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This is a part of the proof of Lemma 4 of "Cobordism of classical knots" by Casson and Gordon. Here, $\widetilde{X}$ denotes a prime-fold cyclic covering of $X$.

Let $h\colon X\to X$ generate the group of covering translations of $X$. Let $G$ be the image of $H_1(\widetilde{X},\mathbb{Z}_p)$ in $H_1(X,\mathbb{Z}_p)$. Since $H_1(X,\mathbb{Z}_p)$ is finite, there is a non-zero integer $r$ such that $(h_\ast)^r(G)=G$. It follows that $h^r\colon X\to X$ lifts to a homeomorphism $\tilde{h}\colon\widetilde{X}\to\widetilde{X}$, and $\widetilde{X}$ is an infinite cyclic covering of the finite complex $\widetilde{X}/\tilde{h}$.

Okay, the existence of $\tilde{h}$ comes from the lifting criterion for covering spaces. But how can we know that it is a homeomorphism?

Sumanta
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NothingInSense
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1 Answers1

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It is not true in general that for any covering map $f : \widetilde X \to X$ of any path connected, locally path connected space, and for any homeomorphism $h : X \to X$ and any lift $\tilde h : \widetilde X \to \widetilde X$ of $h$, the map $\tilde h$ is a homeomorphism.

However, if one adds the hypothesis that the given covering map $f$ has finite degree, which matches your situation, then the conclusion does hold, by an application of the general lifting lemma, as follows.

Pick $p \in X$, and $\tilde p \in \widetilde X$, and let $q=h(p)$ and $\tilde q = \tilde h(\tilde p)$. The fact that a lift $\tilde h : (\widetilde X,\tilde p) \to (\widetilde X,\tilde q)$ of $h$ exists implies that the group isomorphism $h_* : \pi_1(X,p) \to \pi_1(X,q)$ maps the subgroup $f_*(\pi_1(\widetilde X, \tilde p)) < \pi_1(X,p)$ isomorphically into the subgroup $f_*(\pi_1(\widetilde X,\tilde q)) < \pi_1(X,q)$. However, each of the subgroups $f_*(\pi_1(\widetilde X, \tilde p)) < \pi_1(X,p)$ and $f_*(\pi_1(\widetilde X,\tilde q)) < \pi_1(X,q)$ has finite index equal to $\text{degree}(f)$. Also, the group isomorphism $h_*$ preserves index of subgroups. The only way one subgroup of finite index can include into another subgroup of the same finite index is if those two subgroups are equal. Therefore, $h_*$ maps $f_*(\pi_1(\widetilde X, \tilde p))$ is isomorphically onto $f_*(\pi_1(\widetilde X,\tilde q))$.

The inverse homomorphism $(h^{-1})_* : \pi_1(X,q) \to \pi_1(X,p)$ therefore maps $f_*(\pi_1(\widetilde X,\tilde q))$ isomorphically onto $f_*(\pi_1(\widetilde X, \tilde p))$. An application of the general lifting lemma then produces a lift of $f^{-1}$ that takes $\tilde q$ to $\tilde p$, and the uniqueness clause of the general lifting lemma let's one conclude that this lift is a continuous inverse of $\tilde f$, hence $\tilde f$ is a homeomorphism.

Lee Mosher
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