There´s a theorem which says that $h(x)$ is bijective $\iff$ it´s invertible so that $h^{-1}(y)=x \iff y=h(x)$
So should I prove that it exists?
In any case h(x) is injective since $x_1,x_2 \in \mathbb{Z}$ so that $h(x_1)=h(x_2)\Rightarrow (x_1-2)^3-3=(x_2-2)^3-3\Rightarrow x_1=x_2$
But I´m thinking it´s not onto, the image of h should be $\mathbb{Z}$
Let $x=\sqrt[3]{y+3}+2$, so $(\sqrt[3]{y+3}+2-2)^3-3= y$, what about $0 \in \mathbb{Z}$?
$x=\sqrt[3]{0+3}+2\Rightarrow x=\sqrt[3]{3}+2$ which is not in $\mathbb{Z}$
So it shouldn´t have an inverse function, shoud it?
$h \circ h^{-1}= h(\sqrt[3]{x+3}+2)=(\sqrt[3]{x+3}+2)^3-3=x$
$h^-1 \circ h=h^{-1}((x - 2)^3 - 3)=\sqrt[3]{(x - 2)^3 - 3+3}+2=x$
So apparently it should be that one, Does it have or not?
To show that the function is not surjective, it suffices to show that there is some integer $y$ such that $$(x-2)^3-3=y,$$ has no integer solution, or equivalently, such that $y+3$ is not a perfect cube.