The classical inverse function theorem reads as follows:

Theorem 1 ( inverse function theorem)Let be an open set, and let be an continuously differentiable function, such that for every , the derivative map is invertible. Then is a local homeomorphism; thus, for every , there exists an open neighbourhood of and an open neighbourhood of such that is a homeomorphism from to .

It is also not difficult to show by inverting the Taylor expansion

that at each , the local inverses are also differentiable at with derivative

The textbook proof of the inverse function theorem proceeds by an application of the contraction mapping theorem. Indeed, one may normalise and to be the identity map; continuity of then shows that is close to the identity for small , which may be used (in conjunction with the fundamental theorem of calculus) to make a contraction on a small ball around the origin for small , at which point the contraction mapping theorem readily finishes off the problem.

I recently learned (after I asked this question on Math Overflow) that the hypothesis of continuous differentiability may be relaxed to just everywhere differentiability:

Theorem 2 (Everywhere differentiable inverse function theorem)Let be an open set, and let be an everywhere differentiable function, such that for every , the derivative map is invertible. Then is a local homeomorphism; thus, for every , there exists an open neighbourhood of and an open neighbourhood of such that is a homeomorphism from to .

As before, one can recover the differentiability of the local inverses, with the derivative of the inverse given by the usual formula (1).

This result implicitly follows from the more general results of Cernavskii about the structure of finite-to-one open and closed maps, however the arguments there are somewhat complicated (and subsequent proofs of those results, such as the one by Vaisala, use some powerful tools from algebraic geometry, such as dimension theory). There is however a more elementary proof of Saint Raymond that was pointed out to me by Julien Melleray. It only uses basic point-set topology (for instance, the concept of a connected component) and the basic topological and geometric structure of Euclidean space (in particular relying primarily on local compactness, local connectedness, and local convexity). I decided to present (an arrangement of) Saint Raymond’s proof here.

To obtain a local homeomorphism near , there are basically two things to show: local surjectivity near (thus, for near , one can solve for some near ) and local injectivity near (thus, for distinct near , is not equal to ). Local surjectivity is relatively easy; basically, the standard proof of the inverse function theorem works here, after replacing the contraction mapping theorem (which is no longer available due to the possibly discontinuous nature of ) with the Brouwer fixed point theorem instead (or one could also use degree theory, which is more or less an equivalent approach). The difficulty is local injectivity – one needs to preclude the existence of nearby points with ; note that in contrast to the contraction mapping theorem that provides both existence and uniqueness of fixed points, the Brouwer fixed point theorem only gives existence and not uniqueness.

In one dimension one can proceed by using Rolle’s theorem. Indeed, as one traverses the interval from to , one must encounter some intermediate point which maximises the quantity , and which is thus instantaneously non-increasing both to the left and to the right of . But, by hypothesis, is non-zero, and this easily leads to a contradiction.

Saint Raymond’s argument for the higher dimensional case proceeds in a broadly similar way. Starting with two nearby points with , one finds a point which “locally extremises” in the following sense: is equal to some , but is adherent to at least two distinct connected components of the set . (This is an oversimplification, as one has to restrict the available points in to a suitably small compact set, but let us ignore this technicality for now.) Note from the non-degenerate nature of that was already adherent to ; the point is that “disconnects” in some sense. Very roughly speaking, the way such a critical point is found is to look at the sets as shrinks from a large initial value down to zero, and one finds the first value of below which this set disconnects from . (Morally, one is performing some sort of Morse theory here on the function , though this function does not have anywhere near enough regularity for classical Morse theory to apply.)

The point is mapped to a point on the boundary of the ball , while the components are mapped to the interior of this ball. By using a continuity argument, one can show (again very roughly speaking) that must contain a “hemispherical” neighbourhood of inside , and similarly for . But then from differentiability of at , one can then show that and overlap near , giving a contradiction.

The rigorous details of the proof are provided below the fold.

** — 1. Proof — **

Fix . By a translation, we may assume ; by a further linear change of variables, we may also assume (which by hypothesis is non-singular) to be the identity map. By differentiability, we have

as . In particular, there exists a ball in such that

for all ; by rescaling we may take , thus

Among other things, this gives a uniform lower bound

for all , and a uniform upper bound

*Proof:* Let . From (2), we see that the map avoids , and has degree around ; contracting to a point, we conclude that for some , yielding the claim.

Alternatively, one may proceed by invoking the Brouwer fixed point theorem, noting that the map is continuous and maps the closed ball to the open ball by (2), and has a fixed point precisely when .

A third argument (avoiding the use of degree theory or the Brouwer fixed point theorem, but requiring one to replace with the slightly smaller ball ) is as follows: let minimise . From (2) and the hypothesis we see that lies in the interior . If the minimum is zero, then we have found a solution to as required; if not, then we have a stationary point of , which implies that is degenerate, a contradiction. (One can recover the full ball by tweaking the expression to be minimised in a suitable fashion; we leave this as an exercise for the interested reader.)

Corollary 4is an open map: the image of any open set is open.

*Proof:* It suffices to show that for every , the image of any open neighbourhood of is an open neighbourhood of . Proposition 3 handles the case ; the general case follows by renormalising.

Suppose we could show that is injective on . By Corollary 4, the inverse map is also continuous. Thus is a homeomorphism from to , which are both neighbourhoods of by Proposition 3; giving the claim.

It remains to establish injectivity. Suppose for sake of contradiction that this was not the case. Then there exists and such that

For every radius , the set

is closed and contains both and . Let denote the connected component of that contains . Since is non-decreasing in , is non-decreasing also.

Now let us study the behaviour of as ranges from to . The two extreme cases are easy to analyse:

*Proof:* Since is non-singular, we see from differentiability that for all sufficiently close to . Thus is an isolated point of , and the claim follows.

Lemma 6We have for all . In particular, is compact for all , and contains for .

*Proof:* Since , we see that ; since is connected and contains , we conclude that .

Next, if , then by (3) we have , and hence . Thus is disjoint from the sphere . Since lies in the interior of this sphere we thus have as required.

Next, we show that the increase continuously in :

Lemma 7If and , then for sufficiently close to , is contained in an -neighbourhood of .

*Proof:* By the finite intersection property, it suffices to show that . Suppose for contradiction that there is a point outside of that lies in for all . Then lies in for all , and hence lies in . As and lie in different connected components of the compact set (recall that is disjoint from ), there must be a partition of into two disjoint closed sets that separate from (for otherwise the only clopen sets in that contain would also contain , and their intersection would then be a connected subset of that contains both and , contradicting the fact that lies outside ). By normality, we may find open neighbourhoods of that are disjoint. For all on the boundary , one has for all . As is compact and is continuous, we thus have for all if is sufficiently close to . This makes clopen in , and so cannot lie in , giving the desired contradiction.

Observe that contains for , but does not contain for . By the monotonicity of the and least upper bound principle, there must therefore exist a critical such that contains for all , but does not contain for . From Lemma 7 we see that must also contain . In particular, by Lemma 5, .

We now analyse the critical set . By construction, this set is connected, compact, contains both and , contained in , and one has for all .

*Proof:* The openness is clear from the continuity of (and the local connectedness of ). Now we show disconnectedness. Being an open subset of , connectedness is equivalent to path connectedness, and and both lie in , so it suffices to show that and cannot be joined by a path in . But if such a path existed, then by compactness of and continuity of , one would have for some . This would imply that , contradicting the minimal nature of , and the claim follows.

*Proof:* Let be a connected component of ; then is non-empty and contained in . As is open, is also open, and thus by Corollary 4, is open also.

We claim that is in fact all of . Suppose this were not the case. As is connected, this would imply that is not closed in ; thus there is an element of which is adherent to , but does not lie in . Thus one may find a sequence in with converging to . By compactness of (which contains ), we may pass to a subsequence and assume that converges to a limit in ; then . By continuity, there is thus a ball centred at that is mapped to for some ; this implies that lies in and hence in (since ) and thence in (since is strictly less than ). As is adherent to and is connected, we conclude that lies in . In particular lies in and so lies in , a contradiction.

As is equal to , we thus see that contains an element of . However, each element of must be isolated since is non-singular. By compactness of , the set (and hence ) thus contains at most finitely many elements of , and so there are finitely many components as claimed.

*Proof:* If , then . If then and we are done, so we may assume . By differentiability, one has

for all sufficiently close to . If we choose to lie on a ray emenating from such that lies on a ray pointing towards from (this is possible as is non-singular), we conclude that for all sufficiently close to on this ray, . Thus all such points lie in ; since lies in and the ray is locally connected, we see that all such points in fact lie in and thence in . The claim follows.

Corollary 11There exists a point with (i.e. lies outside ) which is adherent to at least two connected components of .

*Proof:* Suppose this were not the case, then the closures of all the connected components of would be disjoint. (Note that an element of one connected component of cannot lie in the closure of another component.) By Lemma 10, these closures would form a partition of by closed sets. By Lemma 8, there are at least two such closed sets, each of which is non-empty; by Lemma 9, the number of such closed sets is finite. But this contradicts the connectedness of .

Next, we prove

Proposition 12Let be such that , and suppose that is adherent to a connected component of . Let be the vector such that(this vector exists and is non-zero since is non-singular). Then contains an open ray of the form for some .

This together with Corollary 11 gives the desired contradiction, since one cannot have two distinct components both contain a ray from in the direction .

*Proof:* As is differentiable at , we have

for all sufficiently small ; we rearrange this using (5) as

In particular, for all sufficiently small positive . This shows that all sufficiently small open rays lie in , hence in (since ), and hence in . In fact, the same argument shows that there is a cone

that will lie in if is small enough. As this cone is connected, it thus suffices to show that intersects this cone.

Let be a small radius to be chosen later. As is non-singular, we see if is small enough that whenever . By continuity, we may thus find such that whenever .

Consider the set

As is adherent to , is non-empty. By construction of , we see that we also have

and so is open. By Corollary 4, is then also non-empty and open. By construction, also lies in the set

We claim that is in fact all of . The proof will be a variant of the proof of Lemma 9. Suppose this were not the case. As is connected, this implies that there is an element of which is adherent to , but does not lie in . Thus one may find a sequence in with converging to . By compactness of (which contains ), we may pass to a subsequence and assume that converges to a limit in ; then . By continuity, there is thus a ball centred at contained in that is mapped to for some ; this implies that lies in and hence in (since ) and thence in (since is strictly less than ). As is adherent to and is connected, we conclude that lies in and thence in . In particular lies in and so lies in , a contradiction.

As , we may thus find a sequence converging to zero, and a sequence , such that

However, if is small enough, we have comparable to (cf. (2)), and so converges to . By Taylor expansion, we then have

and thus

for some matrix-valued error . Since is invertible, this implies that

In particular, lies in the cone (6) for large enough, and the claim follows.

## 26 comments

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12 September, 2011 at 4:31 pm

Leandro CiolettiThanks for sharing this. I read the mathoverflow question and I liked very much this result, but we don’t have in our local library a copy of the papers cited there.

12 September, 2011 at 4:55 pm

PhilippeSmall correction: the person that pointed this proof to you is Julien Melleray (not Malleray).

[Corrected, thanks – T.]12 September, 2011 at 5:54 pm

vineel567what tools do you use to write such a wonderful technical articles with all the special characters and mathematical style preserved.??? Thanks in advance

15 September, 2011 at 3:17 am

Willie WongTake a look at this page: http://terrytao.wordpress.com/about/ especially after the “Some technical remarks” fold.

14 September, 2011 at 7:45 pm

AnonymousSorry I mean everywhere invertibility. Is it necessary? If so why is the local invertibility not sufficient?

15 September, 2011 at 3:20 am

Willie WongThe statement does only require local invertibility, no? It only requires invertibility at every point in the open neighborhood . Or are you asking whether invertibility at a single point is sufficient?

15 September, 2011 at 5:21 am

AnonymousThis result seems counter-intuitive. In the 1D case, if f'(x)<0 for x0 for x>x_0 then it doesn’t seem possible for the function f to be invertible at x_0. Also, the derivative is invertible in a neighborhood of x_0 so it seems to satisfy the conditions of the theorem.

[You will need to use < and > instead of < and > to avoid your comments being mangled by the HTML parser. In any event, in one dimension the result is precisely the contrapositive of Rolle’s theorem. – T.]15 September, 2011 at 5:30 am

AnonymousIt should be f'(x)0 for x > x0.

17 September, 2011 at 7:30 pm

Josh SwansonThank you for typing this up! Reading through it was a nice review of basic real analysis and topology.

The Brouwer fixed point theorem is false with open balls, so the map you mention in the proof of Proposition 3 should be between closed balls. Even so, the fixed point isn’t on the boundary (from (3) scaled by the magnitude of x) so the result still follows easily.

In the same proposition, I wasn’t able to follow the reasoning in your third argument which gives |x| < r (i.e. |x| is not r) without taking |y| < r/3 rather than |y| < r/2. This doesn’t seem to affect anything adversely, though. I really like that argument since it makes the overall proof much more elementary.

I wasn’t able to follow the reasoning in the last few lines–in particular replacing o(||x_n – x_*||) with o(t_n). I was able to show the result anyway, though I had to pass to a further subsequence and use the invertibility of the derivative to do so.

Minor typos/issues:

“point set” -> “point-set”

“A third argument (…) as follows” -> “A third argument (…) is as follows”

“image of any open neighborhood” -> “the image of any open neighborhood”

In the proof of Lemma 6, strictly speaking the B(y, r) should be closed in each case

In Lemma 7, dividing delta by 2 doesn’t seem to serve a purpose

“then f(U_1) is non-empty contained in” -> “then f(U_1) is non-empty and contained in”

“conpactness” -> “compactness”

“This together with Corollary 11 with” -> “This together with Corollary 11″

In defining U’ (and in the restatement right after), there’s a missing vertical bar

[Corrected, thanks – T.]19 September, 2011 at 12:47 pm

AnonymousWhat about the following proof? Is it related to Cernavskii’s or Vaisala’s?

As above, let’s prove that is a local homeomorphism at a point . We may assume and for .

Due to Proposition 3 above, for each , there is a point in such that . Moreover, is compact. Take, in , the point with maximum first coordinate. If there is more than one such point, take the that with maximum second coordinate, and so forth. Let the chosen point equal .

The function is continuous, since

i) Because is open, for every point near there is a pre-image near .

ii) If there is a sequence such that, say, has first coordinate bigger than the first coordinate of with a non-vanishing difference, we can take a subsequence which converges to a point whose first coordinate is bigger than that of , a contradiction.

is obviously injective, and it is defined in a compact set. Thus, is a homeomorphism.

This function is also a left inverse of (that is, in the image of ). The claim then follows from the fact that injective continuous functions from -dimensional sets to -dimensional sets are open.

19 September, 2011 at 1:38 pm

Terence TaoI don’t think the proof of the continuity of g is complete, even in the one-dimensional case. For instance, what prevents from being significantly below in one dimension?

Note that there are plenty of continuous proper maps without continuous left-inverses, e.g. in one dimension. (Note in this case that the function g as defined above is discontinuous.)

19 September, 2011 at 4:09 pm

AnonymousErrata: above, the line “which converges to a point whose first coordinate is bigger than that of , a contradiction” is wrong. Instead of it, define and read “such that converges to a point whose first coordinate is bigger than that of , a contradiction”.

Dear Tao,

indeed that “proof” is wrong. I still think the case is ok:

Take any neighborhood of . If is sufficiently large, , so we have a candidate for in . We may suppose (e.g., by choosing as close to as possible).

In one dimension, by definition, is the biggest element of , so that its lim inf is at least . Then, as above, if does not go to , we achieve a contradiction.

However, in more dimensions, I have overlooked something: the coordinates of the ‘s don’t behave in the same way.

For the sake of clarity, given any , let’s say if and only if or one of the following conditions holds.

.

and .

, and .

Etc.

The failure of the proof is related to the instability of the relation above: if and are sequences such that , and , it is not necessarily true that .

I will try to fix the argument, even though I think success is not probable.

Question: if were continous, doesn’t the conclusion follow? For it is known that injective continuous maps (in this case, ) are local homeomorphisms. So would a homeomorphism between two neighborhoods of , and should be its inverse.

20 September, 2011 at 11:41 am

Twelth Linkfest[…] Tao: The inverse function theorem for everywhere differentiable maps, The Brunn-Minkowski inequality for nilpotent […]

21 September, 2011 at 9:50 am

AnonymousDear Prof. Tao,

in the proof of lemma 7 you ask the reader to “note that is closed”. Can you give me a hint on how to see this? Components are closed in general (which gives the compactness of ), but I do not see, why is also open (relative to ).

Thanks in advance.

22 September, 2011 at 2:12 am

Terence TaoOops, the argument is not quite correct as stated; I’ve rewritten it. The key point is that in a compact set, any two points that don’t lie in the same connected component can be separated from each other by clopen sets (but the connected components themselves need not be clopen).

8 October, 2011 at 10:43 am

Implicit function theorem | Mathitself[…] deep result is that it’s enough that is everywhere differentiable (see Tao’s post), which remains one an other deep result called invariance of […]

9 March, 2012 at 11:26 pm

francescodifuscoReblogged this on FRANCESCO DI FUSCO.

19 March, 2012 at 10:01 am

Alan MacdonaldThe paper “A Strong Inverse Function Theorem” by William J. Knight

(The American Mathematical Monthly, Vol. 95, No. 7, pp. 648-651) gives an improvement of the standard result in a different direction.

9 September, 2012 at 8:26 am

Olaf ZurthA Question: Is there in the first line of Lemma 6 a typo?

should read as or did I misunderstood something?

[Corrected, thanks – T.]8 November, 2012 at 5:35 am

blindmanDear Professor Tao,

the local inverses are also differentiable at

should be replaced by

the local inverses are also differentiable at

[Corrected, thanks – T.]8 November, 2012 at 6:10 am

blindmanDear Professor Terence Tao,

In the book “Mathematical Analysis on Manifold” of Michael Spivak, the author gave a counterexample of Theorem 1 (Exercise 2.39, page 52) if the assumption on the continuity of the derivate is violated.

They consider the function given by

+ if ;

+ if

I would to ask your comments about this situation.

Thank your for your helping.

8 November, 2012 at 6:19 am

Terence TaoThis function has non-zero derivative at 0, but has vanishing derivative at many other points (as can be seen for instance from a plot of the function

), and so does not contradict Theorem 1.

8 November, 2012 at 4:02 pm

blindmanDear Sir. Thank you for your comments and helping.

13 December, 2012 at 3:22 pm

peraAnd what would be counterexample?

28 March, 2014 at 12:11 pm

AlexanderWhy not just to use the Mean value theorem on a segment $[x_1,x_2]$ to obtain injectivity immidiately instead of all these lemmas?

29 March, 2014 at 3:08 am

AlexanderOk, I see: because there’s no such a theorem… sorry!