If $f(x)=\sin x, x \in[-\pi / 2, \pi / 2]$, then which one of the following is not correct?

fof(x) is decreasing on $[-\pi / 2,0]$ and is increasing on $[0, \pi / 2]$

We know that $\sin x$ is increasing on $[-\pi / 2, \pi / 2]$

Now,

$f(x)=\sin x$

$\Rightarrow fof(x)=\sin (\sin x)$

$\Rightarrow \frac{d}{d x}\{fof(x)\}=\cos (\sin x) \cos x \geq 0$ for $x \in[-\pi / 2, \pi / 2]$

$\Rightarrow fof(x)$ is increasing on $[-\pi / 2, \pi / 2]$

Since, fof(x) is strictly increasing on $(-\pi / 2, \pi / 2)$ and fof$\left(\frac{\pi}{2}\right)=\sin 1$, fof $\left(-\frac{\pi}{2}\right)=-\sin 1$

Now, these values are not attained at any other point. Hence, fof is invertible on $[-\pi / 2, \pi / 2]$.