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the open ball of radius 1 is homeomorphic to all of R^n

oblixps

Member
May 20, 2012
38
So i know that if i define [tex] f(x) = \frac{x}{1 + |x|} [/tex] then letting the inverse be [tex] g(y) = \frac{y}{1 - |y|} [/tex] would work, but i'm having trouble with figuring out how to arrive at that inverse.

what i did was set [tex] y = \frac{x}{1 + |x|} [/tex] so that [tex] |y| = \frac{|x|}{1 + |x|} [/tex] and solving for |x| i get [tex] |x| = \frac{1}{1 - |y|} [/tex]. but from here i'm not sure about my logic. i argued that since y and x are basically vectors in [tex] \mathbb{R}^n [/tex] pointing in the same direction, given a vector y, i just need to scale it properly so that it becomes x. since i have the length of x is [tex] |x| = \frac{1}{1 - |y|} [/tex], and since |y| is not 1, we should take [tex] \frac{y}{|y|} [/tex] so that the vector is now of length 1 and pointing in the direction of x, and then multiply by the factor [tex] |x| = \frac{1}{1 - |y|} [/tex] so the inverse function would be [tex] g(y) = \frac{y}{|y|(1 - |y|)} [/tex]. but this doesn't work and i am having trouble figuring out why.

can someone help explain what i did wrong?
 

oblixps

Member
May 20, 2012
38
nevermind. i just realized that i was making the same algebra mistake over and over again. guess that's what fatigue does to you.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
While you certainly are not obligated to do so, would you mind showing what your mistake was and how correcting this led to the solution? This may benefit others who may do an online search and read your topic, and in turn this will help MHB. (Cool)
 

oblixps

Member
May 20, 2012
38
my careless mistake appeared when solving [tex] |y| = \frac{|x|}{1 + |x|} [/tex] for |x|. I thought it was a trivial step so i did it in my head and in the process, I accidentally dropped a factor of |y|. Correctly solving this equation should lead to [tex] |x| = \frac{|y|}{1 - |y|} [/tex] which suggests that we should define the inverse as [tex] g(y) = \frac{y}{1-|y|} [/tex].