# Jacobian and diffeomorphism

What is the relationship between being globally diffeomorphic and the Jacobian of the diffeomorphism?

All I can think of is that if the Jacobian at a point is non-zero, then the map is bijective around that point. For example, if:

$$f(x)=x_0+J(x_0)(x-x_0)$$

where J(x0) is the Jacobian matrix at the point x0, x is the coordinate on one manifold, f(x) is the mapped coordinate to the other manifold, then this can easily be inverted for a non-zero Jacobian:

$$J^{-1}(x_0)[f(x)-x_0]+x_0=x$$

So it seems that having a non-vanishing Jacobian only proves that f is a bijection and that f is C1, but does not prove that f is infinitely differentiable.

## Answers and Replies

Hurkyl
Staff Emeritus
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You meant "non-singular", not "non-zero".

For a differentiable function, the following are equivalent:
• The Jacobian is non-singular at a point
• The function is invertible in a neighborhood of that point

As you noted, the Jacobian tells you nothing about higher-order derivatives.

Also, if the Jacobian is everywhere non-singular, that only tells you the function is a local homeomorphism -- that's not enough to prove it is a homeomorphism.

(e.g. consider the obvious smooth function from the line to the circle)

Also, if the Jacobian is everywhere non-singular, that only tells you the function is a local homeomorphism -- that's not enough to prove it is a homeomorphism.

(e.g. consider the obvious smooth function from the line to the circle)

Thanks.

According to Wikipedia, if U and V are simply connected open subsets of R^n, and f: U->V is smooth, then having the Jacobian non-singular is enough for U and V to be diffeomorphic.

So I guess that's why your function from the line to the circle fails, as the circle is not simply connected.

You wouldn't happen to know if Wikipedia's statement is still true if U and V are still simply connected open subsets, but not of R^n?

Hurkyl
Staff Emeritus
Gold Member
For local homeomorphism to imply injective, I think it is enough that the target space is simply-connected.

The statement you cited is false -- e.g. the inclusion of (0,1) --> R.

One statement I expect to be true (but don't have complete confidence in) is that if U and V are smooth manifolds, V is simply connected, f:U-->V is smooth and surjective, and f' is everywhere nonsingular, then f is a diffeomorphism.

One statement I expect to be true (but don't have complete confidence in) is that if U and V are smooth manifolds, V is simply connected, f:U-->V is smooth and surjective, and f' is everywhere nonsingular, then f is a diffeomorphism.

I think you need to replace 'simply connected' with 'convex'. If V were just simply connected I'm thinking, very hand-wavily, a counter example would be f mapping a "straight piece of dough" into a "doughnut" shape in R^3 where two ends of the doughnut "overlap", preventing 1-1'ness required for the diffeo. (The same idea as mapping the interval into the sphere, but with the extra dimension mere simply-connectedness of the image space is not enough).

Hurkyl
Staff Emeritus
Gold Member
If I understand your map, it either fails to be surjective, or the target space fails to be simply connected.

If I understand your map, it either fails to be surjective, or the target space fails to be simply connected.

Oops, yes of course a "doughnut" in R^3 isn't simply connected, my mistake!

But I'll try again: I think you could map a convex open set in R^3 onto an open set in R^3 homeomorphic to an annulus in such a way that the mapping satisfies your criteria and is not 1-1. E.g., take a blanket and fold the 4 corners over one another so looks like a "ball". Where the corners are "folded over", you could let the function not be 1-1.

Maybe I'll give a more exact description later (or think it through and find a fault). Part of the reason I think convexity of the co-domain is required is I vaguely recall having had a similar question on an exam, and convexity was left out and someone found a counterexample to show the question was ill-posed (asking us to "prove" a false statement).

Hurkyl
Staff Emeritus
Gold Member
Well, I think I have a counter-example anyways, with the source and target both rectangles. I can imagine the image of the rectangle being swept out by a paintbrush -- and the map should be smooth as long as I move the paintbrush smoothly. I should be able to paint a rectangle with smooth strokes that overlap.

The variation I was missing is the following, I think: -- I don't just want f to be locally homeomorphic: I need, every point in the target to have an open neighborhood U whose inverse image is a disjoint union of spaces homeomorphic to U. (With f being the homeomorphism)

This eliminates my paintbrush example; a small disk around a point on the boundary of a stroke would, in its set of inverse images, a half-disk. The idea that was motivating my conjecture was that of a covering map. It turns out I also forgot to make the obvious requirement that the source is connected.

lavinia
Gold Member
Also, if the Jacobian is everywhere non-singular, that only tells you the function is a local homeomorphism

Actually if the function is continuously differentiable then it is a local diffeomorphism in a neighborhood of a non-singular point.

lavinia
Gold Member
One statement I expect to be true (but don't have complete confidence in) is that if U and V are smooth manifolds, V is simply connected, f:U-->V is smooth and surjective, and f' is everywhere nonsingular, then f is a diffeomorphism.

If the manifolds are compact without boundary then this works. If the Jacobian is everywhere non-singular then the map must be a covering. This is not hard to prove. If the target manifold is simply connected then its only covering space is itself.

Also it seem to me that you don't need to assume surjective.
This is because the number of points in the inverse image of any point is locally constant.

Not sure about the non-compact case.

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Hurkyl
Staff Emeritus
Gold Member
I was assuming surjective, because, in general, the number of preimages doesn't have to be locally constant for a local homeomorphism -- but I could believe that compactness solves that problem. (I had been worried about things like the inclusion map from an open subset into a manifold)

lavinia
Gold Member
I was assuming surjective, because, in general, the number of preimages doesn't have to be locally constant for a local homeomorphism -- but I could believe that compactness solves that problem. (I had been worried about things like the inclusion map from an open subset into a manifold)

Around any point in the preimage of a regular value of the function, there is an open neighborhhood that the function maps diffeomorphically into the target manifold. This means that if the domain manifold is compact that there are only a finite number of points in the preimage of a regular value. Since there are only finitely many neighborhoods mapped diffeomorphically onto neighborhoods of the regular value, their images intersect in an open neighborhood. The number of points in the preimage of any point in this intersection is constant.

In the case where all points are regular values and the target manifold is path connected (which is part of being simply connected) any two points can be connected along a path by a chain of open neighborhoods. Each neighborhood can be chosen so that its preimage is a finite collection of open neighborhoods that are mapped diffeomorphically onto it.

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lavinia
Gold Member
I was assuming surjective, because, in general, the number of preimages doesn't have to be locally constant for a local homeomorphism -- but I could believe that compactness solves that problem. (I had been worried about things like the inclusion map from an open subset into a manifold)

Here is an example of how this reasoning is applied.

If one composes stereographic projection of the 2 sphere onto the complex plane with a complex polynomial then follows the polynomial with the inverse of stereographic projection, one obtains a smooth map of the 2 sphere into itself that keeps the north pole fixed.

Since a complex polynomial has finitely many zeros, every point on the sphere except finitely many is a regular value. One can always draw a path on the sphere that avoids the critical values so the number of points in the inverse image of any regular value is constant - and therefore non-empty. At a critical value there is a non-empty set of critical points in its preimage. Therefore the map is surjective. But since the north pole is mapped to itself, the complex polynomial must also be surjective. In particular it must have a zero. This is Milnor's proof of the Fundamental Theorem of Algebra.