I infer the following
f: R →[0, ∞) ∣ f(x)=x² is not injective, so has no left inverse; however, it is surjetive, so has a right inverse g: [0, ∞) → R ∣ g(x)=√x such that f(g(x))=x.
g: [0, ∞) → [0, ∞)∣g(x)=√x is injective, so has a left inverse f: [0, ∞)→ [0, ∞) ∣ f(x)=x², and it is surjective, so has a right inverse f: [0, ∞)→ [0, ∞) ∣ f(x)=x² such that g(f(y))=y.