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

#### oblixps

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

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

can someone help explain what i did wrong?

#### oblixps

##### Member
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

my careless mistake appeared when solving $$|y| = \frac{|x|}{1 + |x|}$$ 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 $$|x| = \frac{|y|}{1 - |y|}$$ which suggests that we should define the inverse as $$g(y) = \frac{y}{1-|y|}$$.