If $X\sim \textbf{Laplace}(0,\lambda)$. Prove that $|X|\sim \textbf{Exp}(\lambda)$ How can I prove this statement?

If $X\sim \textbf{Laplace}(0,\lambda^{-1})$. Prove that $|X|\sim \textbf{Exp}(\lambda)$

My approach: Since that $X\sim \textbf{Laplace}(0,\lambda)$, so the density function is $$f_{X}(x)=\frac{1}{2\lambda^{-1}}e^{-\frac{|x-0|}{\lambda^{-1}}}, x\in \mathbb{R}, \lambda>0$$
Now, let $Y=|X|$, so we have that $Y$ is $\nearrow$ in $[0,\infty)$ and $y=x$, $x\in [0,\infty)$.
So, by transformation theorem and taking $h^{-1}(y)=y$, we have that $$f_{Y}(y)=\frac{1}{2\lambda^{-1}}e^{-\frac{|y|}{\lambda^{-1}}},\quad x\geq 0 $$
and similar, we have that $$f_{Y}(y)=\frac{1}{2\lambda^{-1}}e^{-\frac{|-y|}{\lambda^{-1}}},\quad x<0$$
So, we have that $$f_{|X|}(x)=\frac{1}{2\lambda^{-1}}e^{-\frac{|x|}{\lambda^{-1}}}, x\in \mathbb{R}, \lambda>0$$

But, I think that I can't conclude with that information that $|X|\sim \textbf{Exp}(\lambda)$.
 A: The error in your approach is that $h^{-1}$ is not well-defined: for every input $x$, there are two values $y$ (they are $\pm x$) such that $h(y) = x$. If I remember correctly, whether or not $|X|$ is increasing is not really relevant; instead, what is relevant is whether the transformation $h^{-1}$ is increasing (because this is a sufficient condition for invertibility).
Think of $f_{X} (x) dx = P(X \in [x - dx / 2, x + dx / 2])$ for some infinitesimal $dx$. Then the event $|X| \in [x - dx / 2, x + dx / 2]$ corresponds to $X \in [-x - dx / 2, -x + dx / 2]$ or $X \in [x - dx / 2, x + dx / 2]$, which we know have probabilities $f_X(-x)dx$ and $f_X(x)dx$, respectively. Thus,
$$f_{|X|} (x) = \begin{cases}
2 f_X(x) & x \geq 0 \\
0 & x < 0
\end{cases}$$
You should notice that this is exactly the density function of $\text{Expon}(\lambda)$.
A: The PDF of $Y,f_Y(y)$ has to depend on $y$, not $x$. Since $Y=|X|$ is not a bijection, you run into an issue.
The distribution function of $Y$ is$$F_Y(y)=P(Y\le y)=P(|X|\le y)=\begin{cases}0,&y\le0\\P(-y\le X\le y),&y>0\end{cases}$$The PDF of $Y$ is $0$ for $y<0$. Using $P(-y\le X\le y)=F_X(y)-F_X(-y)$ where $F_X(x)$ is the distribution function of $X$, we get for $y>0$,$$f_Y(y)=f_X(y)+f_x(-y)=2f_x(y)$$which is the PDF of exponential distribution.
