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\begin{theorem*}[Inverse function theorem]
    Let $U \subset \RR^n$ be open and let $f : U
    \to \RR^n$ be a continuously differentiable
    map. Let $p \in U$, $f(p) = q$ and suppose
    $Df|_p$ is invertible. Then there is an open
    neighbourhood $V$ of $q$ and a differentiable
    map $g : V \to \RR^n$, $g(q) = p$ with image
    an open neighbourhood $U' \subset U$ of $p$
    such that $f \circ g = \id_V$ and $g \circ f =
    \id_U$. If $f$ is smooth, then so is $g$.
\end{theorem*}

\begin{hiddenflashcard}[inverse-fn-thm-geom]
\begin{theorem*}[Inverse function theorem for geometry]
    Let $U \subset \RR^n$ be open and let $f : U
    \to \RR^n$ be \cloze{a \fcemph{continuously
    differentiable} map. Let $p \in U$, $f(p) = q$
    and suppose $Df|_p$ \fcemph{invertible}. Then
    there is an open neighbourhood $V$ of $q$ and
    a differentiable map $g : V \to \RR^n$ with
    image an open neighbourhood $U' \subset U$ of
    $p$ such that $f \circ g = \id_V$ and $g
    \circ f = \id_U$. \fcemph{If $f$ is smooth
    then so is $G$.}
    }
\end{theorem*}
\end{hiddenflashcard}

\begin{remark*}
    $Df|_q = (Df|_p)^{-1}$ by the chain rule.
\end{remark*}

\noindent
If we have a map $f : \RR^n \to \RR^m$ where $n >
m$, what can we say if $Df |_p$ is surjective?
\[ Df|_p = \left( \pfrac{f_i}{x_j} \right)_{m
\times n} \]
having full rank means that permuting coordinates
if necessary, we can assume that the first (or
last) $m$ columns are linearly independent.

\begin{theorem*}[Implicit function theorem]
    Let $p = (x_0, y_0) \in U$, $U \subset \RR^k
    \times \RR^l$ open and let $f : U \to \RR^l$,
    $p \mapsto 0$ be a continuously differentiable
    map with $\left( \pfrac{f_i}{y_j} \right)_{l
    \times l}$ an isomorphism at $p$. Then there
    exists an open neighbourhood $x_0 \in V
    \subset \RR^k$ and a continuously
    differentiable map $g : V \to \RR^l$ such that
    if $(x, y) \in U \cap (V \times \RR^l)$, then
    \[ f(x, y) = 0 \iff y = g(x) \]
    If $f$ is smooth so is $g$.
\end{theorem*}

\begin{proof}
    Introduce $F : U \to \RR^k \times \RR^l$, $(x,
    y) \mapsto (x, f(x, y))$. Then
    \[ DF =
    \begin{pmatrix}
        I & * \\
        0 * \pfrac{f_i}{y_j}
    \end{pmatrix}
    \]
    So $DF|_{(x_0, y_0)}$ is an isomorphism. So
    the inverse function theorem says that $F$ is
    locally invertible near
    \[ F(x_0, y_0) = (x_0, f(x_0, y_0)) = (x_0, 0) \]
    Take a product open neighbourhood $(x_0, 0)
    \in V \times V'$, $V \subset \RR^t$ open, $V'
    \subset \RR^l$ open and the continuously
    differentiable inverse $G : V \times V' \to U'
    \subset U \subset \RR^k \times \RR^l$ such
    that $F \circ G = \id_{V \times V'}$. Write
    \[ G(x, y) = (\varphi(v, y), \psi(x, y)) \]
    Then
    \begin{align*}
        F \circ G(x, y)
        &= (\varphi(x, y), f(\varphi(x, y),
        \psi(x, y)) \\
        &= (x, y) \\
        \implies \varphi(x, y)
        &= x \\
        \text{and } f(x, \psi(x, y))
        &= y
    \end{align*}
    when $(x, y) \in V \times V'$. Thus
    \[ f(x, y) = 0 \iff y = \psi(x 0) \]
    Define $g : V \to \RR^l$ as $x \to \psi(x,
    0)$, $v_0 \mapsto y_0$ and does what we want.
\end{proof}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $f : \RR^2 \to \RR$ smooth, $f(x_0, y_0)
        = 0$. Assume $\pfrac{f}{y} |_{(x_0, y_0)}
        \neq 0$. Then there exists smooth $g :
        (x_0 - \eps, x_0 + \eps) \to \RR$ such
        that
        \[ g(x_0) = g(y_0) \]
        and
        \[ f(x, y) = 0 \iff y = g(x) \]
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/f28e2834a23b11ed.png}
        \end{center}
        Since $f(x, g(x)) = 0$, chain rule implies
        \begin{align*}
            \pfrac{f}{x} + \pfrac{f}{y}g'(x)
            &= 0 \\
            \implies g'(x)
            &= -\frac{\pfrac{f}{x}}{\pfrac{f}{y}}
        \end{align*}
    \item Let $f : \RR^3 \to \RR$ smooth and
        $f(x_0, y_0, z_0) = 0$, and assume
        $Df|_{(x_0, y_0, z_0)} \neq 0$. Permuting
        coordinates if necessary we may assume
        that $\left. \pfrac{f}{z} \right|_{(x_0,
        y_0, z_0)} \neq 0$. Then there exists an
        open neighbourhood $(x_0, y_0) \in V
        \subset \RR^2$ and a smooth $g : V \to
        \RR$, $g(x_0, y_0) = z_0$ such that for an
        open set $U \ni (x_0, y_0, z_0)$
        \[ f^{-1}(0) \cap U =
        \operatorname{graph}(g) \]
        ($\{(x, y, g(x, y)) : (x, y) \in V\}$)
\end{enumerate}

\noindent
We return to Theorem 1.7. Recall:

\begin{theorem*}
    For a subset $\Sigma \subset \RR^3$, the
    following are equivalent (TFAE):
    \begin{enumerate}[(a)]
        \item $\Sigma$ is a smooth surface in
            $\RR^3$
        \item $\Sigma$ is locally a graph over a
            coordinate plane.
        \item $\Sigma$ is locally $f^{-1}(0)$, $f$
            smooth and $Df|_p \neq 0$
        \item $\Sigma$ is locally the image of an
            allowable parametrisation $\sigma : V
            \to \Sigma \subset \RR^3$, $\sigma$
            smooth, $\sigma$ homeomorphism onto
            $\sigma(V)$ and $D\sigma$ injective.
    \end{enumerate}
\end{theorem*}

\begin{proof}
    \begin{enumerate}[(1)]
        \item (b) implies all others.

            \myskip
            If $\Sigma$ is locally $\{(x, y, g(x,
            y)) : (x, y) \in V\}$ then we get a
            chart from projection $\Pi_{xy}$ which
            is smooth and defined on an open
            neighbourhood of $\Sigma$, hence (b)
            implies (c).

            \myskip
            Also, it is cut out by
            \[ f(x, y, z) = z - g(x, y) \]
            Clearly $\pfrac{f}{z} = 1 \neq 0$, (b)
            implies (c). Also $\sigma(x, y) = (x,
            y, f(x, y))$ is allowable and smooth:
            \begin{align*}
                \sigma_x
                &= (1, 0, g_x) \\
                \sigma_y
                &= (0, 1, g_y)
            \end{align*}
            are linearly independent, (b) implies
            (d).
        \item (a) implies (d) is part of the
            definition of being a smooth surface
            in $\RR^3$ and hence locally
            diffeomorphic to $\RR^2$ (the inverse
            of the local diffeomorphism is the
            allowable parametrisation).
        \item (c) implies (b) was example number 2
            above for the implicit function
            theorem.
        \item We'll show that (d) implies (b) and
            we're done. Let $p \in \Sigma$,
            $\sigma : V \to U \subset \Sigma$,
            $\sigma(0) = p \in U$.
            \begin{align*}
                \sigma
                &= (\sigma_1(u, v), \sigma_2(u,
                v), \sigma_3(u, v)) \\
                D\sigma
                &=
                \begin{pmatrix}
                    \pfrac{\sigma_1}{u} &
                    \pfrac{\sigma_1}{v} \\
                    \pfrac{\sigma_2}{u} &
                    \pfrac{\sigma_2}{v} \\
                    \pfrac{\sigma_3}{u} &
                    \pfrac{\sigma_3}{v}
                \end{pmatrix}
            \end{align*}
            So there exists 2 rows defining an
            invertible matrix. Suppose first two
            rows
            \[ \Pi_{xy} \circ \sigma : V \to
            \RR^2, D(\Pi_{xy} \circ \sigma)|_0 \]
            \emph{isomorphism}. Then inverse
            function theorem implies locally
            invertible, let $\phi = \Pi_{xy} \circ
            \sigma$. $\Sigma$ is now the graph of
            \[ (x, y, \sigma(\phi^{-1}(x, y)) =
            (\sigma_1(u, v), \sigma_2(u, v),
            \sigma_3(u, v)) \in \Sigma \qedhere \]
    \end{enumerate}
\end{proof}