# How do you construct the lifted topology of a groupoid cover?

If I have a particularly nice space $X$ (Hausdorf, locally path connected, semi-locally 1-connected, I think), then then there is an equivalence of categories between the cover category over $X$ and the groupoid cover category over the fundamental groupoid $\pi X$. One direction of this equivalence is given by the fundamental groupoid functor. The other is given by 'lifting' a groupoid cover $B \rightarrow \pi X$ to a topological cover $B_! \rightarrow X$. What is the topology on $B_!$?

I am traveling and left my copy of Topology and Groupoids at home.

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Do you understand the "classic" pointed case? I.e. the equivalence between covers of $X$ and $\pi_1(X,x)$-sets (when $X$ is also path-connected). –  Martin Brandenburg Jun 8 '14 at 17:31

Given a space $X$ and covering morphism of groupoids $q: G \to \pi_1 X$ the aim is to construct a topology on $Y= Ob(G)$ so that $p=Ob(q): Y \to X$ is a covering map of the kind required. Given a point $x \in X$ one assumes one can choose a neighbourhood $N$ of $x$ such that the induced map of groupoids $\pi_1(N) \to \pi_1(X)$ lifts to a set of morphisms $\pi_1(N) \to G$ such that these map $x \in N$ to all points $y$ in $q^{-1}(x)$. For this one needs that the image of $\pi_1(N,x)$ in $\pi_1(X,x)$ is contained in the images under $p$ of all $G(y)$. This gives one the necessary local condition. Assuming this, the image under these lifts of $N$ in $Y$ gives a basis for the topology of $Y$ which is needed. This is the account in Topology and Groupoids, as it was in the 1968, 1988 editions.