Inreal analysis, a branch ofmathematics, theinverse function theorem is atheorem that asserts that, if areal functionf has acontinuous derivative near a point where its derivative is nonzero, then, near this point,f has aninverse function. The inverse function is alsocontinuously differentiable, and theinverse function rule expresses its derivative as themultiplicative inverse of the derivative off.

The theorem applies verbatim tocomplex-valued functions of acomplex variable. It generalizes to functions fromn-tuples (of real or complex numbers) ton-tuples, and to functions betweenvector spaces of the same finite dimension, by replacing "derivative" with "Jacobian matrix" and "nonzero derivative" with "nonzeroJacobian determinant".

If the function of the theorem belongs to a higherdifferentiability class, the same is true for the inverse function. There are also versions of the inverse function theorem forholomorphic functions, for differentiable maps betweenmanifolds, for differentiable functions betweenBanach spaces, and so forth.

The theorem was first established byPicard andGoursat using an iterative scheme: the basic idea is to prove afixed point theorem using thecontraction mapping theorem.

For functions of a singlevariable, the theorem states that if${\displaystyle f}$ is acontinuously differentiable function with nonzero derivative at the point${\displaystyle a}$; then${\displaystyle f}$ is injective (or bijective onto the image) in a neighborhood of${\displaystyle a}$, the inverse is continuously differentiable near${\displaystyle b=f(a)}$, and the derivative of the inverse function at${\displaystyle b}$ is the reciprocal of the derivative of${\displaystyle f}$ at${\displaystyle a}$:$${\displaystyle (f^{-1}{)}^{\prime }(b)=\frac{1}{f^{\prime }(a)}=\frac{1}{f^{\prime }(f^{-1}(b))}.}$$

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

For functions of more than one variable, the theorem states that if${\displaystyle f}$ is a continuously differentiable function from an open subset${\displaystyle A}$ of${\displaystyle {\mathbb{R}}^n}$ into${\displaystyle {\mathbb{R}}^n}$, and thederivative${\displaystyle f^{\prime }(a)}$ is invertible at a pointa (that is, the determinant of theJacobian matrix off ata is non-zero), then there exist neighborhoods${\displaystyle U}$ of${\displaystyle a}$ in${\displaystyle A}$ and${\displaystyle V}$ of${\displaystyle b=f(a)}$ such that${\displaystyle f(U)\subset V}$ and${\displaystyle f:U\to V}$ is bijective.^[1] Writing${\displaystyle f=(f_1,\dots ,f_n)}$, this means that the system ofn equations${\displaystyle y_i=f_i(x_1,\dots ,x_n)}$ has a unique solution for${\displaystyle x_1,\dots ,x_n}$ in terms of${\displaystyle y_1,\dots ,y_n}$ when${\displaystyle x\in U,y\in V}$. Note that the theoremdoes not say${\displaystyle f}$ is bijective onto the image where${\displaystyle f^{\prime }}$ is invertible but that it is locally bijective where${\displaystyle f^{\prime }}$ is invertible.

Moreover, the theorem says that the inverse function${\displaystyle f^{-1}:V\to U}$ is continuously differentiable, and its derivative at${\displaystyle b=f(a)}$ is the inverse map of${\displaystyle f^{\prime }(a)}$; i.e.,

${\displaystyle (f^{-1}{)}^{\prime }(b)=f^{\prime }(a{)}^{-1}.}$

In other words, if${\displaystyle J{f}^{-1}(b),Jf(a)}$ are the Jacobian matrices representing${\displaystyle (f^{-1}{)}^{\prime }(b),f^{\prime }(a)}$, this means:

${\displaystyle J{f}^{-1}(b)=Jf(a{)}^{-1}.}$

The hard part of the theorem is the existence and differentiability of${\displaystyle f^{-1}}$. Assuming this, the inverse derivative formula follows from thechain rule applied to${\displaystyle f^{-1}\circ f=I}$. (Indeed,${\displaystyle 1=I^{\prime }(a)=(f^{-1}\circ f{)}^{\prime }(a)=(f^{-1}{)}^{\prime }(b)\circ f^{\prime }(a).}$) Since taking the inverse is infinitely differentiable, the formula for the derivative of the inverse shows that if${\displaystyle f}$ is continuously${\displaystyle k}$ times differentiable, with invertible derivative at the pointa, then the inverse is also continuously${\displaystyle k}$ times differentiable. Here${\displaystyle k}$ is a positive integer or${\displaystyle \mathrm{\infty }}$.

There are two variants of the inverse function theorem.^[1] Given a continuously differentiable map${\displaystyle f:U\to {\mathbb{R}}^m}$, the first is

• The derivative${\displaystyle f^{\prime }(a)}$ is surjective (i.e., the Jacobian matrix representing it has rank${\displaystyle m}$) if and only if there exists a continuously differentiable function${\displaystyle g}$ on a neighborhood${\displaystyle V}$ of${\displaystyle b=f(a)}$ such that${\displaystyle f\circ g=I}$ near${\displaystyle b}$,

and the second is

• The derivative${\displaystyle f^{\prime }(a)}$ is injective if and only if there exists a continuously differentiable function${\displaystyle g}$ on a neighborhood${\displaystyle V}$ of${\displaystyle b=f(a)}$ such that${\displaystyle g\circ f=I}$ near${\displaystyle a}$.

In the first case (when${\displaystyle f^{\prime }(a)}$ is surjective), the point${\displaystyle b=f(a)}$ is called aregular value. Since${\displaystyle m=\dim \ker (f^{\prime }(a))+\dim \mathrm{im}(f^{\prime }(a))}$, the first case is equivalent to saying${\displaystyle b=f(a)}$ is not in the image ofcritical points${\displaystyle a}$ (a critical point is a point${\displaystyle a}$ such that the kernel of${\displaystyle f^{\prime }(a)}$ is nonzero). The statement in the first case is a special case of thesubmersion theorem.

These variants are restatements of the inverse functions theorem. Indeed, in the first case when${\displaystyle f^{\prime }(a)}$ is surjective, we can find an (injective) linear map${\displaystyle T}$ such that${\displaystyle f^{\prime }(a)\circ T=I}$. Define${\displaystyle h(x)=a+Tx}$ so that we have:

${\displaystyle (f\circ h{)}^{\prime }(0)=f^{\prime }(a)\circ T=I.}$

Thus, by the inverse function theorem,${\displaystyle f\circ h}$ has inverse near${\displaystyle 0}$; i.e.,${\displaystyle f\circ h\circ (f\circ h{)}^{-1}=I}$ near${\displaystyle b}$. The second case (${\displaystyle f^{\prime }(a)}$ is injective) is seen in the similar way.

Consider thevector-valued function${\displaystyle F:{\mathbb{R}}^2\to {\mathbb{R}}^2}$ defined by:

${\displaystyle F(x,y)=\left[\begin{array}{c}{e}^x\cos y\\ e^x\sin y\end{array}\right].}$

The Jacobian matrix of it at${\displaystyle (x,y)}$ is:

with the determinant:

${\displaystyle detJF(x,y)=e^{2x}{\cos }^{2}y+e^{2x}{\sin }^{2}y=e^{2x}.}$

The determinant${\displaystyle e^{2x}}$ is nonzero everywhere. Thus the theorem guarantees that, for every pointp in${\displaystyle {\mathbb{R}}^2}$, there exists a neighborhood aboutp over whichF is invertible. This does not meanF is invertible over its entire domain: in this caseF is not eveninjective since it is periodic:${\displaystyle F(x,y)=F(x,y+2\pi )}$.

If one drops the assumption that the derivative is continuous, the function is no longer necessarily locally injective. For example${\displaystyle f(x)=x+2x^2\sin (\frac{1}{x})}$ and${\displaystyle f(0)=0}$ has discontinuous derivative${\displaystyle f^{\prime }(x)=1-2\cos (\frac{1}{x})+4x\sin (\frac{1}{x})}$ and${\displaystyle f^{\prime }(0)=1,}$ which vanishes arbitrarily close to${\displaystyle x=0}$. These critical points are local max/min points of${\displaystyle f,}$ so${\displaystyle f}$ is not one-to-one (and not invertible) on any interval containing${\displaystyle x=0}$. Intuitively, the slope${\displaystyle f^{\prime }(0)=1}$ does not propagate to nearby points, where the slopes are governed by a weak but rapid oscillation.

If the derivative is continuous but zero at a point, the function is no longer necessarily locally injective. A real function that islocally constant at a point${\displaystyle x\in \mathbb{R}}$ in the interior of its domain is not locally injective at${\displaystyle x}$ but is trivially continuously differentiable at${\displaystyle x}$.

As an important result, the inverse function theorem has been given numerous proofs. The proof most commonly seen in textbooks relies on thecontraction mapping principle, also known as theBanach fixed-point theorem (which can also be used as the key step in the proof ofexistence and uniqueness of solutions toordinary differential equations).^[2][3]

Since the fixed point theorem applies in infinite-dimensional (Banach space) settings, this proof generalizes immediately to the infinite-dimensional version of the inverse function theorem^[4] (seeGeneralizations below).

An alternate proof in finite dimensions hinges on theextreme value theorem for functions on acompact set.^[5] This approach has an advantage that the proof generalizes to a situation where there is no Cauchy completeness (see§ Over a real closed field).

Yet another proof usesNewton's method, which has the advantage of providing aneffective version of the theorem: bounds on the derivative of the function imply an estimate of the size of the neighborhood on which the function is invertible.^[6]

We want to prove the following:Let${\displaystyle D\subseteq \mathbb{R}}$ be an open set with${\displaystyle x_0\in D,f:D\to \mathbb{R}}$ a continuously differentiable function defined on${\displaystyle D}$, and suppose that${\displaystyle f^{\prime }(x_0)\ne 0}$. Then there exists an open interval${\displaystyle I}$ with${\displaystyle x_0\in I}$ such that${\displaystyle f}$ maps${\displaystyle I}$ bijectively onto the open interval${\displaystyle J=f(I)}$, and such that the inverse function${\displaystyle f^{-1}:J\to I}$ is continuously differentiable, and for any${\displaystyle y\in J}$, if${\displaystyle x\in I}$ is such that${\displaystyle f(x)=y}$, then${\displaystyle (f^{-1}{)}^{\prime }(y)={\displaystyle \frac{1}{f^{\prime }(x)}}}$.

We may without loss of generality assume that${\displaystyle f^{\prime }(x_0)>0}$. Given that${\displaystyle D}$ is an open set and${\displaystyle f^{\prime }}$ is continuous at${\displaystyle x_0}$, there exists${\displaystyle r>0}$ such that${\displaystyle (x_0-r,x_0+r)\subseteq D}$ and

In particular,

This shows that${\displaystyle f}$ is strictly increasing for all. Let${\displaystyle \delta >0}$ be such that. Then${\displaystyle [x-\delta ,x+\delta ]\subseteq (x_0-r,x_0+r)}$. By the intermediate value theorem, we find that${\displaystyle f}$ maps the interval${\displaystyle [x-\delta ,x+\delta ]}$ bijectively onto${\displaystyle [f(x-\delta ),f(x+\delta )]}$. Denote by${\displaystyle I=(x-\delta ,x+\delta )}$ and${\displaystyle J=(f(x-\delta ),f(x+\delta ))}$. Then${\displaystyle f:I\to J}$ is a bijection and the inverse${\displaystyle f^{-1}:J\to I}$ exists. The fact that${\displaystyle f^{-1}:J\to I}$ is differentiable follows from the differentiability of${\displaystyle f}$. In particular, the result follows from the fact that if${\displaystyle f:I\to \mathbb{R}}$ is a strictly monotonic and continuous function that is differentiable at${\displaystyle x_0\in I}$ with${\displaystyle f^{\prime }(x_0)\ne 0}$, then${\displaystyle f^{-1}:f(I)\to \mathbb{R}}$ is differentiable with${\displaystyle (f^{-1}{)}^{\prime }(y_0)={\displaystyle \frac{1}{f^{\prime }(x_0)}}}$, where${\displaystyle y_0=f(x_0)}$ (a standard result in analysis). This completes the proof.

To prove existence, it can be assumed after an affine transformation that${\displaystyle f(0)=0}$ and${\displaystyle f^{\mathrm{\prime }}(0)=I}$, so that${\displaystyle a=b=0}$.

By themean value theorem for vector-valued functions, for a differentiable function${\displaystyle u:[0,1]\to {\mathbb{R}}^m}$,$\Vert u(1)-u(0)\Vert \le \underset{0\le t\le 1}{sup}\Vert u^{\mathrm{\prime }}(t)\Vert$. Setting${\displaystyle u(t)=f(x+t(x^{\mathrm{\prime }}-x))-x-t(x^{\mathrm{\prime }}-x)}$, it follows that

${\displaystyle \Vert f(x)-f(x^{\mathrm{\prime }})-x+x^{\mathrm{\prime }}\Vert \le \Vert x-x^{\mathrm{\prime }}\Vert \underset{0\le t\le 1}{sup}\Vert f^{\mathrm{\prime }}(x+t(x^{\mathrm{\prime }}-x))-I\Vert .}$

Now choose${\displaystyle \delta >0}$ so that for. Suppose that and define${\displaystyle x_n}$ inductively by${\displaystyle x_0=0}$ and${\displaystyle x_{n+1}=x_n+y-f(x_n)}$. The assumptions show that if then

${\displaystyle \Vert f(x)-f(x^{\mathrm{\prime }})-x+x^{\mathrm{\prime }}\Vert \le \Vert x-x^{\mathrm{\prime }}\Vert /2}$.

In particular${\displaystyle f(x)=f(x^{\mathrm{\prime }})}$ implies${\displaystyle x=x^{\mathrm{\prime }}}$. In the inductive schemeand. Thus${\displaystyle (x_n)}$ is aCauchy sequence tending to${\displaystyle x}$. By construction${\displaystyle f(x)=y}$ as required.

To check that${\displaystyle g=f^{-1}}$ is C¹, write${\displaystyle g(y+k)=x+h}$ so that${\displaystyle f(x+h)=f(x)+k}$. By the inequalities above, so that.On the other hand, if${\displaystyle A=f^{\mathrm{\prime }}(x)}$, then. Using thegeometric series for${\displaystyle B=I-A}$, it follows that. But then

${\displaystyle \frac{\Vert g(y+k)-g(y)-f^{\mathrm{\prime }}(g(y){)}^{-1}k\Vert }{\Vert k\Vert }=\frac{\Vert h-f^{\mathrm{\prime }}(x{)}^{-1}[f(x+h)-f(x)]\Vert }{\Vert k\Vert }\le 4\frac{\Vert f(x+h)-f(x)-f^{\mathrm{\prime }}(x)h\Vert }{\Vert h\Vert }}$

tends to 0 as${\displaystyle k}$ and${\displaystyle h}$ tend to 0, proving that${\displaystyle g}$ is C¹ with${\displaystyle g^{\mathrm{\prime }}(y)=f^{\mathrm{\prime }}(g(y){)}^{-1}}$.

The proof above is presented for a finite-dimensional space, but applies equally well forBanach spaces. If an invertible function${\displaystyle f}$ is C^k with${\displaystyle k>1}$, then so too is its inverse. This follows by induction using the fact that the map${\displaystyle F(A)=A^{-1}}$ on operators is C^k for any${\displaystyle k}$ (in the finite-dimensional case this is an elementary fact because the inverse of a matrix is given as theadjugate matrix divided by itsdeterminant).^[1][7] The method of proof here can be found in the books ofHenri Cartan,Jean Dieudonné,Serge Lang,Roger Godement andLars Hörmander.

Here is a proof based on thecontraction mapping theorem. Specifically, following T. Tao,^[8] it uses the following consequence of the contraction mapping theorem.

Basically, the lemma says that a small perturbation of the identity map by a contraction map is injective and preserves a ball in some sense. Assuming the lemma for a moment, we prove the theorem first. As in the above proof, it is enough to prove the special case when${\displaystyle a=0,b=f(a)=0}$ and${\displaystyle f^{\prime }(0)=I}$. Let${\displaystyle g=f-I}$. Themean value inequality applied to${\displaystyle t\mapsto g(x+t(y-x))}$ says:

Since${\displaystyle g^{\prime }(0)=I-I=0}$ and${\displaystyle g^{\prime }}$ is continuous, we can find an${\displaystyle r>0}$ such that

${\displaystyle |g(y)-g(x)|\le 2^{-1}|y-x|}$

for all${\displaystyle x,y}$ in${\displaystyle B(0,r)}$. Then the early lemma says that${\displaystyle f=g+I}$ is injective on${\displaystyle B(0,r)}$ and${\displaystyle B(0,r/2)\subset f(B(0,r))}$. Then

${\displaystyle f:U=B(0,r)\cap f^{-1}(B(0,r/2))\to V=B(0,r/2)}$

is bijective and thus has an inverse. Next, we show the inverse${\displaystyle f^{-1}}$ is continuously differentiable (this part of the argument is the same as that in the previous proof). This time, let${\displaystyle g=f^{-1}}$ denote the inverse of${\displaystyle f}$ and${\displaystyle A=f^{\prime }(x)}$. For${\displaystyle x=g(y)}$, we write${\displaystyle g(y+k)=x+h}$ or${\displaystyle y+k=f(x+h)}$. Now, by the early estimate, we have

${\displaystyle |h-k|=|f(x+h)-f(x)-h|\le |h|/2}$

and so${\displaystyle |h|/2\le |k|}$. Writing${\displaystyle \Vert \cdot \Vert }$ for the operator norm,

${\displaystyle |g(y+k)-g(y)-A^{-1}k|=|h-A^{-1}(f(x+h)-f(x))|\le \Vert A^{-1}\Vert |Ah-f(x+h)+f(x)|.}$

As${\displaystyle k\to 0}$, we have${\displaystyle h\to 0}$ and${\displaystyle |h|/|k|}$ is bounded. Hence,${\displaystyle g}$ is differentiable at${\displaystyle y}$ with the derivative${\displaystyle g^{\prime }(y)=f^{\prime }(g(y){)}^{-1}}$. Also,${\displaystyle g^{\prime }}$ is the same as the composition${\displaystyle \iota \circ f^{\prime }\circ g}$ where${\displaystyle \iota :T\mapsto T^{-1}}$; so${\displaystyle g^{\prime }}$ is continuous.

It remains to show the lemma. First, we have:

${\displaystyle |x-y|-|f(x)-f(y)|\le |g(x)-g(y)|\le c|x-y|,}$

which is to say

${\displaystyle (1-c)|x-y|\le |f(x)-f(y)|.}$

This proves the first part. Next, we show${\displaystyle f(B(0,r))\supset B(0,(1-c)r)}$. The idea is to note that this is equivalent to, given a point${\displaystyle y}$ in${\displaystyle B(0,(1-c)r)}$, find a fixed point of the map

${\displaystyle F:\overline{B}(0,r^{\prime })\to \overline{B}(0,r^{\prime }),x\mapsto y-g(x)}$

where such that${\displaystyle |y|\le (1-c)r^{\prime }}$ and the bar means a closed ball. To find a fixed point, we use the contraction mapping theorem and checking that${\displaystyle F}$ is a well-defined strict-contraction mapping is straightforward. Finally, we have:${\displaystyle f(B(0,r))\subset B(0,(1+c)r)}$ since

${\displaystyle |f(x)|=|x+g(x)-g(0)|\le (1+c)|x|.◻}$

As might be clear, this proof is not substantially different from the previous one, as the proof of the contraction mapping theorem is by successive approximation.

The inverse function theorem can be used to solve a system of equations

i.e., expressing${\displaystyle y_1,\dots ,y_n}$ as functions of${\displaystyle x=(x_1,\dots ,x_n)}$, provided the Jacobian matrix is invertible. Theimplicit function theorem allows to solve a more general system of equations:

for${\displaystyle y}$ in terms of${\displaystyle x}$. Though more general, the theorem is actually a consequence of the inverse function theorem. First, the precise statement of the implicit function theorem is as follows:^[9]

• given a map${\displaystyle f:{\mathbb{R}}^n\times {\mathbb{R}}^m\to {\mathbb{R}}^m}$, if${\displaystyle f(a,b)=0}$,${\displaystyle f}$ is continuously differentiable in a neighborhood of${\displaystyle (a,b)}$ and the derivative of${\displaystyle y\mapsto f(a,y)}$ at${\displaystyle b}$ is invertible, then there exists a differentiable map${\displaystyle g:U\to V}$ for some neighborhoods${\displaystyle U,V}$ of${\displaystyle a,b}$ such that${\displaystyle f(x,g(x))=0}$. Moreover, if${\displaystyle f(x,y)=0,x\in U,y\in V}$, then${\displaystyle y=g(x)}$; i.e.,${\displaystyle g(x)}$ is a unique solution.

To see this, consider the map${\displaystyle F(x,y)=(x,f(x,y))}$. By the inverse function theorem,${\displaystyle F:U\times V\to W}$ has the inverse${\displaystyle G}$ for some neighborhoods${\displaystyle U,V,W}$. We then have:

${\displaystyle (x,y)=F(G_1(x,y),G_2(x,y))=(G_1(x,y),f(G_1(x,y),G_2(x,y))),}$

implying${\displaystyle x=G_1(x,y)}$ and${\displaystyle y=f(x,G_2(x,y)).}$ Thus${\displaystyle g(x)=G_2(x,0)}$ has the required property.${\displaystyle ◻}$

In differential geometry, the inverse function theorem is used to show that the pre-image of aregular value under a smooth map is a manifold.^[10] Indeed, let${\displaystyle f:U\to {\mathbb{R}}^r}$ be such a smooth map from an open subset of${\displaystyle {\mathbb{R}}^n}$ (since the result is local, there is no loss of generality with considering such a map). Fix a point${\displaystyle a}$ in${\displaystyle f^{-1}(b)}$ and then, by permuting the coordinates on${\displaystyle {\mathbb{R}}^n}$, assume the matrix${\displaystyle {\left[\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{x}_j}(a)\right]}_{1\le i,j\le r}}$ has rank${\displaystyle r}$. Then the map${\displaystyle F:U\to {\mathbb{R}}^r\times {\mathbb{R}}^{n-r}={\mathbb{R}}^n,x\mapsto (f(x),x_{r+1},\dots ,x_n)}$ is such that${\displaystyle F^{\prime }(a)}$ has rank${\displaystyle n}$. Hence, by the inverse function theorem, we find the smooth inverse${\displaystyle G}$ of${\displaystyle F}$ defined in a neighborhood${\displaystyle V\times W}$ of${\displaystyle (b,a_{r+1},\dots ,a_n)}$. We then have

${\displaystyle x=(F\circ G)(x)=(f(G(x)),G_{r+1}(x),\dots ,G_n(x)),}$

which implies

${\displaystyle (f\circ G)(x_1,\dots ,x_n)=(x_1,\dots ,x_r).}$

That is, after the change of coordinates by${\displaystyle G}$,${\displaystyle f}$ is a coordinate projection (this fact is known as thesubmersion theorem). Moreover, since${\displaystyle G:V\times W\to U^{\prime }=G(V\times W)}$ is bijective, the map

${\displaystyle g=G(b,\cdot ):W\to f^{-1}(b)\cap U^{\prime },(x_{r+1},\dots ,x_n)\mapsto G(b,x_{r+1},\dots ,x_n)}$

is bijective with the smooth inverse. That is to say,${\displaystyle g}$ gives a local parametrization of${\displaystyle f^{-1}(b)}$ around${\displaystyle a}$. Hence,${\displaystyle f^{-1}(b)}$ is a manifold.${\displaystyle ◻}$ (Note the proof is quite similar to the proof of the implicit function theorem and, in fact, the implicit function theorem can be also used instead.)

More generally, the theorem shows that if a smooth map${\displaystyle f:P\to E}$ is transversal to a submanifold${\displaystyle M\subset E}$, then the pre-image${\displaystyle f^{-1}(M)\hookrightarrow P}$ is a submanifold.^[11]

The inverse function theorem is a local result; it applies to each point.A priori, the theorem thus only shows the function${\displaystyle f}$ is locally bijective (or locally diffeomorphic of some class). The next topological lemma can be used to upgrade local injectivity to injectivity that is global to some extent.

Proof:^[14] First assume${\displaystyle X}$ iscompact. If the conclusion of the theorem is false, we can find two sequences${\displaystyle x_i\ne y_i}$ such that${\displaystyle f(x_i)=f(y_i)}$ and${\displaystyle x_i,y_i}$ each converge to some points${\displaystyle x,y}$ in${\displaystyle A}$. Since${\displaystyle f}$ is injective on${\displaystyle A}$,${\displaystyle x=y}$. Now, if${\displaystyle i}$ is large enough,${\displaystyle x_i,y_i}$ are in a neighborhood of${\displaystyle x=y}$ where${\displaystyle f}$ is injective; thus,${\displaystyle x_i=y_i}$, a contradiction.

In general, consider the set${\displaystyle E=\{(x,y)\in X^2\mid x\ne y,f(x)=f(y)\}}$. It is disjoint from${\displaystyle S\times S}$ for any subset${\displaystyle S\subset X}$ where${\displaystyle f}$ is injective. Let${\displaystyle X_1\subset X_2\subset \cdots }$ be an increasing sequence of compact subsets with union${\displaystyle X}$ and with${\displaystyle X_i}$ contained in the interior of${\displaystyle X_{i+1}}$. Then, by the first part of the proof, for each${\displaystyle i}$, we can find a neighborhood${\displaystyle U_i}$ of${\displaystyle A\cap X_i}$ such that${\displaystyle U_i^2\subset X^2-E}$. Then${\displaystyle U=\bigcup _i{U}_i}$ has the required property.${\displaystyle ◻}$ (See also^[15] for an alternative approach.)

The lemma implies the following (a sort of) global version of the inverse function theorem:

Note that if${\displaystyle A}$ is a point, then the above is the usual inverse function theorem.

There is a version of the inverse function theorem forholomorphic maps.

The theorem follows from the usual inverse function theorem. Indeed, let${\displaystyle J_{\mathbb{R}}(f)}$ denote the Jacobian matrix of${\displaystyle f}$ in variables${\displaystyle x_i,y_i}$ and${\displaystyle J(f)}$ for that in${\displaystyle z_j,{\overline{z}}_j}$. Then we have${\displaystyle det{J}_{\mathbb{R}}(f)=|detJ(f){|}^2}$, which is nonzero by assumption. Hence, by the usual inverse function theorem,${\displaystyle f}$ is injective near${\displaystyle 0}$ with continuously differentiable inverse. By chain rule, with${\displaystyle w=f(z)}$,

${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }{\overline{z}}_j}(f_j^{-1}\circ f)(z)=\sum _k\frac{\mathrm{\partial }{f}_j^{-1}}{\mathrm{\partial }{w}_k}(w)\frac{\mathrm{\partial }{f}_k}{\mathrm{\partial }{\overline{z}}_j}(z)+\sum _k\frac{\mathrm{\partial }{f}_j^{-1}}{\mathrm{\partial }{\overline{w}}_k}(w)\frac{\mathrm{\partial }{\overline{f}}_k}{\mathrm{\partial }{\overline{z}}_j}(z)}$

where the left-hand side and the first term on the right vanish since${\displaystyle f_j^{-1}\circ f}$ and${\displaystyle f_k}$ are holomorphic. Thus,${\displaystyle \frac{\mathrm{\partial }{f}_j^{-1}}{\mathrm{\partial }{\overline{w}}_k}(w)=0}$ for each${\displaystyle k}$.${\displaystyle ◻}$

Similarly, there is the implicit function theorem for holomorphic functions.^[19]

As already noted earlier, it can happen that an injective smooth function has the inverse that is not smooth (e.g.,${\displaystyle f(x)=x^3}$ in a real variable). This is not the case for holomorphic functions because of:

The inverse function theorem can be rephrased in terms of differentiable maps betweendifferentiable manifolds. In this context the theorem states that for a differentiable map${\displaystyle F:M\to N}$ (of class${\displaystyle C^1}$), if thedifferential of${\displaystyle F}$,

${\displaystyle d{F}_p:T_{p}M\to T_{F(p)}N}$

is alinear isomorphism at a point${\displaystyle p}$ in${\displaystyle M}$ then there exists an open neighborhood${\displaystyle U}$ of${\displaystyle p}$ such that

${\displaystyle F{|}_U:U\to F(U)}$

is adiffeomorphism. Note that this implies that the connected components ofM andN containingp andF(p) have the same dimension, as is already directly implied from the assumption thatdF_p is an isomorphism.If the derivative ofF is an isomorphism at all pointsp inM then the mapF is alocal diffeomorphism.

The inverse function theorem can also be generalized to differentiable maps betweenBanach spacesX andY.^[20] LetU be an open neighbourhood of the origin inX and${\displaystyle F:U\to Y}$ a continuously differentiable function, and assume that the Fréchet derivative${\displaystyle d{F}_0:X\to Y}$ ofF at 0 is abounded linear isomorphism ofX ontoY. Then there exists an open neighbourhoodV of${\displaystyle F(0)}$ inY and a continuously differentiable map${\displaystyle G:V\to X}$ such that${\displaystyle F(G(y))=y}$ for ally inV. Moreover,${\displaystyle G(y)}$ is the only sufficiently small solutionx of the equation${\displaystyle F(x)=y}$.

There is also the inverse function theorem forBanach manifolds.^[21]

The inverse function theorem (and theimplicit function theorem) can be seen as a special case of the constant rank theorem, which states that a smooth map with constantrank near a point can be put in a particular normal form near that point.^[22] Specifically, if${\displaystyle F:M\to N}$ has constant rank near a point${\displaystyle p\in M}$, then there are open neighborhoodsU ofp andV of${\displaystyle F(p)}$ and there are diffeomorphisms${\displaystyle u:T_{p}M\to U}$ and${\displaystyle v:T_{F(p)}N\to V}$ such that${\displaystyle F(U)\subseteq V}$ and such that the derivative${\displaystyle d{F}_p:T_{p}M\to T_{F(p)}N}$ is equal to${\displaystyle v^{-1}\circ F\circ u}$. That is,F "looks like" its derivative nearp. The set of points${\displaystyle p\in M}$ such that the rank is constant in a neighborhood of${\displaystyle p}$ is an open dense subset ofM; this is a consequence ofsemicontinuity of the rank function. Thus the constant rank theorem applies to a generic point of the domain.

When the derivative ofF is injective (resp. surjective) at a pointp, it is also injective (resp. surjective) in a neighborhood ofp, and hence the rank ofF is constant on that neighborhood, and the constant rank theorem applies.

If it is true, theJacobian conjecture would be a variant of the inverse function theorem for polynomials. It states that if a vector-valued polynomial function has aJacobian determinant that is an invertible polynomial (that is a nonzero constant), then it has an inverse that is also a polynomial function. It is unknown whether this is true or false, even in the case of two variables. This is a major open problem in the theory of polynomials.

When${\displaystyle f:{\mathbb{R}}^n\to {\mathbb{R}}^m}$ with${\displaystyle m\le n}$,${\displaystyle f}$ is${\displaystyle k}$ timescontinuously differentiable, and the Jacobian${\displaystyle A=\mathrm{\nabla }f(\overline{x})}$ at a point${\displaystyle \overline{x}}$ is ofrank${\displaystyle m}$, the inverse of${\displaystyle f}$ may not be unique. However, there exists a localselection function${\displaystyle s}$ such that${\displaystyle f(s(y))=y}$ for all${\displaystyle y}$ in aneighborhood of${\displaystyle \overline{y}=f(\overline{x})}$,${\displaystyle s(\overline{y})=\overline{x}}$,${\displaystyle s}$ is${\displaystyle k}$ times continuously differentiable in this neighborhood, and${\displaystyle \mathrm{\nabla }s(\overline{y})=A^T(A{A}^T{)}^{-1}}$ (${\displaystyle \mathrm{\nabla }s(\overline{y})}$ is theMoore–Penrose pseudoinverse of${\displaystyle A}$).^[23]