Lifting submanifolds

Question

Let $\Sigma$ be a submanifold of $M$ and let $\pi\colon \widetilde{M} \rightarrow M$ be a covering map. I would like to know if it is always true that $\pi^{-1}(\Sigma)$ is a submanifold of $\widetilde{M}$.

Answer 1

Let $\hat x\in\pi^{-1}(\Sigma)$, there exists an open subset $x\in \hat U$ such that the restriction $\pi_{\mid \hat U}:\hat U\rightarrow\pi(\hat U)=U$ is a diffeomorphism.

Since $\Sigma$ a submanifold, there exits a neighborhood $V$ of $x=p(\hat x)$ a submersion $f:V\rightarrow \mathbb{R}^p$ such that $V\cap \Sigma= f^{-1}(0)$.

Let $\hat V=(\pi_{\mid \hat U})^{-1}(V)$ and $g:\hat V\rightarrow \mathbb{R}^p$ defined by $f\circ\pi_{\mid\hat U}$, $g$ is a submersion and $g^{-1}(0)=\hat V\cap \pi^{-1}(\Sigma)$. This implies that $\pi^{-1}(\Sigma)$ is asubmanifold.