# Universal cover modulo the monodromy action

Let $\tilde{X} \to X$ be the universal cover of a connected, locally path-connected and semi-locally simply connected topological space $X$. Is it always true that the orbit space $$\tilde{X} \;/\; \pi_1(X)$$ is naturally isomorphic to $X$? Or do we need any extra assumptions?

This is the example I have in mind: the torus $T^2$ and it's universal cover $\mathbb{R}^2$. The fundamental group is $\mathbb{Z}^2$, and we indeed have $$\mathbb{R}^2\;/\; \mathbb{Z}^2 \cong T^2$$

## 1 Answer

Let $p\colon\tilde{X}\rightarrow X$ be a covering map. The orbit space is the space of all orbits under the action of $\pi_1(X)$ on $\tilde{X}$ (the action given by lifts of loops at a point). Note however that, if a covering has a deck transformation group which is transitive on fibers, then it is regular, and every universal cover is a regular cover.

The orbit of a point $x\in\tilde{X}$ is then precisely the fiber of the point $p(x)\in X$. So, we have a natural homeomorphism between the quotient space $\tilde{X}/\pi_1(X)$ and $X$ which sends fibers over $p(x)\in X$ (equivalently orbits of the point $x\in\tilde{X}$) to the point $p(x)$. Note, this behaves well because the action of $\pi_1(X)$ on $\tilde{X}$ is properly discontinuous and free.

This is rather succinctly explained here.