Show that a continuously differentiable function is not 1-1

You may want to think about what the derivative of the function ##g(x,y)=f(x,y)-y## is (at a point where ##f_y(x_0,y_0)\ne 0##).Another alternative is to think about how ##f## maps the ##x-## and ##y-## axes. What happens around the point ##(0,0)##?Are you asking about the values of ##\lim_{(x,0)\rightarrow (0,0)} f(x,0)## and ##\lim_{(0,y)\rightarrow (0,0)}...##?No, I'm asking about the limit of the function itself. If you consider the unit circle ##
  • #36
i think you are assuming that "homeo" means homeo to an open set, but all you know is homeo to some set with the induced topology. i.e. your homeo only maps an open set to the intersection of an open set in R^2 with the image set. that is what an open set of the image means. this is very unintuitive i agree, but i think it is unavoidable.

in fact it is not true that a homeo maps interior to interior. take a homeo from an open disc in the plane to an open disc in R^3, which has no interior. it is still a homeo onto its image but the image has empty interior in R^3. you have to prove that cannot happen when the image is in R^2. it is not at all obvious. well it depends what you mean by obvious. it seems obvious but it is not easy to prove.
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  • #37
note in the vector field argument above how sneaky, i.e. brilliant, it was of Borsuk Ulam to assume that no two antipodal points mapped to the same image, rather than just that no two points at all did so. That gave us the ability to construct a map from the unit circle to itself with opposite values on opposite points, which forces the winding number to be non zero. I.e. when you go half way around, the image point must be half way around from the original image point, hence the total angle change has form 2πn + π. And when you go the rest of the way around you must repeat the same values so you must complete the tour around the circle, giving an odd winding number, i.e. angle change of form 4πn + 2π = (2n+1).2π, hence non zero winding number 2n+1. I know this is vague but it seems clear to my visual mind that this trick allows you to actually compute a non zero winding number, something I lamented not being able to do with merely a hypothesis of continuity. Thus it was actually easier to prove the more precise statement that some pair of antipodal points mapped to the same image than to prove that some arbitrary two points did so. this is cleverness personified.

A quick perusal of one of my most sophisticated algebraic topology books, e.g. the lectures of Albrecht Dold, reveal that the theorem that no continuous map from an n sphere to euclidean n space can be injective, is rather advanced, using the full power of homology and neighborhood retracts, etc...occurring on about page 78 of his somewhat condensed and elegant book, in the section on jordan curve theorem and invariance of domain.

by the way, in the differential geometry and differential topology books i have, the discussion of smooth jordan curve theorems seem rather weak, since they only prove that a submanifold of R^n+1 which is diffeomorphic to S^n separates the ambient space into two pieces, each open, one bounded and one not. The weakness is in the assumption that the space is a submanifold, rather than just assuming we have the image of a smooth injective map from S^n into R^n+1. I.e. the hard part seems to be to prove that the image is indeed a submanifold, which these books just assume. Or maybe I have skipped some earlier preliminaries where they prove this but i did not notice it. And these are fine books, by authors far more expert than me, so I could well be wrong, indeed I hope so.

but i think we are learning from this discussion that assumptions on the rank of a smooth map are quite crucial in making arguments on the structure of the image easier, since they are needed for applying the implicit function theorem. i.e. without assuming maximal rank, it is hard to draw conclusions. i suspect that's why the theory of singularities of maps is so difficult.

Yes, I think now one gets very little benefit from assuming just smoothness, and to get a real bump one needs to assume smooth of maximal rank. Maybe I should revise my estimate of the cleverness or usefulness of the OP's original problem, since his prof has found a way to get mileage just out of, no wait a minute, for a map from R^2 to R^1, rank 1 is maximal. So again we have a maximal rank smooth map, which then is goosed up to a rank 2 map from R^2 to R^2 to use the inverse function theorem.

So smoothness does not help much on understanding the structure of the image of a smooth injective map from a circle to the plane unless we also assume the map is an immersion, i.e. rank 1 everywhere. Then one can prove the image curve is a submanifold, hence each point has a nice local "collar", so one can understand the two "sides" of the curve, and what iot means to cross the curve transversally, etc...

well this has helped me a lot to understand why authors make the hypotheses they do. I recall when I taught calc of one variable, I always said (exaggerating somewhat) that "when the derivative is zero, i.e. not of maximal rank, it tells you zero". I.e. you cannot conclude you have a max or min or anything at all really at that point. (Zero derivative at one point tells you little that is, since of course one important case where you do get something useful is when the derivative is zero everywhere.) The same seems to hold in diff top, you need to know your space is a manifold, or your subspace is a submanifold. After all you want to use smoothness to approximate your situation well by a linear situation, and for that you need maximal rank, so that the linear approximating object has the same dimension as the manifold.

If anyone figures out how to use these elementary, i.e. winding number arguments, to prove that an injective continuous map from an open disc to the plane is an open map, i.e. that every injective contin. map R^2-->R^2 is a homeo onto not just some subset, but onto an open subset of R^2, I think that would be interesting. By way of encouragement, this is presumably somewhat easier than the jordan curve theorem, since we are assuming given an injective continuous map on the whole disc. The Jordan curve theorem and Schoenflies theorem together say, if I recall correctly, that every continuous injection from the circle to the plane extends to a continuous injection of the disc to the plane, and the extension is a homeomorphism from the open disc to an open subset of the plane bounded by the image of the circle; but we are giving ourselves this injective extension.

For references, there is actually a presentation, in the wonderfully impressive analysis book by Dieudonne'. Foundations of Modern Analysis, of the Jordan curve theorem, but I think not the Schoenflies theorem, using as a tool only the winding number concept and some complex analysis, in an appendix to his chapter IX, which treats elementary complex analysis. I have never read it, but always meant to, since valid elementary proofs of the Jordan theorem are famously hard to find. I sat through one in grad school many years ago but learned essentially nothing fom it, except that I did not want to specialize in that sort of topology. It may have been taken from the classic book of Newman.
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  • #38
By the way, in reference to the attempted homology arguments above, I cannot find a reference that shows all subsets of R^2 have zero degree 2 homology. It is true however for neighborhood retracts, and the property of being a neighborhood retract is a homeomorphism invariant. Thus since an n sphere is a neighborhood retract in R^(n+1), then also any subspace of any euclidean space that is homeomorphic to a sphere is also a neighborhood retract. It is also possible to give a direct proof using the Tietze extension lemma, for the case of a subspace homeomorphic to a sphere. Thus indeed an injective continuous map from R^3 to R^2 would induce a homeomorphism from the 2 sphere to some neighborhood retract in R^2. This is then a contradiction since the 2 sphere has non zero 2nd homology while the neighborhood retract in R^2 does not.

These arguments depend on rather sophisticated and lengthy arguments at least in the book of Dold, using local homology, retracts, and numerous exact sequences. The vanishing result for retracts has something like 6 steps, spread over some 4 1/2 pages. And this is not a wordy book. This is far more sophisticated than the winding number argument above that no continuous map from the 2 sphere to the plane can be injective, via Borsuk Ulam. Of course you do get more from the homology version, since homology let's you count connected components. Thus Dold also concludes that the complement of a subset of the plane homeomorphic to a disc is connected. and the complement of a subset homeo to a circle has two components. Thus given an injective continuous map of a disc to the plane, the interior of the disc must map onto the unique bounded connected component of the complement of the image of the boundary circle, which must be open. We can then conclude that any injective continuous map R^2 to R^2 is a homeo onto an open subset. But this uses a lot of machinery.
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  • #39
I just reviewed the presentation in the little book of Marvin Greenberg, who benefits from borrowing heavily from the book of Emil Artin, and the argument for these results looks much easier. Apparently that is because he argues only for the case at hand (sets homeo to a sphere or a ball) rather than proving vanishing of homology for all neighborhood retracts as Dold does. So Dold's argument is more difficult and longer but yields a much more general result.

Indeed that special case seems to be covered by just one of Dold's 6 steps, (step 2), which he deals with even more briefly than does Greenberg.
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  • #40
The question of when an injective map is an open embedding is quite interesting and has different answers in different settings. We have been exploring the theorem that an injective continuous map of n dimensional topological manifolds is indeed a homeomorphism onto an open set, in particular the inverse function not only exists but is guaranteed to be continuous.

A more elementary example is the theorem that a linear injection of vector spaces of the same dimension is an isomorphism, also surjective and the inverse function is linear.

On the category of smooth real manifolds, the result is false, since we have the smooth bijective map of R to R taking t to t^3, which is infinitely differentiable and bijective but whose inverse function is not differentiable at t=0.

If we pass to the realm of complex holomorphic manifolds we have a different result. A smooth map of complex manifolds, whose derivatives are everywhere complex linear, i.e. a holomorphic map, is holomorphically invertible if and only if it is bijective.

The same result applies to algebraic varieties over the complex numbers, or any algebraically closed field of characteristic zero; namely an injective regular map of non singular algebraic varieties is an open immersion.

The result is not true for singular varieties however, since the map taking t to (t^2,t^3) is a bijection from C to the curve y^2=x^3 which is not invertible; i.e. the inverse map taking (t^2,t^3) = (x,y) to t = y/x, is not regular at (0,0).

So in general we need both a non singular (or at least "normal") target space and an algebraically closed field of definition, but in the case of continuous maps over R we are good. Notice that similar results are much easier in algebra, e.g. bijective homomorphisms are also isomorphisms for groups and rings. This is also true for Banach spaces but there the requirement the inverse linear map be continuous makes the proof non trivial in infinite dimensions where linear does not automatically imply continuous.
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