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

\begin{proof}
  \cloze{Let $p(z) = z^n + a_1 z^{n - 1} + \cdots + a_n$
  a monic polynomial over $\CC$. Choose
  \[ r > \max(|a_1| + |a_2| + \cdots + |a_n|, 1) \]
  On the circle $|z| = r$, we have
  \begin{align*}
    |z^n|
    &= |z^{n - 1}| r \\
    &> |z^{n - 1}| (|a_1| + \cdots + |a_n|) \\
    &> |a_1 z^{n - 1} + \cdots + a_n|
  \end{align*}
  Therefore for any $t \in [0, 1]$, the polynomial
  \[ p_t(z) = z^n + t(a_1 z^{n - 1} + \cdots +
  a_n) \]
  does not have a root on the circle $|z| = r$.
  Consider the \gls{homotopy} of loops in $S^1
  \subset \CC$:
  \[ F(s, t) = \frac{p_t(r \cdot e^{2\pi i s}) /
  p_t(r)}{|p_t(r \cdot e^{2\pi i s}) / p_t(r)|} .\]
  The divisions is allowed since we know that at
  all times $t$, the polynomial does not have a
  root on the circle $|z| = r$. At $t = 0$, it is
  the loop $s \mapsto e^{2\pi i sn}$. This is $n
  \in \ZZ \cong \pio\bs(S^1, 1)$. At $t = 1$, it
  is the loop
  \[ s \stackrel{f_r}{\to} \frac{p(r \cdot e^{2\pi
  i s}) / p(r)}{|p(r \cdot e^{2\pi i s}) / p(r)|}
  .\]
  This also represents $n \in \ZZ \cong
  \pio\bs(S^1, 1)$. Suppose $p$ has no roots. Then
  $f_n$ is a continuous loop for all $r \in [0,
  \infty)$ and varying $r$ gives a \gls{homotopy}
  from $f_r$ to $f_0$. Note that the loop $f_0$
  is:
  \[ s \mapsto f_0(s) = \frac{p(r) / p(r)}{|p(r) /
  p(r)|} = 1 \]
  This corresponds to the constant loop, i.e. $0
  \in \ZZ \cong \pio\bs(S^1, 1)$.
  Hence we must have $n = 0 \in \ZZ$, so $p$ was
  constant.}
\end{proof}
\end{flashcard}

\subsection{Construction of universal covers}

\textbf{Observation 1:} Let $p : \tilde{X} \to X$
a \gls{unicov}, $x \in X$. Let $U \ni x$ a
neighbourhood which is \gls{evencov} (i.e.
$p^{-1}(U) = \coprod V_\alpha$). Let $\gamma : I
\to U$ be a loop in $U$ based at $x_0$. This
\glspl{lift} to a
\[ \begin{tikzcd}
  \tilde{\gamma} : I \ar[rr] \ar[rd, "\gamma",
  swap]
  && V_\alpha \subset \tilde{X} \\
  & U \ar[ur, "p|_{V_\alpha}^{-1}", swap]
\end{tikzcd} \]
This is \gls{homotopic} to a constant loop in
$\tilde{X}$, as $\tilde{X}$ is \gls{simpconn}.
Applying $p$ gives $\gamma \homotopic
\constpath{x_0}$ in $X$, i.e. ``every $x \in X$ has
a neighbourhood $U \ni x$ such that the \gls{map} $\pio\bs(U,
x) \stackrel{\ps{(\inc)}}{\to} \pio\bs(X, x)$ is
trivial''. We call this property ``semi-locally
simply connected''.

\begin{flashcard}[semi-locally-simply-connected-defn]
\begin{definition*}[Semi-locally simply-connected]
  \glsadjdefn{slsc}{semi-locally
  simply-connected}{space}
  \cloze{We say a space $X$ is \emph{semi-locally
  simply-connected} if every $x \in X$ has a
  neighbourhood $U \ni x$ such that
  \[ \pio\bs(U, x) \stackrel{\ps{(\inc)}}{\to}
  \pio\bs(X, x) \]
  is trivial.}
\end{definition*}
\end{flashcard}

\begin{example*}
  The ``Hawaiian earring'' is not \gls{slsc}:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/f4504ddd9cf3469a.png}
  \end{center}
  The circles all share a common tangency point,
  and we have a circle of radius $\frac{1}{n}$ for
  each $n$.
\end{example*}

\textbf{Observation 2:} Suppose $p : \tilde{X} \to
X$ is a \gls{unicov} and $x_0 = p(\tilde{x}_0)$ is
a base point. Any $y \in \tilde{X}$ has a
\emph{unique} \gls{path} $\alpha$ from
$\tilde{x}_0$ to $y$ (up to \gls{homotopy}). Then
\[ y = \text{the end point of the lift of $p \circ
\alpha$ starting at $\tilde{x}_0$.} ,\]
i.e. there is a \emph{bijection}
\begin{align*}
  \tilde{X}
  &\rightarrow \left\{
  \substack{
    \text{\gls{homotopy} classes of} \\
    \text{\glspl{path} in $X$ starting} \\
    \text{at $x_0$}
  } \right\} \\
  y
  &\mapsto p \circ \alpha \\
  \tilde{\gamma}(1)
  &\mapsfrom [\gamma]
\end{align*}

\begin{flashcard}[slsc-implies-unicov-existence-thm]
\begin{theorem*}[Existence of universal covers]
  \label{unicov_existence}
  \cloze{Let $X$ be \gls{pathconn}, \gls{locpathconn},
  and \gls{slsc}. Then it has a \gls{unicov}.}
\end{theorem*}
\end{flashcard}

\begin{proof}[Proof (non-examinable)]
  As a set let
  \[ \tilde{X} \defeq \left\{
  \substack{
    \text{\gls{homotopy} classes of} \\
    \text{\glspl{path} in $X$ starting} \\
    \text{at $x_0$}
  } \right\}
  ,\]
  and
  \begin{align*}
    p : \tilde{X}
    &\to X \\
    [\gamma]
    &\mapsto \gamma(1)
  \end{align*}
  Want:
  \begin{enumerate}[(i)]
    \item Make a topology on $\tilde{X}$.
    \item Show $p$ is continuous.
    \item Show $p$ is a \gls{covmap}.
    \item Show $\tilde{X}$ is \gls{simpconn}.
  \end{enumerate}
  Consider
  \[ \mathcal{U} = \{U \subset X \st \text{$U$
  open, \gls{pathconn} and $\pio\bs(U, x) \to
  \pio\bs(X, x)$ is trivial $\forall
  x \in U$}\} \]
  \textbf{Claim:} This is a basis for the topology
  on $X$.

  Proof: Let $V \ni x$ be an open neighbourhood.
  Then:
  \begin{enumerate}[(i)]
    \item $X$ \gls{slsc} $\implies \exists U' \ni
      x$ such that $\pio\bs(U', x) \to \pio(X, x)$
      is trivial.
    \item As $X$ is \gls{locpathconn}, can fine $V
      \cap U' \supset U \ni x$ which is
      \gls{pathconn}.
    \item The map
      \[ \begin{tikzcd}[ampersand replacement=\&]
        \pio\bs(U, x) \ar[rr]
        \&\& \pio\bs(X, x) \\
        \& \pio\bs(U', x) \ar[ur, "\mathrm{triv}"]
      \end{tikzcd} \]
      is trivial.
    \item Let $y \in U$ be another point, and $u
      : I \to U$ be a \gls{path} from $x$ to $y$.
      Then the following diagram commutes
      \[ \begin{tikzcd}[ampersand replacement=\&]
        \pio\bs(U, y) \ar[r]
        \& \pio\bs(X, y) \\
        \pio\bs(U, x) \ar[u, "\simeq" description,
        "\pathisom u"] \ar[r, "\mathrm{trivial}"]
        \& \pio\bs(X, x) \ar[u, "\simeq"
        description, "\pathisom u"]
      \end{tikzcd} \]
      which shows the top \gls{map} is trivial.
  \end{enumerate}
  This completes the proof of the claim.

  For $[\alpha] \in \tilde{X}$ and a $U \in
  \mathcal{U}$ such that $\alpha(1) \in U$, define
  \[ ([\alpha], U) = \{[\beta] \in \tilde{X} \st
  [\beta] = [\alpha \concat \alpha'] \text{ for some
  \gls{path} $\alpha'$ in $U$}\} .\]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/8db9732ec2e54f9e.png}
  \end{center}
  \textbf{Claim:} These sets form a basis for a
  topology on $\tilde{X}$.

  Proof: Let $[\beta] \in ([\alpha_0], U_0) \cap
  (\alpha_1, U_1)$. As in the figure, there are
  $\alpha_0', \alpha_1'$ with $[\alpha \concat
  \alpha_0'] = [\beta] = [\alpha_1 \concat
  \alpha_1']$. Let $\beta(1) \in W \subset U_0
  \cap U_1$ with $W \in \mathcal{U}$. Want to show
  \[ ([\beta], W) \subset ([\alpha_0], U_0) \cap
  ([\alpha_1], U_1) \]
  If $[\gamma] \in ([\beta], W)$, then there is a
  \gls{path} $\delta$ in $W$ with
  \[ [\gamma] = [\beta \concat \delta] = [\alpha_0
  \concat \ub{\alpha_0' \concat \delta}_{\in U_0}] \in
  ([\alpha_0], U_0) .\]
  Similarly
  \[ [\gamma] = [\beta \concat \delta] = [\alpha_1
  \concat \ub{\alpha_1' \concat \delta}_{\in U_1}]
  \in ([\alpha_1], U_1) .\]
  This finishes the proof of the claim.