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?
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$\begingroup$ What does path homotopic mean here, can a path be path homotopic to a loop without being a loop? $\endgroup$ – Connor Malin Sep 4 '20 at 1:14
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$\begingroup$ 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 $\Gamma(0,s) = \gamma(s), \Gamma(1, s) = \gamma'(s), \Gamma(t, 0) = x₀, \Gamma(t, 1) = x₁$. $\endgroup$ – Jos van Nieuwman Sep 4 '20 at 1:17
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1$\begingroup$ 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. $\endgroup$ – Connor Malin Sep 4 '20 at 1:19
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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 $.
