Why Can't the Inverse Function Theorem Be Applied to a Function from R^n to R^m?

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yifli
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I know it is impossible to apply inverse function theorem on a function from R^n to R^m because the Jacobian is not a square matrix.
Is there any other reason why this is impossible?
 
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Hi yifli! :smile:

If the inverse function theorem were applicable then we would obtain a homeomorphism between an open set of [itex]\mathbb{R}^n[/itex] and an open set of [itex]\mathbb{R}^m[/tex]. But it's a result from topology that such an open set can not be homeomorphic.<br /> <br /> For example if n=1 and m=2, then there would exist a homeomorphism between an interval ]a,b[ and an open disk, but such an homeomorphism would never exist.<br /> <br /> Does this answer your question?[/itex]
 
micromass said:
Hi yifli! :smile:

If the inverse function theorem were applicable then we would obtain a homeomorphism between an open set of [itex]\mathbb{R}^n[/itex] and an open set of [itex]\mathbb{R}^m[/tex]. But it's a result from topology that such an open set can not be homeomorphic.<br /> <br /> For example if n=1 and m=2, then there would exist a homeomorphism between an interval ]a,b[ and an open disk, but such an homeomorphism would never exist.<br /> <br /> Does this answer your question?[/itex]
[itex] <br /> Yes. thank you so much![/itex]
 
micromass said:
Hi yifli! :smile:

If the inverse function theorem were applicable then we would obtain a homeomorphism between an open set of [itex]\mathbb{R}^n[/itex] and an open set of [itex]\mathbb{R}^m[/itex]. But it's a result from topology that such an open set can not be homeomorphic.

Actually I have another question regarding your answer:
A homeomorphism f is a bijective mapping which is continuous and has a continuous inverse. Inverse function theorem guarantees that f is continuous and so is its inverse, but what about bijectivity?
 
yifli said:
Actually I have another question regarding your answer:
A homeomorphism f is a bijective mapping which is continuous and has a continuous inverse. Inverse function theorem guarantees that f is continuous and so is its inverse, but what about bijectivity?

If it has an inverse, it is automatically bijective. In fact, this is an equivalence.
 
yifli said:
I know it is impossible to apply inverse function theorem on a function from R^n to R^m because the Jacobian is not a square matrix.
Is there any other reason why this is impossible?

If you don't want to use the chain rule then try to show that the map can not be 1 to 1 from the higher dimensional space to the lower dimensional space.