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% II Algebraic Topology
% Oscar Randal-Williams (or257@cam.ac.uk)

\subsubsection*{Why study Algebraic Topology?}

The main kind of problem we will study is:

\textbf{Extension problem:} If $X$ is a space, $A
\subset X$ a subspace and $f : A \to Y$ a
continuous function, is there a continuous
function $F : X \to Y$ such that $F|_A = f$.

Specific examples look like:

\begin{theorem*}
  There is no continuous $F : D^n \to S^{n - 1}$
  such that
  \[ S^{n - 1}
  \stackrel{\text{incl}}{\hookrightarrow} D^n
  \stackrel{F}{\to} S^{n - 1} \]
  is the identity.
\end{theorem*}

\begin{theorem*}
  There is no group homomorphism $F : \{0\} \to
  \ZZ$ such that
  \[ \ZZ \to \{0\} \stackrel{F}{\to} \ZZ \]
  is the identity.
\end{theorem*}

\begin{theorem*}
  $\RR^n \cong \RR^m \iff n = m$.
\end{theorem*}

\begin{theorem*}[Fundamental theorem of algebra]
  Any non-constant polynomial over $\CC$ has a
  root in $\CC$.
\end{theorem*}

\subsubsection*{Recollections}

\glsnoundefn{map}{map}{maps}%
We will call a continuous function a \emph{map}.

\begin{flashcard}[gluing-lemma]
\begin{lemma*}[Gluing lemma]
  \label{gluing_lemma}
  \cloze{
  Let $f : X \to Y$ be a function between
  topological spaces, and let $C, K \subset X$ be
  closed sets such that $X = C \cup K$. Then $f$
  is continuous if and only if $f|_C$, $f|_K$ are
  continuous.}
\end{lemma*}
\end{flashcard}

\begin{proof}
  Exercise.
\end{proof}

\begin{flashcard}[lebesgue-number-lemma]
\begin{lemma*}[Lebesgue number lemma]
  \label{lebesgue_number_lemma}
  \cloze{Let $(X, d)$ be a metric space, and assume it is
  compact. For any open cover $\mathcal{U} =
  \{U_\alpha \subset X\}$, there is $\delta > 0$
  such that each $B_\delta(x)$ is contained in
  some $U_\alpha$.}
\end{lemma*}
\end{flashcard}

\subsubsection*{Cell complex}

\begin{flashcard}[attaching-n-cell-definition]
\begin{definition*}[Attaching an $n$-cell]
  \glssymboldefn{fcellcup}{cupf}{cupf}
  \glsverbdefn{attncell}{attaching an $n$
  cell}{attaching an $n$ cell}
  \glsnoundefn{cellstruc}{cell structure}{cell
  structures}
  \cloze{For a space $X$ and a \gls{map} $f : S^{n - 1} \to X$,
  the space obtained by attaching an $n$-cell to
  $X$ is
  \[ X \fcellcup D^n = (X \ci D^n) / (z \in S^{n -
  1} \subset D^n \tilde f(z) \in X .\]
  }
  \fcscrap{
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/e8fa91cdfb804475.png}
  \end{center}
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[finite-cell-complex-defn]
\begin{definition*}[Finite cell complex]
  \cloze{A (finite) \emph{cell complex} is a space $X$
  obtained by the following recipe:
  \begin{enumerate}[(i)]
    \item Start with a finite set $X^0$ with the
      discrete topology.
    \item If $X^{n - 1}$ has been defined, form
      $X^n$ by \glsref[attncell]{attaching a finite collection of
      $n$-cells} along some \glspl{map} $\{f_\alpha : S^{n
      - 1} \to X^{n - 1}\}$. This $X^n$ is called
      the $n$-skeleton.
    \item Stop with $X = X^k$. $k$ is called the
      dimension of $X$.
  \end{enumerate}
  }
\end{definition*}
\end{flashcard}

\setcounter{section}{2}
\subsection{Homotopy}

\begin{flashcard}[homotopy-defn]
\begin{definition*}[Homotopy]
  \glssymboldefn{homotopic}{$\simeq$}{$\simeq$}%
  \glsnoundefn{homotopy}{homotopy}{N/A}
  \glsadjdefn{homotopic}{homotopic}{homotopic}
  \cloze{Let $f, g : X \to Y$ be \glspl{map}. A
  \emph{homotopy} from $f$ to $g$ is a map $H : X
  \times I \to Y$ such that
  \begin{align*}
    H(x, 0)
    &= f(x) \\
    H(x, 1)
    &= g(x)
  \end{align*}
  If such an $H$ exists, say \emph{$f$ is
  homotopic to $g$}, and write $f \homotopic g$.

  \glssymboldefn{homrel}{$f \simeq g ~rel~ A$}{$f \simeq g ~rel~ A$}
  If $A \subset X$ is a subspace, say $H$ is a
  \emph{homotopy relative to $A$} if $H(a, t) =
  H(a, 0)$ for all $t \in I$, $a \in A$. Write $f
  \homotopic g \homrel A$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[homotopic-rel-A-is-eq-rel]
\begin{proposition*}
  Being homotopic $\homrel A$ is an equivalence
  relation on the set of \glspl{map} from $X$ to
  $Y$.
\end{proposition*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $f \homotopic f$ via $H(x, t) = f(x)$.
    \item If $f \simeq g$ via $H$, let $H'(x, t) =
      H(x, 1 - t)$. This is a \gls{homotopy} from
      $g$ to $f$.
    \item If $f \homotopic g$ via $H$, and $g
      \homotopic h$ via $H'$ then let
      \[ H''(x, t) = \begin{cases}
        H(x, 2t) & 0 \le t \le \half \\
        H(x, 2t - 1) & \half \le t \le 1
      \end{cases} \]
      (well-defined as $H(x, 1) = g(x) = H'(x, 0)$).
      This is continuous on $X \times \left[ 0,
      \half \right]$ and on $X \times \left[ \half
      \times 1\right]$, so is continuous on $X
      \times I$ by \nameref{gluing_lemma}. So $f
      \homotopic h$.
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[homotopy-equivalence-defn]
\begin{definition*}[Homotopy equivalence]
  \glsnoundefn{homequivce}{homotopy
  equivalence}{homotopy equivalences}%
  \glsnoundefn{homequivt}{homotopy
  equivalent}{homotopy equivalent}
  \glssymboldefn{homoteq}{$\simeq$}{$\simeq$}
  \cloze{A \gls{map} $f ; X \to Y$ is a \emph{homotopy
  equivalence} if there is a $g : Y \to X$ such
  that $f \circ g \homotopic \id_Y$ and $g \circ f
  \homotopic \id_X$.

  Say $X$ is \emph{homotopy equivalent to $Y$},
  written $X \homoteq Y$, if a \gls{homequivce}
  $f : X \to Y$ exists.}
\end{definition*}
\end{flashcard}

\begin{example*}
  Let $X = S^1$, $Y = \RR^2 - \{0\}$. Let $i : X \to
  Y$ be the inclusion. Let $r : Y \to X$,
  continuous, given by $x \mapsto \frac{x}{|x|}$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/7f275121a970457b.png}
  \end{center}
  Now $r \circ i = \id_X$. On the other hand, $i
  \circ r : Y \to Y$ is not the identity. But
  \[ H(x, t) = \frac{x}{t + |x|(1 - t)} \]
  is a \gls{homotopy} from $\id_Y$ to $i \circ r$.
  So $\RR^2 - \{b\} \homoteq S^1$.
\end{example*}

\begin{flashcard}[contractible-defn]
\begin{definition*}[Contractible]
  \glsadjdefn{contractible}{contractible}{contractible}%
  \cloze{
  $X$ is called \emph{contractible} if $X
  \homotopic \{*\}$ (here, $\{*\}$ denotes a 1-point
  space).}
\end{definition*}
\end{flashcard}

\begin{lemma*}
  Let $f_0, f_1 : X \to Y$ and $g_0, g_1 : Y \to
  Z$ be \gls{homotopic} \glspl{map}. Then $g_0
  \circ f_0$ and $g_1 \circ f_1$ are
  \gls{homotopic}.
\end{lemma*}

\begin{proof}
  Lets show $g_0 \circ f_0 \homotopic g_0 \circ
  f_1 \homotopic g_1 \circ f_1$.

  To show $g_0 \circ f_0 \homotopic g_0 \circ
  f_1$: If $H$ is a \gls{homotopy} $f_0$ to $f_1$,
  then $g_0 \circ H : X \times I \to Y \to Z$ is a
  \gls{homotopy}.

  To show $g_0 \circ f_1 \homotopic g_1 \circ
  f_1$: If $G$ is a \gls{homotopy} $g_0$ to $g_1$,
  then $G \circ (f_1 \times \id_I) : X \times I
  \to Y \times I \to Z$ is a
  \gls{homotopy}.
\end{proof}
