Is it true that $F_U(U(\bar\omega)) = F_X(z)$ implies $z=X(\bar\omega)$? For a statistics class, I have to prove a result which leads me to the following question. If I can show that it is true, my proof is done. So here is the question.
Suppose that $U: \Omega \rightarrow [0,1]$ is a random variable uniformly distributed on $[0, 1]$ and $X: \Omega \rightarrow \mathbb{R}$ is continuously distributed as some distribution $F_X(·)$ (where $\Omega$ is the sample space). Is it true that  

$F_U(U(\bar{\omega})) = F_X(z_{\bar{\omega}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

or equivalently, 

$P \{\omega \in \Omega~|~ U(\omega) \leq U(\bar{\omega})\} = P \{\omega \in \Omega~|~ X(\omega) \leq z_{\bar{\omega}}\}$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

? I suspect that it is true, but I haven't been able to prove or to disprove it.
 A: For any random variable $Y$ with the distribution function $F_Y$ the random variable $F_Y(Y)$ has a uniform distribution in the interval $[0,1]$. Hence your first statement should be wrong, if we ignore the fact that it is not mathematically correct. 
You equate two different mathematical objects. $F_U(U(\omega))$ is a random variable, $F_X(z)$ is a real function. They cannot be compared without giving precise definition of what you mean by $=$ in this case.
A: Following mpiktas answer, things became clearer to me and I think I can know answer the question. The first statement

$F_U(U(\bar{\omega})) = F_X(z_{\bar{w}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

Is in fact equivalent to (and better understood as)

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega)=X({\omega})$, where $\sim$ means "distributed as".

Now, this turns out not to be quite true (see mpiktas comment below). What is true however is 

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega) =^d X({\omega})$, where $=^d$ means that the two random variables have the same distribution.

As noticed by mpiktas, $F_U(U({\omega})) \sim U(0,1)$ ( a proof can be found here). Hence we must have $F_X(Z(\omega)) \sim U(0,1)$. But then we must have $Z(\omega) =^d X(\omega)$. All that is needed for the proof of this last fact can also be found here. But in the link, the statement is reversed ($F_X(X(\omega)) \sim U(0,1)$ instead of $F_X(Z(\omega)) \sim U(0,1) \Rightarrow Z(\omega) =^d X(\omega)$ ) so I figured it might be helpful to reformulate it here.


*

*Assume $F_X(Z(\omega)) \sim U(0,1)$.

*Then $F_{F_X(Z(\omega))}(a) = a$

*Notice that \begin{align} a &= F_X(F^{-1}_X(a))\\
                               &= P(X\leq F^{-1}_X(a))\\ 
                               &= P(F_X(X) \leq a)\\
                               &= F_{ F_X(X)} (a) \end{align} where I omitted the $\omega$'s for notational convenience.

*So we have $F_{F_X(Z(\omega))}(a) = a = F_{ F_X(X(\omega))} (a)$, which in turn implies $Z(\omega) =^d X(\omega)$. 

