composition $f\circ p:X\to (X/{\simeq})\to Y$ is continuous. \end{ex} We will show this constructions in several special cases. Let $A$ be a subspace of $X$. The \emph{quotient}\index{quation} $X/A$ is the space $X/{\simeq}$ where $x\simeq y$ iff $x=y$ or both $x$ and $y$ are elements of $A$. This space is often considered as a based space with base point determined by $A$. If $A=\emptyset$ we put $X/\emptyset=X\cup \{\ast\}$. \begin{ex} Prove that $D^n/S^{n-1}$ is homeomorphic to $S^n$. For it consider $f:D^n\to S^n$ \begin{equation*}f(x_1,x_2,\dots,x_n)=(2\sqrt{1-\Vert x\Vert^2}x,2\Vert x\Vert^2-1). \end{equation*} \end{ex} \emph{Disjoint union}\index{disjoint union} of spaces $X$ and $Y$ will be denoted $X\sqcup Y$. Open sets are unions of open sets in $X$ and in $Y$. Let $A$ be a subspace of $X$ and let $f:A\to Y$ be a map. Then $X\cup_f Y$ is the space $(X\sqcup Y)/{\simeq}$ where the equivalence is generated by relations $a\simeq f(a)$. The \emph{mapping cylinder}\index{mapping cone} of a map $f:X\to Y$ is the space $$M_f=X\times I\cup_{f\times{1}}Y$$ which arises from $X\times I$ and $Y$ after identification of points $(x,1)\in X\times I$ and $f(x)\in Y$. %vlozeno 20130226 \begin{figure}[htb] \centering \def\svgwidth{400pt} \input{img/01_mcyl.pdf_tex} \caption{Mapping cylinder} \end{figure} %vlozeno 20130226 \begin{ex} We have two inclusions $i_X:X=X\times \{0\}\hookrightarrow M_f$ and $i_Y:Y\hookrightarrow M_f$ and a retraction $r:M_f\to Y$. How is $r$ defined? $$ \xymatrix{ & X \ar[ld]_-f \ar[d]^{i_X} \ar[rd]^-f\\ Y\ar[r]_-{i_Y} & M_f\ar[r]_-r &Y }$$ Prove that \begin{enumerate} \item $Y$ is a deformation retract of $M_f$, \item $i_X\circ r=f$, \item $i_Y\circ f\sim i_X$. \end{enumerate} \end{ex} The \emph{mapping cone}\index{mapping cone} of a mapping $f:X\to Y$ is the space $$C_f=M_f/(X\times \{0\}).$$ A special case of a mapping cone is the \emph{cone}\index{cone} of a space $X$ $$CX=X\times I/(X\times\{0\})=C_{\id_X}.$$ The \emph{suspension}\index{suspension} of a space $X$ is the space $$SX=CX/(X\times\{1\}).$$ \begin{ex} Show that $SS^n=S^{n+1}$. For it consider the map $f:S^n\times I\to S^{n+1}$ $$f(x,t)=(\sqrt{1-(2t-1)^2}x, 2t-1).$$ \end{ex} The \emph{join}\index{join} of spaces $X$ and $Y$ is the space $$X\star Y=X\times Y\times I/{\simeq}$$ where $\simeq$ is the equivalence generated by $(x,y_1,0)\simeq(x,y_2,0)$ and $(x_1,y,1)\simeq(x_2,y,1)$. \begin{ex} Show that the join operation is associative and compute the joins of two points, two intervals, several points, $S^0\star X$, $S^n\star S^m$. \end{ex} \end{cislo} \begin{cislo}{\bf Basic constructions in $\top_*$ and $\top^2$.} Let $X$ be a space with a base point $x_0$. The \emph{reduced suspension}\index{reduced suspension} of $X$ is the space $$\Sigma X=SX/(\{x_0\}\times I)$$ with base point determined by $x_0\times I$. In the next section in \ref{CW1crit} we will show that $\Sigma X$ is homotopy equivalent to $SX$. The space $$(X,x_0)\lor (Y,y_0)=X\times\{y_0\}\cup \{x_0\}\times Y$$ with distinquished point $(x_0,y_0)$ is called the \emph{wedge}\index{wedge} of $X$ and $Y$ and usually denoted only as $X\lor Y$. The \emph{smash product}\index{smash product} of spaces $(X,x_0)$ and $(Y,y_0)$ is the space $$X\land Y=X\times Y/(X\times\{y_0\}\cup \{x_0\}\times Y)=X\times Y/X\lor Y.$$ Analogously, the smash product of pairs $(X,A)$ and $(Y,B)$ is the pair $$(X\times Y, A\times Y\cup X\times B).$$ \begin{ex} Show that $S^m\land S^n=S^{n+m}$. One way how to do it is to prove that $$X/A\land Y/B\cong X\times Y/A\times Y\cup X\times B.$$ \end{ex} \end{cislo} \begin{cislo}{\bf Homotopy extension property.} We say that a pair of topological spaces $(X,A)$ has the \emph{homotopy extension property}\index{homotopy extension property} (abbreviation HEP\index{HEP}) if any map $f:X\to Y$ and any homotopy $h:A\times I\to Y$ such that $h(a,0)=f(a)$ for $a\in A$, and $$f\cup h:X\times\{0\}\cup A\times I\to Y$$ is continuous, can be extended to a homotopy $H:X\times I\to Y$ such that $H(x,0)=f(x)$ and $H(a,t)=h(a,t)$ for all $x\in X$, $a\in A$ and $t\in I$, i.e. $H$ is an arrow making the diagram $$ \xymatrix{ X\times \{0\}\cup A\times I \ar[r]^-{f\cup h}\ar[d] & Y\\ X\times I \ar@{-->}[ur]_-{H}& }$$ commutative. If the pair $(X,A)$ satisfies HEP, we call the inclusion $A\hookrightarrow X$ a \emph{cofibration}\index{cofibration}. %vlozeno 20130226 \begin{figure}[htb] \centering \def\svgwidth{400pt} \input{img/01_fig-2.pdf_tex} \caption{Homotopy extension property} \end{figure} %vlozeno 20130226 \newpage \begin{thm*} A pair $(X,A)$ has HEP if and only if $X\times\{0\}\cup A\times I$ is a retract of $X\times I$. \end{thm*} %vlozeno 20130226 \begin{figure}[htb] \centering \def\svgwidth{300pt} \input{img/01_fig-3.pdf_tex} \caption{Retraction $X\times I \rightarrow X\times \{0\}\cup A\times I$} \end{figure} %vlozeno 20130226 \begin{ex} Using this Theorem show that the pair $(D^n,S^{n-1})$ satisfies HEP. Many other examples will be given in the next section. \end{ex} %vlozeno 20130226 \begin{figure}[htb] \centering \def\svgwidth{300pt} \input{img/01_fig-4.pdf_tex} \caption{Retraction $D^1\times I \rightarrow D^1\times \{0\}\cup S^0\times I$} \end{figure} %vlozeno 20130226 \newpage \begin{proof} [Proof of Theorem] Let $(X,A)$ has HEP. Put $Y=X\times \{0\}\cup A\times I$ and consider $f\cup h:X\times \{0\}\cup A\times I\to X\times\{0\}\cup A\times I$ to be an identity. Its extension $H:X\times I\to X\times \{0\}\cup A\times I$ is a retraction. Let $r:X\times I \to X\times \{0\}\cup A\times I$ be a retraction. Given a map $f$ and a homotopy $h$ as in the definition which together determine a continuous map $F=(f\cup h):X\times \{0\}\cup A\times I\to Y$, then $H=F\circ r$ is an extension of $f\cup h$. \end{proof} \begin{ex} Let a pair $(X,A)$ satisfy HEP and consider a map $g:A\to Y$. Prove that $(X\cup_gY,Y)$ also satisfies HEP. \end{ex} \begin{ex} Let $X$ be a Hausdorff compact space and let an inclusion $A\hookrightarrow X$ is a cofibration. Prove that $A$ is a closed subset of $X$. \end{ex} \begin{ex} Consider the closed subset set $A=\{1/n\in\mathbb R; n=0,1,2,\dots\}\cup \{0\}$ of the interval $[0,1]$. However, the inclusion $A\hookrightarrow [0,1]$ is not a cofibration. Prove it. \end{ex} \begin{ex} Let $M_f$ be a mapping cylinder of a map $f:X\to Y$. Show that the inclusion $i_X:X\hookrightarrow M_f$ is a cofibration. In particular, the map $f:X\to Y$ can be factored into the composition $r\circ i_X$ of the cofibration $i_X$ and the homotopy equivalence $r$. (See the exercise after the definition of the mapping cylinder.) \end{ex} \end{cislo}