# Path-homotopy in covering space implies path-homotopy in original space?

If the lift of a path (in a covering space) is path-homotopic with the constant loop, is then the 'original path' (in the codomain of the covering map) necessarily path-homotopic to the constant loop there?

• What does path homotopic mean here, can a path be path homotopic to a loop without being a loop? – Connor Malin Sep 4 '20 at 1:14
• Two paths $\gamma, \gamma' ∈ P(X; x₀, x₁)$ are path-homotopic if there exists a continuous map $\Gamma : [0, 1] × [0,1] → X$ such that for all $t, s ∈ [0,1]$ we have $\Gamma(0,s) = \gamma(s), \Gamma(1, s) = \gamma'(s), \Gamma(t, 0) = x₀, \Gamma(t, 1) = x₁$. – Jos van Nieuwman Sep 4 '20 at 1:17
• So in particular, if you are path homotopic to the constant map you are a loop to begin with. So if you just project the homotopy to the base space then you have a homotopy to the constant loop. – Connor Malin Sep 4 '20 at 1:19

If $$α,β$$ are two paths in space X which are path homotopic, then their continuous image is also path homotopic. The continuous function need not be a covering map.
Hence your claim is true because lift $$\bar f$$ of a path $$f$$ is a path such that $$f$$ is the continuous image of $$\bar f$$.