When can we directly invert the function across the inequality sign? I am new to this site.
I would like to ask when can we invert the function in the inequality?
Question:
Let X be a continuous random variable with pdf . Suppose () is strictly monotonic, differentiable function of x. The random variable =() has the PDF.
$$
{f_Y} (y)=
\begin{cases}
{f_X}[g^{-1}(y)]  \left|\dfrac{d}{dy}g^{-1}(y)\right|, & \text{if y = g(x) for some x}\\[2ex]
0, &\text{if y $\ne$g(x) for all x}
\end{cases}
$$
Proof. Suppose that y=g(x) for some x. Then, with Y=g(X),
\begin{aligned}
F_Y(y) & =P\{g(X)\le y\}\\ 
& = P\{X\le g^{-1}(y)\}\\
&=F_X(g^{-1}(y))
\end{aligned}
I am not sure why function $g(x)$ can be converted to $g^{-1}{(x)}$ over the inequality sign. Is it because $g(x)$ is strictly monotonic so it definitely has the inverse function? 
Many thanks for your help.
Question and proof
 A: What we need is that $g$ is a strictly increasing function.
This implies that for any numbers $x_1$ and $x_2$ in the domain of $g,$
we will have $g(x_1) \leq g(x_2)$ if and only if $x_1 \leq x_2.$
Now put $X$ for $x_1$ and $g^{-1}(y)$ for $x_2.$
So we have $g(X) \leq g(g^{-1}(y))$ if and only if $X \leq g^{-1}(y).$
That is, the event that $g(X) \leq y$
is exactly the same event as the event that $X \leq g^{-1}(y)$
(you cannot have an outcome that satisfies one of those inequalities and not the other), and it has the same probability.
If  $g$ were strictly monotonic but decreasing rather than increasing,
then we would have $g(x_1) \leq g(x_2)$ if and only if $x_1 \geq x_2,$
and when making the substitutions $X$ for $x_1$ and $g^{-1}(y)$ for $x_2$
we end up with
$g(X) \leq y$ if and only if $X \geq g^{-1}(y),$
which is not what we need for the theorem.
According to the way I learned it (something like this),
monotonic does not imply increasing.
But perhaps the author of the theorem believed it does;
or perhaps they had implied earlier that they would use the term "monotonic" to describe only monotonically increasing functions, never decreasing functions.
