Evaluate absolute expression in a probability density function

Question

From my understanding:

$\frac{1}{\sqrt{2\pi}}$ $e^{{-(-\sqrt{y})^2/2}}$ |-$\frac{1}{2\sqrt{y}}$| = $\frac{1}{\sqrt{2\pi}}$ $e^{y / 2}$ $\frac{1}{2\sqrt{y}}$

$\frac{1}{\sqrt{2\pi}}$ $e^{{-(\sqrt{y})^2/2}}$ |$\frac{1}{2\sqrt{y}}$| = $\frac{1}{\sqrt{2\pi}}$ $e^{-y / 2}$ $\frac{1}{2\sqrt{y}}$

If so,

$\frac{1}{\sqrt{2\pi}}$ $e^{{-(-\sqrt{y})^2/2}}$ |-$\frac{1}{2\sqrt{y}}$| + $\frac{1}{\sqrt{2\pi}}$ $e^{{-(\sqrt{y})^2/2}}$ |$\frac{1}{2\sqrt{y}}$| $\neq$ $\frac{1}{\sqrt{2\pi}\sqrt{y}}$ $e^{-y/2}$

Consequently, I can't understand how the highlighted equality in the following text could be derived. Could someone please explain? Thank you.

Answer 1

$e^{-(-\sqrt y)^{2}/2}=e^{-y/2}$ not $e^{y/2}$.

When you square, the inside minus sign goes away and the outside minus sign remains.