Editing Inverse function theorem (section)

==Statements==
For functions of a single [[Variable (mathematics)|variable]], the theorem states that if <math>f</math> is a [[continuously differentiable]] function with nonzero derivative at the point <math>a</math>; then <math>f</math> is injective (or bijective onto the image) in a neighborhood of <math>a</math>, the inverse is continuously differentiable near <math>b=f(a)</math>, and the derivative of the inverse function at <math>b</math> is the reciprocal of the derivative of <math>f</math> at <math>a</math>:
<math display=block>\bigl(f^{-1}\bigr)'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}.</math><!-- Not sure the meaning of the following alternative version; if the function is already injective, the theorem gives nothing: An alternate version, which assumes that <math>f</math> is [[Continuous function|continuous]] and [[Locally injective function|injective near {{Mvar|a}}]], and differentiable at {{Mvar|a}} with a non-zero derivative, will also result in <math>f</math> being invertible near {{Mvar|a}}, with an inverse that's similarly continuous and [[Injective function|injective]], and where the above formula would apply as well. -->

It can happen that a function <math>f</math> may be injective near a point <math>a</math> while <math>f'(a) = 0</math>. An example is <math>f(x) = (x - a)^3</math>. In fact, for such a function, the inverse cannot be differentiable at <math>b = f(a)</math>, since if <math>f^{-1}</math> were differentiable at <math>b</math>, then, by the chain rule, <math>1 = (f^{-1} \circ f)'(a) = (f^{-1})'(b)f'(a)</math>, which implies <math>f'(a) \ne 0</math>. (The situation is different for holomorphic functions; see [[#Holomorphic inverse function theorem]] below.)

For functions of more than one variable, the theorem states that if <math>f</math> is a continuously differentiable function from an open subset <math>A</math> of <math>\mathbb{R}^n</math> into <math>\R^n</math>, and the [[total derivative|derivative]] <math>f'(a)</math> is invertible at a point {{Mvar|a}} (that is, the determinant of the [[Jacobian matrix and determinant|Jacobian matrix]] of {{Mvar|f}} at {{Mvar|a}} is non-zero), then there exist neighborhoods <math>U</math> of <math>a</math> in <math>A</math> and <math>V</math> of <math>b = f(a)</math> such that <math>f(U) \subset V</math> and <math>f : U \to V</math> is bijective.<ref name="Hörmander">Theorem 1.1.7. in {{cite book|title=The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis|series=Classics in Mathematics|first=Lars|last= Hörmander|author-link=Lars Hörmander|publisher=Springer|year= 2015|edition=2nd|
isbn= 978-3-642-61497-2}}</ref> Writing <math>f=(f_1,\ldots,f_n)</math>, this means that the system of {{Mvar|n}} equations <math>y_i = f_i(x_1, \dots, x_n)</math> has a unique solution for <math>x_1, \dots, x_n</math> in terms of <math>y_1, \dots, y_n</math> when <math>x \in U, y \in V</math>. Note that the theorem ''does not'' say <math>f</math> is bijective onto the image where <math>f'</math> is invertible but that it is locally bijective where <math>f'</math> is invertible.

Moreover, the theorem says that the inverse function <math>f^{-1} : V \to U</math> is continuously differentiable, and its derivative at <math>b=f(a)</math> is the inverse map of <math>f'(a)</math>; i.e.,
:<math>(f^{-1})'(b) = f'(a)^{-1}.</math>
In other words, if <math>Jf^{-1}(b), Jf(a)</math> are the Jacobian matrices representing <math>(f^{-1})'(b), f'(a)</math>, this means:
:<math>Jf^{-1}(b) = Jf(a)^{-1}.</math>
The hard part of the theorem is the existence and differentiability of <math>f^{-1}</math>. Assuming this, the inverse derivative formula follows from the [[chain rule]] applied to <math>f^{-1}\circ f = I</math>. (Indeed, <math>1=I'(a) = (f^{-1} \circ f)'(a) = (f^{-1})'(b) \circ f'(a).</math>) Since taking the inverse is infinitely differentiable, the formula for the derivative of the inverse shows that if <math>f</math> is continuously <math>k</math> times differentiable, with invertible derivative at the point {{Mvar|a}}, then the inverse is also continuously <math>k</math> times differentiable. Here <math>k</math> is a positive integer or <math>\infty</math>.

There are two variants of the inverse function theorem.<ref name="Hörmander" /> Given a continuously differentiable map <math>f : U \to \mathbb{R}^m</math>, the first is
*The derivative <math>f'(a)</math> is surjective (i.e., the Jacobian matrix representing it has rank <math>m</math>) if and only if there exists a continuously differentiable function <math>g</math> on a neighborhood <math>V</math> of <math>b = f(a)</math> such that <math>f \circ g = I</math> near <math>b</math>,
and the second is
*The derivative <math>f'(a)</math> is injective if and only if there exists a continuously differentiable function <math>g</math> on a neighborhood <math>V</math> of <math>b = f(a)</math> such that <math>g \circ f = I</math> near <math>a</math>.

In the first case (when <math>f'(a)</math> is surjective), the point <math>b = f(a)</math> is called a [[regular value]]. Since <math>m = \dim \ker(f'(a)) + \dim \operatorname{im}(f'(a))</math>, the first case is equivalent to saying <math>b = f(a)</math> is not in the image of [[Critical point (mathematics)#Critical point of a differentiable map|critical points]] <math>a</math> (a critical point is a point <math>a</math> such that the kernel of <math>f'(a)</math> is nonzero). The statement in the first case is a special case of the [[submersion theorem]].

These variants are restatements of the inverse functions theorem. Indeed, in the first case when <math>f'(a)</math> is surjective, we can find an (injective) linear map <math>T</math> such that <math>f'(a) \circ T = I</math>. Define <math>h(x) = a + Tx</math> so that we have:
:<math>(f \circ h)'(0) = f'(a) \circ T = I.</math>
Thus, by the inverse function theorem, <math>f \circ h</math> has inverse near <math>0</math>; i.e., <math>f \circ h \circ (f \circ h)^{-1} = I</math> near <math>b</math>. The second case (<math>f'(a)</math> is injective) is seen in the similar way.