The Implicit Function Theorem

Contents

Statement
The contraction
Existence and uniqueness of the implicit map
Continuity, differentiability and the derivative formula
The two consequences the treatise uses
Reading the result

This appendix proves the one input that Theorem 17.59 of Differentiable Manifolds, Tensors, and Curvature takes from multivariable calculus and that Real Analysis does not state: where the derivative of a system of equations with respect to one block of variables is invertible, the system can be solved locally for that block as a differentiable function of the remaining variables, and the solution is as smooth as the system. It is the theorem that turns a level set into a graph, and through the regular value theorem it is what makes a sphere, a hyperboloid, a constraint surface or a mass shell a manifold at all.

What is assumed as already available is exactly the several-variable differential calculus of Section 11.10.2: partial derivatives and the gradient (Definition 11.97), the class $C^{1}$ (Definition 11.98), differentiability at a point with the uniqueness of the differential (Definitions 11.99 and 11.102), the theorem that a $C^{1}$ map is differentiable (Theorem 11.100), the chain rule (Proposition 11.104), the one-variable mean value theorem (Theorem 11.35), the Cauchy criterion (Theorem 11.8) and the geometric series (Proposition 11.46). No fixed-point theorem is imported: the contraction is carried out explicitly, in the manner of the Picard iteration of Theorem 13.8, with the geometric estimate written out. Nothing else is quoted — see Remark A28.10.

Throughout, $\abs{\cdot}$ is the Euclidean length (Equation (10.12)) and, for a real matrix $A$,

\begin{equation}\tag{A28.1} \norm{A} = \left(\sum_{i,j}\left(A_{ij}\right)^{2}\right)^{1/2}\ec \qquad \abs{Av}\le\norm{A}\abs{v}\ec \end{equation}

by the Cauchy–Schwarz inequality of Linear Algebra and Representation Theory applied row by row. Points of $\R^{m}\times\R^{n}$ are written $(x,y)$ with $x\in\R^{m}$, $y\in\R^{n}$; for a map $F$ into $\R^{n}$ we write

\begin{equation}\tag{A28.2} \left(D_{x}F\right)^{i}{}_{j} = \pdv{F^{i}}{x^{j}}\ec\qquad \left(D_{y}F\right)^{i}{}_{j} = \pdv{F^{i}}{y^{j}}\ec \end{equation}

the first an $n\times m$ matrix, the second an $n\times n$ matrix.

Statement

Theorem A28.1 (Implicit function theorem).

Let $A\subseteq\R^{m}\times\R^{n}$ be open, let $F:A\longrightarrow\R^{n}$ be of class $C^{k}$ with $k\ge1$, and let $(a,b)\in A$ satisfy

\begin{equation}\tag{A28.3} F(a,b) = 0\ec\qquad \det\left[D_{y}F(a,b)\right]\neq 0\ep \end{equation}

Then there are open sets $W\subseteq\R^{m}$ with $a\in W$ and $V\subseteq\R^{n}$ with $b\in V$, satisfying $W\times V\subseteq A$, and a map $h:W\longrightarrow V$ such that

\begin{equation}\tag{A28.4} \set{(x,y)\in W\times V\mid F(x,y)=0} = \set{\bigl(x,h(x)\bigr)\mid x\in W}\ec \qquad h(a)=b\ep \end{equation}

The map $h$ is unique: it is the only map $W\longrightarrow V$ whose graph is the zero set. It is of class $C^{k}$, and

\begin{equation}\tag{A28.5} Dh(x) = -\left[D_{y}F\bigl(x,h(x)\bigr)\right]^{-1} D_{x}F\bigl(x,h(x)\bigr)\ec\qquad x\in W\ep \end{equation}

If $F$ is smooth, so is $h$. Rests on Theorem 11.100, Proposition 11.104 and Definition 11.99.

The proof is in four movements: the equation is recast as a fixed-point problem whose iteration contracts (The contraction); the iteration is run and its limit shown to be the unique solution (Existence and uniqueness of the implicit map); the solution is shown Lipschitz and then differentiable, with Equation (A28.5) (Continuity, differentiability and the derivative formula); and the smoothness is bootstrapped to $C^{k}$ there as well. The two consequences the treatise uses draws the two corollaries the treatise uses.

The contraction

Lemma A28.2 (The Newton map contracts).

Under the hypotheses of Theorem A28.1, put

\begin{equation}\tag{A28.6} C = \left[D_{y}F(a,b)\right]^{-1}\ec\qquad T(x,y) = y - C\,F(x,y)\ep \end{equation}

Then there are $r>0$ and $\rho\in(0,r]$ such that, writing $\overline{B}_{r}$ for closed balls, $\overline{B}_{\rho}(a)\times\overline{B}_{r}(b)\subseteq A$ and:

for every $y\in\overline{B}_{r}(b)$ and $x\in\overline{B}_{\rho}(a)$, $F(x,y)=0$ if and only if $T(x,y)=y$;
$\abs{T(x,y)-T(x,y')}\le\tfrac{1}{2}\abs{y-y'}$ for all $y,y'\in\overline{B}_{r}(b)$ and $x\in\overline{B}_{\rho}(a)$;
$\abs{T(x,b)-b}\le r/4$ for every $x\in\overline{B}_{\rho}(a)$;
$T(x,\cdot)$ maps $\overline{B}_{r}(b)$ into itself; and
$\det\left[D_{y}F(x,y)\right]\ne0$ throughout $\overline{B}_{\rho}(a)\times\overline{B}_{r}(b)$.

Rests on Equation (A28.3), Theorem 11.35 and Definition 11.98.

Proof.

Derives Lemma A28.2. $C$ exists by Equation (A28.3). Assertion (1) is immediate from Equation (A28.6): $T(x,y)=y$ says $C\,F(x,y)=0$, and $C$ is invertible, so this says $F(x,y)=0$.

Differentiating Equation (A28.6) in $y$,

\begin{equation}\tag{A28.7} D_{y}T(x,y) = \identity - C\,D_{y}F(x,y)\ec \end{equation}

which vanishes at $(a,b)$ by the definition of $C$. The entries of $D_{y}F$ are continuous, because $F$ is $C^{1}$ (Definition 11.98), hence so are those of $D_{y}T$ and so is $\det\left[D_{y}F\right]$. Choose $r>0$ with $\overline{B}_{r}(a)\times\overline{B}_{r}(b)\subseteq A$ and, by that continuity together with $D_{y}T(a,b)=0$ and $\det[D_{y}F(a,b)]\ne0$, small enough that

\begin{equation}\tag{A28.8} \norm{D_{y}T(x,y)}\le\frac{1}{2\sqrt{n}}\ec\qquad \det\left[D_{y}F(x,y)\right]\ne0\ec \end{equation}

on the whole of $\overline{B}_{r}(a)\times\overline{B}_{r}(b)$. That is assertion (5).

For (2), fix $x$ and $y,y'$ in the convex set $\overline{B}_{r}(b)$. Applying Theorem 11.35 to $\lambda\longmapsto T^{i}\left(x,y'+\lambda(y-y')\right)$ on $[0,1]$ gives $\theta_{i}\in(0,1)$ with

\begin{equation*} T^{i}(x,y)-T^{i}(x,y') = \nabla_{y}T^{i}\bigl(x,\xi_{i}\bigr)\cdot(y-y')\ec \qquad \xi_{i}=y'+\theta_{i}(y-y')\in\overline{B}_{r}(b)\ec \end{equation*}

so that $\abs{T^{i}(x,y)-T^{i}(x,y')} \le\norm{D_{y}T(x,\xi_{i})}\,\abs{y-y'} \le\abs{y-y'}/(2\sqrt{n})$, the row length being at most the whole matrix norm Equation (A28.1). Squaring and summing the $n$ components gives (2).

For (3): $T(a,b)=b$ because $F(a,b)=0$, and $x\longmapsto T(x,b)$ is continuous, so there is $\rho\in(0,r]$ with $\abs{T(x,b)-b}\le r/4$ for $\abs{x-a}\le\rho$.

For (4), combine (2) and (3): for $y\in\overline{B}_{r}(b)$ and $\abs{x-a}\le\rho$,

\begin{equation}\tag{A28.9} \abs{T(x,y)-b} \le\abs{T(x,y)-T(x,b)}+\abs{T(x,b)-b} \le\frac{\abs{y-b}}{2}+\frac{r}{4} \le\frac{3r}{4}<r\ep \end{equation}

∎

Existence and uniqueness of the implicit map

Lemma A28.3 (The iteration converges).

With $r,\rho$ as in Lemma A28.2, fix $x\in\overline{B}_{\rho}(a)$ and define

\begin{equation}\tag{A28.10} y_{0} = b\ec\qquad y_{j+1} = T(x,y_{j})\ep \end{equation}

Then every $y_{j}$ lies in $\overline{B}_{r}(b)$, the sequence converges to a limit $h(x)$ with

\begin{equation}\tag{A28.11} \abs{h(x)-b}\le\frac{r}{2}\ec \end{equation}

and $h(x)$ is the unique point of $\overline{B}_{r}(b)$ with $F\bigl(x,h(x)\bigr)=0$. Rests on Lemma A28.2, Theorem 11.8 and Proposition 11.46.

Proof.

Derives Lemma A28.3. The iterates stay in $\overline{B}_{r}(b)$ by Equation (A28.9) and induction. By assertion (2) of Lemma A28.2,

\begin{equation}\tag{A28.12} \abs{y_{j+1}-y_{j}} = \abs{T(x,y_{j})-T(x,y_{j-1})} \le\frac{1}{2}\abs{y_{j}-y_{j-1}} \le\cdots\le 2^{-j}\abs{y_{1}-y_{0}} \le 2^{-j}\,\frac{r}{4}\ec \end{equation}

the last step by assertion (3). For $l>j$ the triangle inequality (Proposition 11.3) and the geometric series (Proposition 11.46) give

\begin{equation*} \abs{y_{l}-y_{j}} \le\sum_{s=j}^{l-1}\abs{y_{s+1}-y_{s}} \le\frac{r}{4}\sum_{s\ge j}2^{-s} = \frac{r}{2^{\,j+1}}\ec \end{equation*}

so each of the $n$ coordinate sequences is a Cauchy sequence of reals and converges (Theorem 11.8); call the limit $h(x)$, which lies in the closed set $\overline{B}_{r}(b)$. Letting $l\to\infty$ with $j=0$ in the display gives $\abs{h(x)-b}\le r/2$, which is Equation (A28.11). Since $T(x,\cdot)$ is continuous (indeed Lipschitz, by assertion (2)), passing to the limit in Equation (A28.10) gives $T(x,h(x))=h(x)$, i.e.\ $F(x,h(x))=0$ by assertion (1).

Uniqueness in $\overline{B}_{r}(b)$: if $y$ and $y'$ both satisfy $F(x,\cdot)=0$ there, both are fixed points of $T(x,\cdot)$, and assertion (2) gives $\abs{y-y'}\le\tfrac12\abs{y-y'}$, hence $y=y'$.

∎

Corollary A28.4 (The zero set is a graph).

Put $W=B_{\rho}(a)$ and $V=B_{r}(b)$, the open balls. Then $W\times V\subseteq A$, $h(W)\subseteq V$, $h(a)=b$, and Equation (A28.4) holds; moreover $h$ is the only map $W\longrightarrow V$ whose graph is the zero set of $F$ in $W\times V$. Rests on Lemma A28.3.

Proof.

Derives Corollary A28.4. $h(x)\in V$ by Equation (A28.11), since $r/2<r$; and $h(a)=b$ because $y\equiv b$ is then a fixed point, unique by Lemma A28.3. If $(x,y)\in W\times V$ has $F(x,y)=0$ then $y\in\overline{B}_{r}(b)$, so $y=h(x)$ by the uniqueness clause; conversely $F(x,h(x))=0$ for every $x\in W$. That is Equation (A28.4). Any other map $g:W\longrightarrow V$ with the same graph would satisfy $F(x,g(x))=0$ with $g(x)\in\overline{B}_{r}(b)$, hence $g=h$.

∎

Continuity, differentiability and the derivative formula

Lemma A28.5 ($h$ is Lipschitz).

There is $K\ge0$ with

\begin{equation}\tag{A28.13} \abs{h(x)-h(x')}\le K\,\abs{x-x'}\ec\qquad x,x'\in W\ep \end{equation}

In particular $h$ is continuous. Rests on Lemma A28.3 and Theorem 11.35.

Proof.

Derives Lemma A28.5. The entries of $D_{x}F$ are continuous, hence bounded on the compact set $\overline{B}_{\rho}(a)\times\overline{B}_{r}(b)$ (Theorem 11.24); let $\mu$ bound $\norm{D_{x}F}$ there. Exactly as in step (2) of Lemma A28.2, the mean value theorem along the segment from $x'$ to $x$ — which lies in the convex ball $\overline{B}_{\rho}(a)$ — gives

\begin{equation}\tag{A28.14} \abs{F(x,y)-F(x',y)}\le\sqrt{n}\,\mu\,\abs{x-x'}\ec \qquad y\in\overline{B}_{r}(b)\ep \end{equation}

Now, using that $h(x)$ and $h(x')$ are fixed points of $T(x,\cdot)$ and $T(x',\cdot)$,

\begin{align*} \abs{h(x)-h(x')} &= \abs{T\bigl(x,h(x)\bigr)-T\bigl(x',h(x')\bigr)}\\ &\le \abs{T\bigl(x,h(x)\bigr)-T\bigl(x,h(x')\bigr)} + \abs{T\bigl(x,h(x')\bigr)-T\bigl(x',h(x')\bigr)}\\ &\le \frac{1}{2}\abs{h(x)-h(x')} + \norm{C}\,\abs{F\bigl(x,h(x')\bigr)-F\bigl(x',h(x')\bigr)}\ec \end{align*}

the last term because $T(x,y)-T(x',y) = -C\left[F(x,y)-F(x',y)\right]$. Absorbing the first term on the left and inserting Equation (A28.14) gives Equation (A28.13) with $K = 2\sqrt{n}\,\mu\,\norm{C}$.

∎

Lemma A28.6 ($h$ is differentiable, with the stated derivative).

$h$ is differentiable at every $x\in W$ and its differential is Equation (A28.5); the right-hand side of Equation (A28.5) is continuous on $W$, so $h$ is $C^{1}$. Rests on Lemma A28.5, Theorem 11.100 and Definition 11.99.

Proof.

Derives Lemma A28.6. Fix $x\in W$, write $y=h(x)$, and let $s\in\R^{m}$ be small enough that $x+s\in W$. Put

\begin{equation*} \Delta = h(x+s)-h(x)\ec\qquad\text{so}\qquad \abs{\Delta}\le K\abs{s} \end{equation*}

by Equation (A28.13). Since $F$ is $C^{1}$ it is differentiable at $(x,y)$ (Theorem 11.100), and its differential is the pair of blocks Equation (A28.2) (Definition 11.102); so by Equation (11.84) there is a function $\vect{\varepsilon}$ with $\abs{\vect{\varepsilon}(s,\Delta)} /\left(\abs{s}+\abs{\Delta}\right)\longrightarrow0$ and

\begin{equation}\tag{A28.15} 0 = F(x+s,\,y+\Delta) - F(x,y) = D_{x}F(x,y)\,s + D_{y}F(x,y)\,\Delta + \vect{\varepsilon}(s,\Delta)\ec \end{equation}

both values of $F$ vanishing by Equation (A28.4). Because $\abs{\Delta}\le K\abs{s}$, the remainder obeys $\abs{\vect{\varepsilon}}\le(1+K)\abs{s}\cdot \abs{\vect{\varepsilon}}/(\abs{s}+\abs{\Delta})$, so it is $o\!\left(\abs{s}\right)$. The matrix $D_{y}F(x,y)$ is invertible by assertion (5) of Lemma A28.2, so solving Equation (A28.15) for $\Delta$,

\begin{equation}\tag{A28.16} \Delta = -\left[D_{y}F(x,y)\right]^{-1}D_{x}F(x,y)\,s - \left[D_{y}F(x,y)\right]^{-1}\vect{\varepsilon}\ep \end{equation}

The second term is $o\!\left(\abs{s}\right)$ as well, the matrix being a fixed one. Hence $h(x+s)-h(x) - \left(-\left[D_{y}F\right]^{-1}D_{x}F\right)s = o\!\left(\abs{s}\right)$, which by Definition 11.99 says that $h$ is differentiable at $x$ with the differential Equation (A28.5); the differential being unique, no other candidate exists.

Continuity: the entries of $\left[D_{y}F\right]^{-1}$ are, by Cramer's rule (Linear Algebra and Representation Theory), quotients of polynomials in the entries of $D_{y}F$ by the determinant, which does not vanish on $\overline{B}_{\rho}(a)\times\overline{B}_{r}(b)$; those entries are continuous, and so is $h$ (Lemma A28.5), so the composite $x\longmapsto -\left[D_{y}F(x,h(x))\right]^{-1} D_{x}F(x,h(x))$ is continuous. By Definition 11.98, $h$ is $C^{1}$.

∎

Proof of Theorem A28.1. Derives Theorem A28.1. Corollary A28.4 supplies $W$, $V$, $h$, Equation (A28.4) and the uniqueness; Lemma A28.6 supplies Equation (A28.5) and the class $C^{1}$. It remains to raise the smoothness. Write

\begin{equation}\tag{A28.17} Dh(x) = G\bigl(x,h(x)\bigr)\ec\qquad G(x,y) = -\left[D_{y}F(x,y)\right]^{-1}D_{x}F(x,y)\ep \end{equation}

If $F$ is $C^{k}$ then $D_{x}F$ and $D_{y}F$ are $C^{k-1}$, and so is $\left[D_{y}F\right]^{-1}$, its entries being quotients of polynomials in $C^{k-1}$ functions with non-vanishing denominator; hence $G$ is $C^{k-1}$. We prove by induction on $j$ that $h$ is $C^{j}$ for $1\le j\le k$. The case $j=1$ is Lemma A28.6. Suppose $h$ is $C^{j}$ with $j<k$. Then $x\longmapsto\bigl(x,h(x)\bigr)$ is $C^{j}$, and $G$ is $C^{k-1}$ with $k-1\ge j$; a composition of a $C^{j}$ map with a $C^{j}$ map is $C^{j}$, because by the chain rule Equation (11.91) each first partial derivative of the composite is a sum of products of first partial derivatives of the two factors, and an induction on $j$ carries this to order $j$. So $Dh$ is $C^{j}$ by Equation (A28.17), that is, $h$ is $C^{j+1}$. After $k-1$ steps $h$ is $C^{k}$. If $F$ is smooth the argument applies for every $k$, so $h$ is smooth.

∎

The two consequences the treatise uses

Corollary A28.7 (Inverse function theorem).

Let $U\subseteq\R^{n}$ be open, $f:U\longrightarrow\R^{n}$ of class $C^{k}$ with $k\ge1$, and $\alpha\in U$ with $\det\left[Df(\alpha)\right]\ne0$. Then there are open sets $U_{0}\subseteq U$ with $\alpha\in U_{0}$ and $V_{0}\subseteq\R^{n}$ with $f(\alpha)\in V_{0}$ such that $f$ maps $U_{0}$ bijectively onto $V_{0}$, the inverse $g:V_{0}\longrightarrow U_{0}$ is $C^{k}$, and

\begin{equation}\tag{A28.18} Dg(v) = \left[Df\bigl(g(v)\bigr)\right]^{-1}\ep \end{equation}

Rests on Theorem A28.1 and Proposition 11.104.

Proof.

Derives Corollary A28.7. Apply Theorem A28.1 to

\begin{equation*} \Phi:\R^{n}\times U\longrightarrow\R^{n}\ec\qquad \Phi(v,u) = v - f(u)\ec \end{equation*}

at the point $\left(f(\alpha),\alpha\right)$, where $\Phi=0$ and $D_{u}\Phi = -Df(\alpha)$ is invertible. It supplies open sets $V_{0}\ni f(\alpha)$ and $V\ni\alpha$ and a $C^{k}$ map $g:V_{0}\longrightarrow V$ whose graph is the zero set of $\Phi$ in $V_{0}\times V$; that is,

\begin{equation}\tag{A28.19} \text{for } (v,u)\in V_{0}\times V:\qquad f(u)=v \iff u=g(v)\ep \end{equation}

Put $U_{0} = V\cap f^{-1}(V_{0})$, open because $f$ is continuous, and note $\alpha\in U_{0}$. If $u\in U_{0}$ then $(f(u),u)\in V_{0}\times V$, so $u=g(f(u))$ by Equation (A28.19): $f$ is injective on $U_{0}$. If $v\in V_{0}$ then $g(v)\in V$ and $f(g(v))=v\in V_{0}$, so $g(v)\in U_{0}$ and $v\in f(U_{0})$: $f$ maps $U_{0}$ onto $V_{0}$, with $g$ as its two-sided inverse. Finally, differentiating $g\circ f=\id$ on $U_{0}$ by the chain rule Equation (11.91) gives $Dg\bigl(f(u)\bigr)Df(u)=\identity$, which is Equation (A28.18) at $v=f(u)$.

∎

Corollary A28.8 (Solving one scalar equation for one coordinate).

Let $A\subseteq\R^{n}$ be open, $F:A\longrightarrow\R$ of class $C^{k}$ with $k\ge1$, and let $P\in A$ satisfy $F(P)=c$ and $\pp F/\pp x^{n}(P)\neq0$. Write $P=\left(P',P^{n}\right)$ with $P'\in\R^{n-1}$. Then there are an open set $W\subseteq\R^{n-1}$ containing $P'$ and an open interval $I$ containing $P^{n}$, with $W\times I\subseteq A$, and a $C^{k}$ function $\varsigma:W\longrightarrow I$ with $\varsigma(P')=P^{n}$ such that, inside $W\times I$,

\begin{equation}\tag{A28.20} F\left(x^{1},\ldots,x^{n}\right)=c \quad\iff\quad x^{n} = \varsigma\left(x^{1},\ldots,x^{n-1}\right)\ec \end{equation}

and

\begin{equation}\tag{A28.21} \pdv{\varsigma}{x^{i}}\left(x'\right) = -\left.\frac{\pp F/\pp x^{i}}{\pp F/\pp x^{n}} \right|_{\left(x',\varsigma(x')\right)}\ec \qquad i=1,\ldots,n-1\ep \end{equation}

Rests on Theorem A28.1.

Proof.

Derives Corollary A28.8. Apply Theorem A28.1 with $m=n-1$, target dimension $1$, and $G(x',x^{n}) = F(x',x^{n})-c$ at the point $\left(P',P^{n}\right)$: $G=0$ there, and $D_{y}G$ is the $1\times1$ matrix $\pp F/\pp x^{n}(P)\ne0$, whose determinant is itself. The theorem returns $W$, an open ball $V\subseteq\R$ — that is, an open interval $I$ — and $\varsigma$ of class $C^{k}$ with Equation (A28.4), which is Equation (A28.20). Equation (A28.5) reads, for $1\times1$ blocks, $D\varsigma = -\left(\pp F/\pp x^{n}\right)^{-1} \left(\pp F/\pp x^{1},\ldots,\pp F/\pp x^{n-1}\right)$, which is Equation (A28.21).

∎

Example A28.9 (The sphere, made explicit).

Take $n=3$ and $F(x,y,z)=x^{2}+y^{2}+z^{2}$, $c=R^{2}>0$, the level set of Example 17.61. At a point of the sphere with $z\neq0$ one has $\pp F/\pp z = 2z\ne0$, so Corollary A28.8 applies with the whole of $\R^{3}$ as $A$; the function it produces is the one the chapter writes down by hand, $\varsigma(x,y)=+\sqrt{R^{2}-x^{2}-y^{2}}$ on the northern hemisphere, and Equation (A28.21) returns $\pp\varsigma/\pp x = -x/z$ and $\pp\varsigma/\pp y = -y/z$, which is what differentiating the explicit square root gives. The point of the corollary is that the same conclusion holds when no explicit solution can be written: at a point of the sphere with $z=0$ one of the other two partial derivatives is non-zero, and the corollary — applied after relabelling the coordinates — produces a graph there too. If all three partial derivatives could vanish on the level set, no such chart would be guaranteed; that is exactly the regularity hypothesis of Theorem 17.59, and on the sphere it fails only at the origin, which is not on the level set. Rests on Corollary A28.8 and Example 17.61.

Reading the result

Remark A28.10 (Nothing is quoted here).

This section imports no theorem that the treatise does not prove. In particular it does not use a fixed-point theorem: the Banach contraction principle is not in this book, and rather than quote it the iteration Equation (A28.10) is run explicitly and its convergence read off from the geometric estimate Equation (A28.12) and the Cauchy criterion Theorem 11.8 — the same device, and the same estimate, by which Theorem 13.8 constructs the solution of an initial value problem. The only other analytic inputs are the mean value theorem Theorem 11.35, used twice to convert a bound on a derivative into a Lipschitz bound; the theorem that a $C^{1}$ map is differentiable (Theorem 11.100), used once to expand $F$ to first order in Equation (A28.15); and the chain rule Proposition 11.104. From linear algebra, only the invertibility criterion $\det\neq0$ and Cramer's rule are used.

Remark A28.11 (Both hypotheses are needed, and the conclusion is local).

The invertibility of $D_{y}F(a,b)$ cannot be dropped. On $\R\times\R$, $F(x,y)=y^{2}-x$ has $F(0,0)=0$ and $\pp F/\pp y(0,0)=0$; the zero set near the origin is the parabola $x=y^{2}$, which is not the graph of any function of $x$ — two solutions for $x>0$, none for $x<0$. The conclusion is also irreducibly local: $F(x,y)=x^{2}+y^{2}-1$ on $\R\times\R$ satisfies the hypotheses at $(0,1)$, and the implicit function $\varsigma(x)=\sqrt{1-x^{2}}$ exists only for $\abs{x}<1$ and cannot be continued past $x=\pm1$, where $\pp F/\pp y$ vanishes. Finally, the smoothness of $h$ can be no better than that of $F$: taking $m=n=1$ and $F(x,y)=y-\phi(x)$ with $\phi$ of class $C^{k}$ but not $C^{k+1}$ makes $h=\phi$. Rests on Theorem A28.1.

Remark A28.12.

The Implicit Function Theorem discharges the derivation owed at Remark 17.60 of Differentiable Manifolds, Tensors, and Curvature. The statement used there is Corollary A28.8: in the proof of the regular value theorem Theorem 17.59 a chart is chosen in which $\pp F/\pp x^{n}\ne0$ at the point, and Equation (A28.20) is exactly the assertion Equation (17.146) that the level set meets the coordinate box in the graph of a smooth $h$ — from which the chapter reads off the chart $u\longmapsto(u,h(u))$ of the hypersurface. The regular value theorem in turn underlies Example 17.61 and Definition 17.58 and every later construction of a submanifold by one equation. The inverse function theorem Corollary A28.7, proved here as a companion, is what licenses a change of chart whenever a Jacobian determinant is non-zero, and is used in that role in The General Stokes Theorem for Differential Forms.