# Does isotopy in a covering space imply isotopy in a base space?

If $p:\tilde{X}\rightarrow X$ is a regular covering space of finite degree, why is it not obvious that if two curves $\gamma$ and $\delta$ are isotopic in $\tilde{X}$ their images are isotopic in $X$? By my understanding, this is a nontrivial theorem (including the statement that the cover is possibly branched) in a paper by Birman and Hilden, and it further requires the condition that the group of deck transformations is solvable.

It seems to me that this statement should follow from the fact that induced map on fundamental group $p_*$ is well defined. What am I missing? Does anyone have a counterexample (preferably one involving surfaces, rather than anything higher dimensional)?

If this is in fact nontrivial, will it be nontrivial with the added condition of the path lifting property?

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Your question isn't properly formulated. If $\gamma$ and $\delta$ are curves in $\tilde X$ they are not curves in $X$ and their projections may not be embedded. – Ryan Budney Apr 13 '11 at 23:41

If two paths are isotopic then they are homotopic via a homotopy $h_t$ that is injective for each $t$. Projecting this to the base doesn't guarantee that $h_t$ will remain injective.