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Suppose that $p: X \to Y$ is the universal covering of some connected and locally path connected space $Y$, and that $\phi$ is a deck transformation. Is $\phi$ homotopic to the identity on $X$? If so, why?

I'm asking because this fact is used in this question and I'm not sure why it should be true.

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Balerion_the_black
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1 Answers1

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Not necessarily. The antipodal map is a deck transformation of $S^n \to \Bbb R P^n$, but it's not homotopic to the identity for even $n$.

Still, the claim is true in the question you link to. The lifts in the question fix base points. Under the given conditions, if two lifts agree on one point, they agree everywhere. See Hatcher's Algebraic Topology, propositions 1.34 and 1.37.

Ayman Hourieh
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  • Well, the claim is true, but the reference to the fact that a deck transformation is homotopic to the identity is sure incorrent then? – Balerion_the_black Jun 21 '14 at 12:42
  • @Balerion_the_black Yeah, I think we're both saying the same thing. That proof can be salvaged by using the correct reason for why $\tilde f \tilde g$ is the identity. – Ayman Hourieh Jun 21 '14 at 13:09
  • @AymanHourieh That's not true though. $X$ and $Y$ are not assumed to be pointed spaces, so $f$ and $g$ shouldn't have to preserve any basepoints. And I'm not sure we can simply assume that $f$ has a homotopy inverse which sends $f(x_0)$ to $x_0$ for some $x_0\in X$. Consequently, I don't think propositions 1.34 and 1.37 of Hatcher apply. – Anonymous Dec 18 '19 at 23:27