# Homotopic equivalence $\iff$ Covering map?

What is the relation between the concepts of homotopic equivalent mappings and covering map.

More precisely,

A continuous map $f:X \to Y$ is an homotopic equivalence $\iff \exists g:Y \to X$ continuous such that $g \circ f = Id \land f \circ g = Id$.

A covering of a space $X$ is $(\hat X,\pi)$ where $\pi:\hat X \to X$ is continuous, surjective and verifies that for each point $x \in X$ there exists an open, path-connected neighborhood set $U$ such that the path-connected components of $\pi^{-1}(U)$ are homeomorphic to $U$.

My question is:

If $f$ is homotopic equivalence we would have $(X,f)$ covering of $Y$? Reciprocally, if $(\hat X,\pi)$ is a covering then $\pi$ is an homotopic equivalence?

• No to both. For the second, $\mathbb{R}$ is not even close to homotopy equivalent to the circle. – Randall Dec 18 '17 at 17:16
• Sure. The real line and a point are homotopy equivalent by a constant map. – Randall Dec 18 '17 at 17:19
• I'm also pretty sure you can prove that homotopy theory would have been dead in 1945 if this were a true equivalence. – Randall Dec 18 '17 at 17:22

The first is false. The real line is contractible, so there is a homotopy equivalence $\mathbb{R} \to \{0\}$. This cannot be a cover because the fiber isn't discrete.
The second is also false. The universal cover $\mathrm{exp:} \mathbb{R} \to S^1$ is not even close to a homotopy equivalence.