Given his definition of parametric surface
let $\Omega \subseteq \mathbb{R}^2$ open, homeomorphic to an open disk and let $ P : \Omega \to \mathbb{R}^3 $ a smooth injective function s.t. the jacobian in every point of $\Omega $ has maximal rank, then $ P $ is a parametric surface.
I need to prove that the sphere is not a parametric surface (image of a a parametric surface). On my notes there is written "it can't be, just because a sphere is not homeomorphic to a disk of $ \mathbb{R}^2 $".
But I don't see why homeomorphisms are involved here, because P, in general, is not an homeo on its image (there are counterexamples) instead it is a local homeo, but this is ok to me. Maybe this fact involves some notions about fundamental groups or homology groups but I don't know results that don't involve homeomorphisms or deformation retracts between two topological spaces (and according to me this is not the case). Just to make clear, I know where the other usual parametrizations fails in this case, but i can't see a general scheme to help me point out the main reason.
Your question is indeed a subtle one. Usually, the definition of a parametric surface in $\mathbb{R}^3$ one encounters is (the image of) a smooth injective map $P \colon \Omega \rightarrow \mathbb{R}^3$ such that $dP_p \colon \mathbb{R}^2 \rightarrow \mathbb{R}^3$ has maximal rank (here, 2) at each $p \in \Omega$ and that $P$ is a homeomorphism onto its image (endowed with the subspace topology). See for example Do Carmo's Curves and Surfaces. Such a parametric surface is called an embedded parametric surface and $P$ is called an embedding. It is clear that $S^2$ cannot be an embedded parametric surface because for example $S^2$ is compact while $\Omega$ is not. However, if one doesn't require the condition that $P$ is a homeomorphism onto its image, one is left with what is sometimes called an immersed parametric surface (and $P$ is called an immersion). For example, if we look at $P \colon (-\pi, \pi) \times (-1,1) \rightarrow \mathbb{R}^3$, given by $P(t,s) = (\sin(2t), \sin(t),s)$, we get the following surface:
This surface is not an embedded parametric surface because the map $P$ is not a homeomorphism onto its image (for example, because the image is not simply connected) but it is an immersed parametric surface. Generally, it is not trivial to decide whether a certain subset of $\mathbb{R}^3$ can be an immersed parametric surface or not because one precisely cannot use "naive" topological arguments.
However, one can show that $S^2$ cannot be an immersed parametric surface as follows. Consider your parametrization $P \colon \Omega \rightarrow \mathbb{R}^3$ as a smooth map between manifolds and note that $P(\Omega) = S^2$ on the level of sets. Considering $S^2 \subseteq \mathbb{R}^3$ as an embedded submanifold of $\mathbb{R}^3$ (or a regular surface in $\mathbb{R}^3$), one can show that the by restricting the codomain of $P$ to $S^2$, one is still left with a smooth map $P \colon \Omega \rightarrow S^2$ between manifolds. This is Corollary 5.30 in Lee's Introduction to Smooth Manifolds. Using the chain rule, one can see that the rank of $dP_p \colon T_p\Omega \rightarrow T_{P(p)}(S^2)$ is two for all $p \in \Omega$. That is, $P \colon \Omega \rightarrow S^2$ is a smooth bijection with full rank and thus by the inverse function theorem must be a local diffeomorphism and (since $P$ is injective and surjective) a diffeomorphism. In particular, $P$ is a homeomorphism. But $S^2$ is compact while $\Omega$ is not - a contradiction.
More generally, if $A \subseteq M$ is a subset of a smooth manifold such that with the induced subset topology $\tau$, $(A,\tau)$ has the structure of a smooth manifold, then the set $A$ has a unique topology and a unique smooth structure such that the inclusion $A \subseteq M$ is an immersion (that is, such that $A$ is an immersed submanifold of $M$). For more details, see Warner's book Foundations of Differentiable Manifolds where this is discussed in the context of manifolds.