Is there a long exact sequence associated to a ramified covering?

Question

A covering map $p:X\to Y$ between topological spaces can be viewed as a fiber bundle $\Sigma\to X\to Y$ with a discrete group $\Sigma=Gal(X/Y)$ as fiber. Such a fiber bundle leads to a long exact sequence of homotopy groups. In this case, if $Y$ is contractible then, of course, so is $X$.

I'm wondering what happens if the covering map $p$ is ramified. Is there any relation between the homotopy groups of $\pi_n(X),\pi_n(Y)$ and $\Sigma$? I'm guessing that perhaps the fixed set $X^\Sigma$ might be involved.

I'm particularly interested in two cases:

1) When $\Sigma=\Sigma_2$, the two-element group. This occurs often in toric topology.

2) What conditions can force $Y$ to be contractible (or just weakly null-homotopic).

Answer 1

If by ramified cover, you mean branched cover, then I am skeptical. For example, consider the suspension $\Sigma S^{\infty} \to \Sigma \Bbb RP^{\infty}$ of the double cover $S^\infty \to \Bbb RP^{\infty}$. This is a branched cover with branch locus $S^0$. The total space is contractible, but the homotopy groups of the base space are not completely known.

By the way, there's a localization sequence (in the sense of fixed point theory) in homology for branched coverings associated with group actions. See for example:

Cohomological Methods in Transformation Groups (Cambridge Studies in Advanced Mathematics)
by Christopher Allday, Volker Puppe

Answer 2

There is a paper "on the homology of double branched covers" by Lee, which is kind of related to your question.