Let $f : \Sigma^k \to M^n$ be an immersion between differentiable manifolds and let $\pi : \tilde{M} \to M$ be a finite-to-one covering map. Let $\tilde{\Sigma} = \pi^{-1}(f(\Sigma))$. Is it true that $\tilde{\Sigma}$ is immersed somehow into $\tilde{M}$?
If $f$ is an embedding, I know how to prove that $\tilde{\Sigma}$ is embedded into $\tilde{M}$, since we can use the fact that $\pi$ is transversal to $f(\Sigma) \cong \Sigma$ (as $\pi$ is a submersion) to conclude that $\pi^{-1}(f(\Sigma))$ is a submanifold of $\tilde{M}$. What about the general case?
You can take the pullback of $\tilde{M}$ along $f$, as depicted in the following diagram: $$\begin{array}{ccc}f^*\tilde{M}&\longrightarrow&\tilde{M}\\\downarrow&\quad&\downarrow\\\Sigma&\xrightarrow{f}&M\end{array}.$$ The pullback can be described in a few different ways. In any case, as the above diagram and the corresponding diagram of tangent bundles commute, it follows that $f^*\tilde{M}\to\tilde{M}$ is an immersion with image $\pi^{-1}(f(\Sigma)).$
Edit: We explain why $f^*\tilde{M}\to\tilde{M}$ is an immersion. Both downwards-pointing arrows in the diagram are covering maps and, in particular, local diffeomorphisms. Hence, the composition $$f^*\tilde{M}\to\Sigma\to M$$ is an immersion. As the diagram commutes, the composition $$f^*\tilde{M}\to\tilde{M}\to M$$ is also an immersion, and consequently, so is the left-hand arrow in it.