How do you show $F$ has a well-defined inverse function $G : (0,1) \to \mathbb{R}$, which is strictly increasing [closed]

If $X$ be a random variable with cdf $F$, and $F$ is continuous and strictly increasing on $\mathbb{R}$ how do you show $F$ has a well-defined inverse function $G : (0,1) \to \mathbb{R}$, which is strictly increasing.

closed as off-topic by jameselmore, Davide Giraudo, Harish Chandra Rajpoot, Em., Yagna PatelJan 26 '16 at 20:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – jameselmore, Davide Giraudo, Harish Chandra Rajpoot, Em., Yagna Patel
If this question can be reworded to fit the rules in the help center, please edit the question.

To put you on the right track: for $u\in(0,1)$ "define" $G(u):=\inf\{x\in\mathbb R\mid F(x)\geq u\}$.
• Is this indeed a well-defined function $\mathbb (0,1)\to\mathbb R$?
• Does this function serve as an inverse of $F$?
Alternatively you can show that for every $u\in(0,1)$ there is a unique $x_u\in\mathbb R$ such that $F(x_u)=u$. In order to prove that $F(x)=u$ for some $x\in\mathbb R$ make use of the continuity of $F$ and apply the intermediate value theorem. Its uniqueness follows from the fact that $F$ is strictly increasing. Now define $G$ by prescription $u\mapsto x_u$.