Finding inverse of F in Munkres' Topology Ch.2 EX 5 pg 106

[Math Amateur]

In Munkres book "Topology" (Second Edition), Munkres proves that a function F is a homeomorphism ...

I need help in determining how to find the inverse of \(\displaystyle F\) ... so that I feel I have a full understanding of all aspects of the example ...

Example 5 reads as follows:View attachment 4193Wishing to understand all aspects of the problem I tried to see how given

\(\displaystyle F(x) = \frac{x}{1 - x^2} \)

one could determine the inverse of \(\displaystyle F\) (and then come up with \(\displaystyle G\), as Munkres did ... ... somehow ??) ... ...I think I proceed by putting

\(\displaystyle y = \frac{x}{1 - x^2}\)

and solving for \(\displaystyle x\) ... ... BUT how exactly do I proceed (I got nowhere with this problem!)Can someone please help?NOTE... I realize that being able to determine the inverse of F and show it is G is not strictly necessary in showing that F is a homeomorphism ... BUT ... I feel very dissatisfied that I cannot see exactly how this works ... so, again, I hope someone can help ...

Peter

 

[Opalg]

I tried to see how given

\(\displaystyle F(x) = \frac{x}{1 - x^2} \)

one could determine the inverse of \(\displaystyle F\) (and then come up with \(\displaystyle G\), as Munkres did ... ... somehow ??) ... ...I think I proceed by putting

\(\displaystyle y = \frac{x}{1 - x^2}\)

and solving for \(\displaystyle x\)

If \(\displaystyle y = \frac{x}{1 - x^2}\) then $yx^2 + x - y = 0$. Solving that quadratic equation for $x$, you get $x = \dfrac{-1 \pm\sqrt{1+4y^2}}{2y}$. You want the solution for which $x\in (-1,1)$, which means that you need the $+$ sign for the square root: $x = \dfrac{-1 + \sqrt{1+4y^2}}{2y}$. That looks different from the formula that Munkres gives for $G(y)$, but if you rationalise the denominator of his formula (multiplying top and bottom of his fraction by $-1 + \sqrt{1+4y^2}$) then you see that the two solutions are equivalent.

 

[Math Amateur]

If \(\displaystyle y = \frac{x}{1 - x^2}\) then $yx^2 + x - y = 0$. Solving that quadratic equation for $x$, you get $x = \dfrac{-1 \pm\sqrt{1+4y^2}}{2y}$. You want the solution for which $x\in (-1,1)$, which means that you need the $+$ sign for the square root: $x = \dfrac{-1 + \sqrt{1+4y^2}}{2y}$. That looks different from the formula that Munkres gives for $G(y)$, but if you rationalise the denominator of his formula (multiplying top and bottom of his fraction by $-1 + \sqrt{1+4y^2}$) then you see that the two solutions are equivalent.

Thanks Opalg ... quite straightforward really... Another case of "easy when you know how" ...

Thanks again for your support and help with this problem ...

Peter