8. Diffeomorphisms and theorems of inverse and implicit function

Tangent lines and planes

Notice that an equation f(x,y)=a describes a curve on the plane. Imagining the graph of function f we see that it is its level line. The direction of this level line in the given point (x_0,y_0) such that f(x_0,y_0)=a is the direction in which the directional derivative is zero (thus the level is constant). Since the directional derivative at (x_0,y_0) in direction (x,y) equals \frac{\partial f}{\partial x}(x_0,y_0)x+\frac{\partial f}{\partial y}(x_0,y_0)y, it means that the equation


where A=\frac{\partial f}{\partial x}(x_0,y_0) and B=\frac{\partial f}{\partial y}(x_0,y_0), describes the line intersecting (0,0) and parallel to the tangent line to the considered curve at (x_0,y_0). To get the equation of the actual tangent line we obviously have to change it:


where D=Ax_0+By_0.

In the three-dimensional case an equation f(x,y,z)=a describes a surface, and analogously the equation of the plane tangent to it at point (x_0,y_0,z_0) such that f(x_0,y_0,z_0)=a, is


where A=\frac{\partial f}{\partial x}(x_0,y_0,z_0), B=\frac{\partial f}{\partial y}(x_0,y_0, z_0), C=\frac{\partial f}{\partial z}(x_0,y_0, z_0) and D=Ax_0+By_0+Cz_0.


A diffeomorphism f\colon U\to V between open sets U and V is a function such that V is its image (it is ,,onto” V), and is one-to-one and C^1 (differentiable, and with continuous first order partial derivatives), and its inverse function is of C^1 class as well.

Intuitively, you can think of a diffeomorphism as of a transformation of an open such which may stretch it but not reap it and squeeze it but not glue any points together (continuity), not making and ,,folds” (continuous derivative).

A standard example of a diffeomorphism is the radial mapping F mapping the inside of the rectangle with vertices (0,0), (1,0), (0,2\pi) and (1,2\pi) onto the inside of a unit circle with centre in (0,0) without the radius from (0,0) to (1,0) by the formula

    \[F(x,y)=(x\cos y,x\sin y)\]

Theorem of inverse function

If F\colon U\to \mathbb{R}^n, where U\subseteq \mathbb{R}^n is open and F is of C^1 class, and for a\in U, \det F'(a)\neq 0, then there exists \delta>0 and an open set V\subseteq \mathbb{R}^n, such that F|_{B}\colon B\to V, where B=B(a,\delta)\subseteq U is a bijection (if x_1\neq x_2\in B then F(x_1)\neq F(x_2) and for every y\in V there exists x\in B, such that F(x)=y) and

  • the mapping F^{-1}\colon V\to B is of C^1 class on V,
  • (F^{-1})'(y)= (F'(x))^{-1}, where x\in B, y\in V and F(x)=y.

In other words if the matrix of the derivative is invertible at a given point, the function itself has an inverse on a neighborhood of this point. Yet, in other words, the function restricted to this neighborhood is a diffeomorphism.

Indeed, to understand the theorem consider function f(x,y)=(xy^2,x). Generally there is no inverse function — it is not possible to get out of (xy^2,x) uniquely (x,y), because even though x can be determined uniquely, we do not know the sign of y. We can write that (a,b)\mapsto (b,\pm\sqrt{a/b}), but we cannot determine the sign.

But in a neighborhood of a given point it may be possible. E.g. consider the point (1,1). In its neighborhood y>0, so for B((1,1),1/2), we get f^{-1}(a,b)=(b,\sqrt{a/b}) — and it works. Looking at the Theorem, indeed,

    \[f'=\left[\begin{array}{cc}y^2& 2xy\\1&0\end{array}\right],\]


    \[f'(1,1)=\left[\begin{array}{cc}1& 2\\1&0\end{array}\right]\]

and \det f'(1,1)=-2\neq 0. Moreover, indeed

    \[(f')^{-1}=\left[\begin{array}{cc}0& 1\\1/2xy&-y/2x\end{array}\right]\]

is the same as

    \[(f^{-1}(a,b))'=\left(b,\sqrt{a/b}\right)'=\left[\begin{array}{cc}0& 1\\1/2xy&-y/2x\end{array}\right].\]

On the other hand at (x,0) (for any x) there is not inverse function since

    \[f'(x,0)=\left[\begin{array}{cc}0& 0\\1&0\end{array}\right]\]

and so \det f'(x,0)=0.

Theorem of implicit function

This theorem describes when an equation of the form F(x_1,\ldots, x_n)=0 defines some variables as function of the other variables.

Let assume that a function F takes n+m arguments (x_1,\ldots, x_n, y_1,\ldots, y_m) and gives m values. We consider an equation (actually a system of euqations!) F(x_1,\ldots, x_n, y_1,\ldots, y_m)=(0,\ldots, 0) and we ask, whether knowing x, y are uniquely determined. The theorem describes when it is possible. Notice that F' is a m\times n+m matrix. We denote by F'_y the m\times m matrix consisting of m right columns of F' and by F'_x the m\times n matrix consisting of n left columns of F' .

Let W\subseteq \mathbbb{R}^{n+m} be open and let (a_1,\ldots, a_n, b_1,\ldots, b_m)\in U. Let F\colon W\to \mathbb{R}^m be of C^1 class and let F(a_1,\ldots, a_n, b_1,\ldots, b_m)=(0,\ldots, 0) and also \det F'_y(a_1,\ldots, a_n, b_1,\ldots, b_m)\neq 0. Then there exist open U\subseteq \mathbb{R}^n and V\subseteq \mathbb{R}^m, such that (a_1,\ldots, a_n)\in U and (b_1,\ldots, b_m)\in V, U\times V \subseteq W and crucially a function H\colon U\to \mathbb{R}^m of class C^1, which determines y from x in a neighborhood of (a_1,\ldots, a_n, b_1,\ldots, b_m), meaning for any (x_1,\ldots, x_n)\in U and (y_1,\ldots, y_m)\in V we have F(x_1,\ldots, x_n, y_1,\ldots, y_m)=(0,\ldots, 0) if and only if (y_1,\ldots, y_m)= H(x_1,\ldots, x_n). Moreover, H'=-(F'_y)^{-1}\cdot F'_x.

E.g. let us consider the following system of equations

    \[\begin{cases} x+y_1y_2^2=0\\ x+y_1=0\end{cases},\]

which corresponds to F(x,y_1,y_2)=(0,0), where F(x,y_1,y_2)=(x+y_1y_2^2, x+y_1). The question is whether we can determine y_, y_2 from x in a neighborhood of this point.
Let us consider neighborhood of (-1,1,1) and try to solve the problem manually. We have that y_1=-x. We also know that y_1\neq 0 (neighborhood of 1) so y_2^2=-x/y_1. We know that y_2 is positive (neighborhood of 1), thus y_2=\sqrt{-x/y_1}=\sqrt{1}=1. Then H(x)=(-x,1).

We check whether it is consistent with the theorem. We have








and the determinant equals 1, so the assumptions are met. Indeed, H exists. Moreover,

    \[-(F'_y)^{-1}\cdot F'_x=-\left[\begin{array}{cc}y_2^2&y_1\\1&0\end{array}\right]^{-1}\cdot \left[\begin{array}{c}1\\1\end{array}\right]=-\left[\begin{array}{cc}0&1\\1/2y_1y_2&-y_2/2y_1\end{array}\right]\cdot \left[\begin{array}{c}1\\1\end{array}\right]=\]

    \[= \left[\begin{array}{c}-1\\\frac{-1+y_2^2}{2y_1y_2}\end{array}\right]=\left[\begin{array}{c}-1\\0\end{array}\right],\]

(because since H(x)=(-x,1) to y_2=1), which is exactly the same as


Meanwhile at (0,0,1)


has zero determinant and manually we also will not be able to determine y from x, since y_1 may be equal to zero.