$\mathbb{R}P^n$ can't be submanifold of $\mathbb{R}^n$ I need to prove that can't exist a function $f:\mathbb{R}P^n \rightarrow \mathbb{R}^n$ such that $(\mathbb{R}P^n,f)$ is a submanifold of $\mathbb{R}^n$.
I can prove that for the case of $n$ even because if such $f$ exist, we have that $df_p$ is an isomorphism $\forall p \in \mathbb{R}P^n$ (because $\mathbb{R}P^n$ and $\mathbb{R}^n$ have the same dimension), then for the inverse function theorem have that $f$ is a local diffeomorphism, wich is an contradiction from the fact that $\mathbb{R}^n$ is orientable and $\mathbb{R}P^n$ is not orientable if $n$ even.
I think that is possible to do for any $n$ thinking in the fact that $\mathbb{R}P^n$ is compact and then $f$ has to be an embedding.
 A: There does not exist any smooth embedding $f: \Bbb R P^n \to \Bbb R^n$ because, since the dimensions of the domain and codomain are equal, this embedding would be an open map by the inverse function theorem. Since $\Bbb R P^n$ is also compact, it would follow that $f(\Bbb R P^n)$ is a compact (hence closed) and open subset of the connected space $\Bbb R^n$, so $f(\Bbb R P^n) = \Bbb R^n$. But $\Bbb R^n$ is not compact, so no such embedding can exist. 
The exact same argument shows that no compact connected manifold (without boundary) embeds in a noncompact connected manifold of the same dimension, and this implies the specific answer above. 
Generalizing in a different direction which also implies the specific answer above, there is no immersion from a non-[simply connected] compact connected manifold (without boundary) to a simply connected manifold of the same dimension. This is because same-dimension and compactness imply that the immersion is a covering map with simply connected codomain, and the theory of covering spaces implies that the domain of such a covering map must be simply connected.
