An equation $F(x,y)=0$ describes a set. The theorem says when that set is locally the graph of a function — when you can solve for $y$ in terms of $x$ without ever writing the solution down. It is the inverse function theorem wearing different clothes, and it is the reason a level set deserves to be called a surface.
Most interesting sets are given by equations, not parametrisations: a sphere is $\|x\|^2=1$, an energy level is $H=E$. To do calculus on such a set you need to know it is a graph locally, and this theorem is the only general tool that says so.
Comparative statics in economics is this theorem: differentiate the equilibrium condition implicitly to see how the equilibrium moves when a parameter does. Constraint solvers, kinematic chains in robotics, and continuation methods for nonlinear systems all sit on it.
Dini gave the standard modern proof in the 1870s, which is why the theorem carries his name in Italian texts. The regularity question — how little smoothness suffices — is still active, with Lipschitz and set-valued versions developed through the twentieth century for optimisation and control.
Write points of $\mathbb{R}^{n+m}$ as $(x,y)$ with $x\in\mathbb{R}^n$ and $y\in\mathbb{R}^m$. For $F:\mathbb{R}^{n+m}\to\mathbb{R}^m$, split the Jacobian into blocks $JF=[\,\partial F/\partial x\ \ \partial F/\partial y\,]$ of sizes $m\times n$ and $m\times m$. The square block is the one that has to be invertible.
Let $F$ be $C^1$ near $(a,b)$ with $F(a,b)=0$, and suppose $\partial F/\partial y(a,b)$ is invertible. Then there are open sets $U$ containing $a$ in $\mathbb{R}^n$ and $W$ containing $(a,b)$, and a $C^1$ map $g:U\to\mathbb{R}^m$ with $g(a)=b$, such that
Moreover, differentiating $F(x,g(x))=0$ gives
with both blocks evaluated at $(x,g(x))$. The count is worth noticing: $m$ equations remove $m$ of the $n+m$ variables, leaving $n$ free.
The two theorems are the same theorem. Which one is "the" result is a matter of taste; several texts prove the implicit version first and deduce the inverse one.
$F(x,y)=x^2+y^2-1$ has $\partial F/\partial y=2y$, invertible exactly when $y\ne0$. So near every point of the circle except $(\pm1,0)$ the circle is a graph $y=g(x)$, and indeed $g(x)=\pm\sqrt{1-x^2}$ with the sign fixed by which point you started at.
At $(\pm1,0)$ the circle is vertical and no function of $x$ can describe it. But $\partial F/\partial x=2x=\pm2\ne0$ there, so it is a graph $x=g(y)$ instead. The circle is a curve everywhere; only the choice of axis changes.
Let $F(x,y)=y^5+xy-1$. Then $\partial F/\partial y=5y^4+x$, which at $(0,1)$ is $5\ne0$, so near $x=0$ there is a $C^1$ function $g$ with $g(0)=1$. Differentiating,
The quintic has no solution formula. The derivative does.
If $F:\mathbb{R}^{n+m}\to\mathbb{R}^m$ is $C^1$ and $JF(p)$ has rank $m$ at every $p$ in the level set $F^{-1}(0)$, then that level set is locally a graph over $n$ of the coordinates, and its tangent space at $p$ is $\ker JF(p)$, of dimension $n$.
Full rank means some $m\times m$ block is invertible; relabel the coordinates so it is the last one and apply the theorem. This statement is the definition of a manifold that Unit 13 will adopt.
$F(x,y)=y^2-x^3$ has $\nabla F=(-3x^2,2y)$, vanishing at the origin, which is on the curve. And the curve $y^2=x^3$ has a cusp there — it is not locally a graph in either variable in a $C^1$ way. Rank failure is not a technicality; it is where the geometry actually breaks.
• Checking $\nabla F\ne0$ when $m>1$. The condition is that the
$m\times m$ block is invertible, that is $JF$ has full rank $m$.
• Expecting a global solution. The graph exists on some neighbourhood;
the circle is the standing reminder that it cannot be all of the level
set.
• Solving for the wrong variables and concluding the level set is
singular. Try the other block first.
• Forgetting to evaluate the derivative formula at $(x,g(x))$ rather than
at $(x,b)$.
• Assuming rank failure implies a singular point. $F(x,y)=y^2$ has
$\nabla F=0$ along its zero set, which is a perfectly smooth line. The
condition is sufficient, not necessary.
Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.
For $F(x,y)=x^2+y^3-xy-3$, where can you solve for $y$ near a solution point?
With $x^2+y^2+z^2=3$, find $\partial z/\partial x$ at $(1,1,1)$.
Given $x+y+z=0$ and $x^2+y^2+z^2=2$, can you solve for $(y,z)$ as functions of $x$ near $(0,1,-1)$?
Why does the theorem fail for $y^2=x^3$ at the origin, and what does the set look like?
Find the tangent plane to $x^2+2y^2+3z^2=6$ at $(1,1,1)$.
For $x^2+y^2=1$ with $y>0$, compute $g''$ by implicit differentiation.
Suppose equilibrium is defined by $F(q,p)=0$ with $\partial F/\partial q\ne0$. How does $q$ respond to $p$?
Show $F(x,y)=y^2$ has a smooth zero set even though $\nabla F$ vanishes on it.
Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:
Spivak deduces the implicit theorem from the inverse one exactly as here; Munkres treats both at length with the block algebra written out. Check the assigned text and exercises against the current syllabus.
Inverse function theorem gives implicit function theorem gives the local graph description of regular level sets, which is what makes a level set a manifold in Unit 13 and what makes Lagrange multipliers legitimate in Unit 8.
Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.
Solving $F(x,y)=0$ for $y$, which block must be invertible?
$m$ equations in $n+m$ unknowns leaves what dimension?
What is the tangent space to a regular level set?
Give a level set that is a graph in neither variable near a point.