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\begin{flashcard}[f-star-isomorphism-thm]
\begin{theorem*}
  If $f : X \to Y$ is a \gls{homequivce}, $x_0 \in X$, then
  \[ \ps f : \pio(X, x_0) \to \pio(Y, f(x_0)) \]
  is an isomorphism.
\end{theorem*}

\begin{proof}
  Let $g ; Y \to X$ be a \gls{homotopy} inverse,
  $f \circ g \homotopic_H \id_Y$, $g \circ f
  \homotopic_{H'} \id_X$. Let $u' : I \to
  X$ be
  \[ u'(t) = H'(x_0, 1 - t) ,\]
  a \gls{path} from $x_0$ to $g \circ f(x_0)$. The
  previous lemma gives an isomorphism
  \[ \pathisom{u'} = \ps{(g \circ f)} \cdot
  \pio\bs(X, x_0) \stackrel{\ps
  f}{\leftrightarrow} \pio\bs(Y, f(x_0)) \stackrel{\ps
  g}{\twoheadrightarrow} \pio\bs(X_1, gf(x_0)) \]
  We want to show $\leftrightarrow$ is an
  isomorphism, and to do this it is enough to show
  $\twoheadrightarrow$ is injective. To show $\ps
  g$ is injective, consider $u(t) = H(g(x_0), 1 -
  t)$ and get an isomorphism
  \[ \pathisom u = \ps{(f \circ g)} \cdot
  \pio\bs(Y, f(x_0)) \stackrel{\ps
  g}{\hookrightarrow} \pio\bs(X, gf(x_0))
  \stackrel{\ps f}{\twoheadrightarrow} \pio\bs(Y,
  fgf(x_0)) \]
  So $\ps g$ is injective, so the $\ps f$ above is
  $\ps{g^{-1}} \circ \pathisom{u'}$, an
  isomorphism.
\end{proof}
\end{flashcard}

\begin{flashcard}[simply-connected-defn]
\begin{definition*}[Simply connected]
  \glsadjdefn{simpconn}{simply-connected}{space}
  \cloze{A space $X$ is \emph{simply-connected} if it is
  \gls{pathconn} and $\pio\bs(X, x_0) = \{e\}$ for
  some (hence all) $x_0 \in X$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  A \gls{contractible} space is \gls{simpconn}. $X
  \homoteq \{*\}$, so $\piz(X)$ and $\pio\bs(X,
  x_0)$ are trivial.
\end{example*}

\begin{flashcard}[simply-connected-iff-lemma]
\begin{lemma*}
  $X$ is \gls{simpconn} if and only if $\forall
  x_0, x_1 \in X$, there is a unique
  \gls{homotopy} class of \glspl{path} form $x_0$
  to $x_1$.
\end{lemma*}

\begin{proof}
  \cloze{Let $X$ be \gls{simpconn}, and $x_0, x_1 \in X$.
  As $X$ is \gls{pathconn}, there \emph{exists} a
  \gls{path} from $x_0$ to $x_1$. If $\gamma,
  \gamma'$ are two such \glspl{path}, then
  $\invpath \gamma \concat \gamma$ is a loop based
  at $x_1$, so $[\invpath \gamma \gamma'] \in
  \pio\bs(X, x_1) = \{e\}$, so $\invpath \gamma
  \concat \gamma' \homotopic c_{x_1}$ relative to
  $x_1$. So $\gamma' \homotopic \gamma \concat
  \invpath \gamma \concat \gamma' \homotopic
  \gamma \concat \constpath{x_1} \homotopic
  \gamma$ relative to end points.

  Conversely, if $X$ has the stated property then:
  \begin{enumerate}[(i)]
    \item It is \gls{pathconn}.
    \item Any loop based at $x_0$ is
      \gls{homotopic} to $\constpath{x_0}$ as
      loops.
  \end{enumerate}
  Hence $\pio\bs(X, x_0) = \{e\}$.}
\end{proof}
\end{flashcard}

\subsubsection*{Covering Spaces}

\begin{flashcard}[covering-map-defn]
\begin{definition*}[Covering map]
  \glsnoundefn{covmap}{covering map}{covering maps}
  \glsnoundefn{covsp}{covering space}{covering
  spaces}
  \cloze{A \emph{covering map} $p : \tilde{X} \to X$ is a
  continuous \gls{map} such that for any $x \in X$
  there exists an open neighbourhood $U \ni
  x$ such that
  \[ p^{-1}(U) = \coprod_{\alpha \in I} V_\alpha \]
  and $p|_{V_\alpha} \to U$ is a homeomorphism.}
\end{definition*}
\end{flashcard}

\begin{example*}
  Let $S^1 \subset \CC$ be the unit complex
  numbers, and
  \begin{align*}
    p : \RR
    &\to S^1 \\
    t
    &\mapsto e^{2\pi i t}
    &= (\cos(2\pi t), \sin(2\pi t))
  \end{align*}
  Let $U_{y > 0} = \{x + iy \in S^1 \st y > 0\}$.
  Then
  \[ p^{-1}(U_{y > 0}) = \coprod_{j \in \ZZ} \left(j, j
  + \half\right) \]
  Now
  \begin{align*}
    p|_{\left( j, j + \half \right)} : \left( j, j
    + \half\right)
    &\to U_{y > 0} \\
    j + \frac{\arccos(x)}{2\pi}
    &\mapsfrom x + iy
  \end{align*}
  Similarly for $U_{y < 0}$, $U_{x < 0}$, $U_{x >
  0}$. So $p$ is a \gls{covmap}.
\end{example*}

\begin{example*}
  For some $n > 0$, let
  \begin{align*}
    p : S^1
    &\to S^1 \\
    z
    &\mapsto z^n
  \end{align*}
  Let $y \in S^1$ and consider $p^{-1}(y) =
  \{\text{$n$-th roots of $y$}\}$. Choosing a root
  $\xi$ and letting $\eta = e^{\frac{2\pi i}{n}}$,
  this can be written as
  \[ p^{-1}(y) = \{\xi, \eta \xi, \eta^2 \xi,
  \ldots, \eta^{n - 1}\xi\} \]
  Then $S^1 - \{y\}$ is open and
  \[ p^{-1}(S^1 - \{y\}) = S^1 - \{\xi, \eta \xi,
  \eta^2 \xi, \ldots, \eta^{n - 1} \xi\} \]
  $V_0 = \left\{z \in S^1 \st 0 < \arg \left(
  \frac{z}{\xi} \right) < \frac{2\pi}{n}\right\}$
  and $V_i = \eta^i \cdot V_0$. Each $x \neq y \in
  S^1$ has a unique $n$-th root in each $V_i$, so
  $p|_{V_i} : V_i \to S^1 - \{y\}$ is a bijection:
  in fact a homeomorphism.
\end{example*}

\begin{example*}
  $S^2 \subset \RR^3$ be the unit vectors. Let
  \[ \RR \PP^2 = S^2 / x \sim -x .\]
  Have
  \begin{align*}
    p : S^2
    &\to \RR \PP^2 \\
    x
    &\mapsto [x]
  \end{align*}
  Let $V = \{(x, y, z) \in S^2 \st z \neq 0\}$,
  and $U = p(V)$. Then $p^{-1}(U) = V$ is open, so
  $U$ is open in $\RR \PP^2$. Now $p^{-1}(U) = V =
  V_{z > 0} \amalg V_{z < 0}$.

  \textbf{Claim:} $p|_{V_{z > 0}} : V_{z > 0} \to
  U$ and $p|_{V_{z > 0}}$ are homeomorphisms.

  \begin{proof}
    To construct an inverse $g : U \to V_{z > 0}$,
    use definition of quotient topology. Consider
    \begin{align*}
      t : V
      &\to V_{z > 0} \\
      (x, y, z)
      &\mapsto (x, y, z) \text{ if $z \ge 0$} \\
      &\mapsto (x, y, -z) \text{ if $z < 0$}
    \end{align*}
    a continuous map. Note $t$ descends to $\ol{t}
    : U \to V_{z > 0}$ as a map of sets, so a
    continuous map by definition of quotient
    toplogy. It is inverse to $p|_{V_{z > 0}}$.
  \end{proof}

  Do same with $x$-coordinate and $y$-coordinate
  to show that this is a \gls{covmap} of $\RR
  \PP^2$.
\end{example*}