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
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 ; thus maps to .
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 4 is 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.
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 6 We 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 :
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.
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
(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
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
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.