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About 1-1 differentiable functions

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eof
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Hi,

I've been thinking about a problem in Spivak's Calculus on Manifolds and noticed that it can be proven quite cleanly if the following is true:

Let g:R^n->R^n be a differentiable 1-1 function. Then we can find a point s.t. det g'(x) != 0.

Geometrically this means that the best linear approximation for the function at at least one point is injective. However I can't seem to be able to lift this up to mean non-injectivity of the function itself if the linear approximation is always non-injective. I don't even know whether or not this is true.

Anyone know whether this is true or not and how to prove it or give a counter example?

Thanks.
 

Answers and Replies

HallsofIvy
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First, this is NOT an "if and only if" statement. It says exactly what it means. If g is differentiable and 1-1, then there exist a point such that det g'(x)= 0. It says nothing about what happens if there is a point such that det g'(x)= 0. It should be obvious that the converse is not true: Take g: R1->R1 with g(x)= x2. That is definitely NOT a one-to-one statement.
 
eof
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I think you misunderstood my question. My question was really that _if_ we know that a function is one to one and differentiable, does this mean that det g'(x) != 0 at some point? The other way around is of course trivial.

Thus a counterexample would be to find a 1-1 differentiable function g:R^n->R^n s.t. det g'(x) = 0 at every point. This is of course trivially impossible in R^1 (mean value theorem), but I don't know about R^n, because it really complicates matters.

It would of course seem somewhat logical that a function whose linear approximation at every point is non-injective would be non-injective, but I'm not going to bet on my intuition.

If this would be true, then assuming that the function is continuously differentiable would guarantee that it has a differentiable inverse locally at some point (i.e. we could find a point where we could apply the inverse function theorem). This is actually what would make this interesting... I haven't seen any book address this, so it seems non-trivial.
 
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Dick
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det(g'(x)) is the Jacobian of g. If the image of a domain D with nonzero volume has nonzero volume then the Jacobian is definitely nonzero lots of places. I would think 1-1 would guarantee that. After all, g is continuous.
 
eof
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You're right. If a function is 1-1 and continuous, we can just take an open nonempty set (which means that the volume is nonzero) and it's preimage would be an open set which is nonempty and thus has nonzero volume. So we would map a set with nonzero volume to a set with nonzero volume.

Do you know how to prove this statement about the Jacobian?
 
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eof
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Ok, I found the proof. This statement follows from Sard's theorem.
 
Dick
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You could think of how g changes a volume at a point x. Dg maps the unit tangents e1...en to Dg(e1)...Dg(en). This is sort of an extension of finding the area of a parallelogram using the cross product. The image 'volume' is |det(Dg)|.
 
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