I just came across the following remark on Terry Tao's blog which I found cryptic:
If $M$ is a connected topological manifold, and $p$ is a point in $M$, the (topological) fundamental group $\pi_1(M,p)$ of $M$ at $p$ is traditionally defined as the space of equivalence classes of loops starting and ending at $p$, with two loops considered equivalent if they are homotopic to each other .... As the name suggests, it is one of the most basic topological invariants of a manifold, which among other things can be used to classify the covering spaces of that manifold. Indeed, given any such covering $\phi: N \rightarrow M$, the fundamental group $\pi_1(M,p)$ acts (on the right) by monodromy on the fibre $\phi^{-1}(\{p\})$, and conversely given any discrete set with a right action of $\pi_1(M,p)$, one can find a covering space with that monodromy action (this can be done by “tensoring” the universal cover with the given action, as illustrated below the fold). In more category-theoretic terms: monodromy produces an equivalence of categories between the category of covers of $M$, and the category of discrete $\pi_1(M,p)$-sets.
I see how the fundamental group acts on fibers via monodromy, what I don't see is the converse statement stated here. How does one "tensor" the universal cover to obtain a covering? Could this equivalence of categories be stated more explicitly?
"...as illustrated below the fold."
"Let $\tilde \phi_1: \tilde M_1 \rightarrow M_1$ be a universal cover for $M_1$, thus $\tilde M_1$ is simply connected, and if we pick a base point $x_1 \in \tilde\phi_1^{-1}(\{p\})$, then every other point $y$ in that fibre there is a unique element $g_1 \in \pi_1(M_1,p)$ for which $y = x_1 g$, where $x_1 g$ is the (right) action of $g$ by monodromy on $x_1$. This gives a left action of $\pi_1(M_1,p)$ on $M_1$ by deck transformations $D_h: \tilde M_1 \rightarrow \tilde M_1$ for each $h \in \pi_1(M_1,p)$, which maps $x_1 g$ to $x_1 hg$ for any $g \in \pi_1(M_1,p)$: $D_h (x_1 g) = x_1 hg$."
"The fibres of the universal cover $\tilde \phi_1$ are copies of $\pi_1(M_1,p)$. We can now form a new cover $\phi_1: N_1 \rightarrow M_1$ whose fibres are copies of $G$, by first forming the Cartesian product $G \times \tilde M_1$ (which still covers $M_1$) and then quotienting out by the equivalence $(g f_1(g_1), x) \equiv (g, D_{g_1} x)$ for all $g \in G$, $x \in \tilde M_1$, $h \in \pi_1(M_1,p)$."