I am having a composition of two maps: $$ f:\mathbb{R}->\mathbb{R_0^+},f(x)=x^2 $$ $$ g:\mathbb{R_0^+}->\mathbb{\mathbb{N}},g(x)=\lfloor x\rfloor $$
$$h=g\circ f:\mathbb{R}->\mathbb{N_0}$$
Now I need a right inverse of the map h. I think that the composition is this map:
$$h=\lfloor x^2\rfloor$$ => So I think, that it is surjective and NOT injective. So there must be a right inverse for the map h. (and NO left inverse!)
What is a right inverse of h and how can I find it? ( I know the definition of right inverse, but I can not find a right inverse)
For a right inverse of $h$, we want a function, say $\phi : \mathbb{N} \rightarrow \mathbb{R}$, such that $h(\phi(y)) = y\;$ for all $y \in \mathbb{N}$.
Try $\phi(y) = \sqrt{y} \qquad$ (just because it seems a likely candidate).
For any $y\in \mathbb{N},\quad h(\phi(y)) = h(\sqrt{y}) = \lfloor (\sqrt{y})^2 \rfloor = \lfloor y \rfloor = y.$
So $\phi$ is a right inverse of $h$.