Inverse Transform Method for a pdf Given a pdf:
$$f(x)=\tau x \exp\left(\frac{-\tau x^2}{2}\right); \quad x, \tau > 0$$
So I found the corresponsing cdf:
$$F(x)=1 - \exp\left(\frac{-\tau x^2}{2}\right)$$
Then I got given a value for tau: $\tau=0.2$ and I derived the inverse function $F^{-1}_X(u).$:
$$F^{-1}_X(u)=\sqrt{\color{red}{\frac{2}{\tau}}10 \log\left(\frac{1}{1-u}\right)}; \quad u\sim U[0,1]$$
Now from what I have been told, $F^{-1} \sim F$ but I just can't see it. I may be drawing wrong conclusions but $$0< F<1$$
and $$-\infty<F^{-1}<\infty$$
Is what I did correct?
EDIT: My lecture notes:


The statement is

If $u \sim U[0,1]$ then $F_X^{-1}(u)\sim f$

To show you this I sample a uniform distribution and calculate 
$$
x = F^{-1}_X(u) = \sqrt{-\frac{2}{\tau}\ln(1 - u)}
$$
Then I make a histogram and overplot the distribution
$$
f_X(x) = \tau x e^{-\tau x^2/2}
$$

