I have the following situation: $K$ is a finite simplicial complex and $\widetilde{K}\to K$ is "a covering space corresponding to the subgroup $H$ of $G=\pi_1(K)$". Can anyone say what exactly $\widetilde{K}$ is? My only idea is that it is such a covering space that $p_*(\pi_1(\widetilde{K}))$ lies in the conjugacy class of $H$ in $G$. Is it so?
Also, if $H$ acts on $K$ (compactly, freely etc., whatever you want) can we say something about connection of $\pi_1(K/H)$ and $G$?
Thank you.
The classification theorem of covering spaces tells you how to realise $\widetilde{K}$: this says that for $K$ a semi-locally simply connected space (eg a simplicial complex), subgroups of the fundamental group correspond to (connected) covering spaces of $K$. In particular, there is a simply-connected covering space $X$ (corresponding to the trivial subgroup of $\pi_1 (K)$), called "the universal covering of $K$"). It has the property that any subgroup of $\pi_1 (K)$ acts on this space (actually $\pi_1 (K) \cong \mathrm{Aut}_K X \subset \mathrm{Homeo} (X)$.
The theorem states that the corresponding covering space $\widetilde{K}$ corresponding to $H$ is precisely $X/H$, and it satisfies $\pi_1 (X/H) \cong H$.
A good reference to learn more on covering theory is Massey's "A basic course in Algebraic Topology" (chapter 5).