# 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
I have been presenting since 1968 the case that the use of covering morphisms of groupoids makes this theory easier to understand, since maps of spaces are modelled by morphisms of groupoids. – Ronnie Brown Mar 19 at 15:14

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.