%! TEX root = AlgT.tex
% vim: tw=50
% 29/01/2024 11AM

\begin{flashcard}[lift-defn]
\begin{definition*}[Lift]
  \glsnoundefn{lift}{lift}{lifts}
  \cloze{Let $p : \tilde{X} \to X$ be a \gls{covmap}, $f
  : Y \to X$ be a \gls{map}. A \emph{lift} of $f$
  along $p$ is a $\tilde{F} : Y \to \tilde{X}$
  such that $p \circ \tilde{f} = f$.
  \[ \begin{tikzcd}[ampersand replacement=\&]
    \& \tilde{X} \ar[d, "p"] \\
    Y \ar[ur, dashed, "\tilde{f}"] \ar[r, "f"]
    \& X
  \end{tikzcd} \]
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[evenly-covered-defn]
\begin{definition*}[Evenly-covered]
  \glsadjdefn{evencov}{evenly-covered}{set}
  \cloze{Given $p : \tilde{X} \to X$ a
  \gls{covmap}, we say that $U \subset X$ is evenly-covered by
  $\{V_\alpha : \alpha \in I\}$
  if $p^{-1}(U) = \coprod_{\alpha \in I} V_\alpha$
  and $p|_{V_\alpha} : V_\alpha
  \stackrel{\cong}{\to} U_\alpha$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[uniqueness-of-lifts-lemma]
\begin{lemma*}[Uniqueness of Lifts]
  \label{uniqueness_of_lifts}
  \cloze{If $\tilde{f}_0$ and $\tilde{f}_1$ are
  \glspl{lift} of $f : Y \to X$ along a
  \gls{covmap} $p : \tilde{X} \to X$ then
  \[ S \defeq \{ y \in Y \st \tilde{f}_0(y) =
  \tilde{f}_1(y) \} \]
  is both open and closed. In particular, if $Y$
  is connected, then $S = \emptyset$ or $S = Y$.}
\end{lemma*}

\begin{proof}
  \cloze{First show $S$ is open. Let $s \in S$ and let $U
  \ni f(s)$ be an open neighbourhood which is
  \gls{evencov} ($p^{-1}(U) = \bigsqcup V_\alpha$).
  Now $\tilde{f}_0(s)$ and
  $\tilde{f}_1(s)$ agree so live in the same
  $V_\alpha$. Then on $N =
  \tilde{f}_0^{-1}(V_\alpha) \cap
  \tilde{f}_1^{-1}(V_\alpha)$ we have
  \[ p|_{V_\alpha} \circ \tilde{f}_0|_N = f|_N =
  p|_{V_\alpha} \circ \tilde{f}_1|_N ,\]
  but $p|_{V_\alpha}$ is a homeomorphism, so
  $\tilde{f}_0|_N = \tilde{f}_1|_N$. So $s \in N
  \subset S$, so $S$ is open.

  Now we show $S$ is closed. Let $y \in \ol{S}$
  and $\tilde{f}_0(y) \neq \tilde{f}_1(y)$. Let $U
  \ni f(y)$ be an open neighbourhood that is
  \gls{evencov}. Then $\tilde{f}_0(y) \in
  V_\beta$ and $\tilde{f}_1(y) \in V_\gamma$ with
  $\beta \neq \gamma$ (as $\tilde{f}_0(y) \neq
  \tilde{f}_1(y)$). So $\tilde{f}_0^{-1}(V_\beta)
  \cap \tilde{f}_1^{-1}(V_\gamma)$ is an open set,
  containing $y \in \ol{S}$. By definition of closure,
  it intersects $S$. But then $V_\beta$ and
  $V_\gamma$ must intersect, contradiction.}
\end{proof}
\end{flashcard}

\begin{flashcard}[homotopy-lifting-lemma]
\begin{lemma*}[Homotopy lifting lemma]
  \label{homotopy_lifting_lemma}
  \cloze{Let $p : \tilde{X} \to X$ be a \gls{covmap}, $H
  : Y \times I \to X$ from $f_0$ to $f_1$, and
  $\tilde{f}_0$ be a \gls{lift} of $f_0$. Then
  there exists a \emph{unique} homotopy $\tilde{H}
  : Y \times I \to \tilde{X}$ such that:
  \begin{enumerate}[(i)]
    \item $\tilde{H}(\bullet, 0) =
      \tilde{f}_0(\bullet)$.
    \item $p \circ \tilde{H} = H$.
  \end{enumerate}}
\end{lemma*}

\begin{proof}
  \cloze{Let $\{U_\alpha\}_{\alpha \in I}$ be an open
  cover of $X$ by sets which are \gls{evencov},
  i.e. $p^{-1}(U_\alpha) = \coprod_{\beta \in
  I_\alpha} V_\beta$ and $p|_{V_\beta} : V_\beta
  \stackrel{\cong}{\to} U_\alpha$. Now
  $\{H^{-1}(U_\alpha)\}$ is an open cover of $Y
  \times I$, and for each $y_0 \in Y$ it gives an
  open cover of $\{y_0\} \times I$. By the
  \nameref{lebesgue_number_lemma}, there is a $N =
  N(y_0)$ such that each path
  \[ H|_{\{y_0\} \times \left[ \frac{i}{N},
  \frac{i + 1}{N} \right]} : \{y_0\} \times \left[
  \frac{i}{N}, \frac{i + 1}{N} \right] \to X \]
  has image inside some $U_\alpha$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/d84b39865bb54297.png}
  \end{center}
  In fact, as $\{y_0\} \times I$ is compact, there
  is an open $W_{y_0} \ni y_0$ such that each
  $H\left(W_{y_0} \times \left[ \frac{i}{N}, \frac{i +
  1}{N} \right]\right)$ lies in some $U_\alpha$.
  We can construct a lift $\tilde{H}|_{W_{y_0}
  \times I}$ as follows:
  \begin{enumerate}[(i)]
    \item $H|_{W_{y_0} \times \left[ 0,
      \frac{1}{N} \right]} \to U_\alpha \subset X$
      and have $\tilde{f}_0 |_{W_{y_0}} : W_{y_0}
      \to \tilde{X}$ with image in some $V_\beta$
      lying in $U_\alpha$.

      Define $\tilde{H}|_{W_{y_0} \times \left[ 0,
      \frac{1}{N} \right]} \stackrel{H|}{\to}
      U_\alpha
      \stackrel[\cong]{p|_{V_\beta}^{-1}}{\to}
      V_\beta \subset \tilde{X}$.
    \item Proceed in the same way,
      \glsref[lift]{lifting} $H|_{W_{y_0} \times
      \left[ \frac{1}{N}, \frac{2}{N} \right]}$
      starting at $\tilde{H}|_{W_{y_0} \times
      \left\{ \frac{1}{N} \right\}}$ etc.
  \end{enumerate}
  At the end of this process, we get a
  $\tilde{H}|_{W_{y_0} \times I}$
  \glsref[lift]{lifting} $H|_{W_{y_0} \times I}$
  and extending $\tilde{f}_0$ at time $0$. We can
  do this for each $y_0 \in Y$, so it is enough to
  check that on
  \[ (W_{y_0} \times I) \cap (W_{y_1} \times I) =
  (W_{y_0} \cap W_{y_1}) \times I ,\]
  the two \glspl{lift} constructed agree.

  For a $y_2 \in W_{y_0} \cap W_{y_1}$, the two
  choices give \glspl{lift} of $H|_{\{y_2\} \times
  I}$ which agree with $\tilde{f}_0(y_2)$ at time
  $0$. By the \nameref{uniqueness_of_lifts}, these
  \glspl{lift} must agree on the whole of $\{y_2\}
  \times i$. So they agree.}
\end{proof}
\end{flashcard}

\begin{flashcard}[path-lifting-coro]
\begin{corollary*}[Path lifting]
  \label{path_lifting}
  \cloze{If $p : \tilde{X} \to X$ is a \gls{covmap}, then
  $\gamma : I \to X$ is a \gls{path} from $x_0$ to
  $x_1$, and $\tilde{x}_0 \in \tilde{X}$ such that
  $p(\tilde{x}_0) = x_0$. Then there is a unique
  \gls{path} $\tilde{\gamma} : I \to \tilde{X}$
  such that $\tilde{\gamma}(0) = \tilde{x}_0$ and
  $p \circ \tilde{\gamma} = \gamma$.}
\end{corollary*}
\end{flashcard}

\begin{flashcard}[homotopic-lifted-paths-coro]
\begin{corollary*}
  Let $p : \tilde{X} \to X$ be a \gls{covmap},
  $\gamma, \gamma' : I \to X$ be \glspl{path} from
  $x_0$ to $x_1$, and $\tilde{\gamma},
  \tilde{\gamma}' : I \to \tilde{X}$ be their
  \glspl{lift} starting at $\tilde{x}_0 \in
  p^{-1}(x_0)$. If $\gamma \homotopic \gamma'$ as
  \glspl{path}, then $\tilde{\gamma} \homotopic
  \tilde{\gamma}'$ as \glspl{path}. In particular,
  $\tilde{\gamma}(1) = \tilde{\gamma}'(1)$.
\end{corollary*}

\begin{proof}
  \cloze{Let $H : I \times I \to X$ be a \gls{homotopy}
  from $\gamma$ to $\gamma'$ relative to
  endpoints.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/f488f96ce9634d61.png}
  \end{center}
  \nameref{homotopy_lifting_lemma} gives a
  $\tilde{H} : I \times I \to \tilde{X}$, which is
  a \gls{homotopy} of \glspl{path}.}
\end{proof}
\end{flashcard}

\begin{flashcard}[path-conn-covmap-coro]
\begin{corollary*}
  Let $p : \tilde{X} \to X$ be a \gls{covmap} and
  $X$ be \gls{pathconn}. Then \cloze{the sets
  $p^{-1}(x)$ are all in bijection with each
  other.}
\end{corollary*}

\begin{proof}
  \cloze{Let $\gamma : I \to X$ be a \gls{path} from
  $x_0$ to $x_1$. Define
  \begin{align*}
    \gamma_* : p^{-1}(x_0)
    &\to p^{-1}(x_0) \\
    y_0
    &\mapsto \tilde{\gamma}(1)
  \end{align*}
  for $\tilde{\gamma}$ the \gls{lift} of $\gamma$
  starting at $y_0$. Similarly $(\invpath
  \gamma)_*  : p^{-1}(x_1) \to p^{-1}(x_0)$. Then
  \begin{align*}
    (\invpath \gamma)_* \gamma_*(y_0)
    &= \text{end point of the \gls{lift} of $\invpath\gamma
    \concat \gamma$ which starts at $y_0$} \\
    &= \text{end point of $\constpath{y_0}$} \\
    &= y_0 \qedhere
  \end{align*}}
\end{proof}
\end{flashcard}