Let $\pi\colon\tilde{X}\to X$ be a p-fold (regular) cyclic covering(p:prime) and $\mathcal{A} = \mathrm{Im}(\pi_* )$, where $\pi_* \colon H_1(\tilde{X};\mathbb{Z}_p) \to H_1(X;\mathbb{Z}_p)$ is induced by $\pi\colon\tilde{X}\to X$. Suppose $f \colon X\to X$ is a homoemorphism satisfying $f_*(\mathcal{A})=\mathcal{A}$, where $f_* \colon H_1(X;\mathbb{Z}_p)\to H_1(X;\mathbb{Z}_p)$ induced by $f\colon X\to X$. Then, does there exists a map $g\colon \tilde{X}\to \tilde{X}$ satisfying $\pi\circ g =f\circ \pi$? I know the lifting criteria of covering space in terms of fundamental group. I hope that somebody tell me why above condition implies the lifting criteria in terms of fundamental group.
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3$\begingroup$ $H_1(X;\mathbb Z/p)$ is the fundamental group of $X$ modulo commutators and $p$'th powers. Hence, a cyclic cover of degree $p$ is completely determined by $\mathcal A$. In particular $f$ lifts iff $f_*$ preserves $\mathcal A$. $\endgroup$– Torsten EkedahlCommented Jul 20, 2010 at 15:51
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$\begingroup$ Thank you for your friendly comment. I was only checking algebraic criterion. Geometrically, it seems obvious that $f$ lifts iff $f_*$ preserves $\mathcal{A}$. $\endgroup$– TopologieeeCommented Jul 20, 2010 at 16:30
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$\begingroup$ And I have another question. Is $H_1(X)/\pi_*(H_1(\tilde{X}))=\pi_1(X)/\pi_*(\pi_1(\tilde{X}))$? $\endgroup$– TopologieeeCommented Jul 20, 2010 at 16:46
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1$\begingroup$ Yes, as $f_*(\pi_1(\tilde X))$ contains commutators and $p$'th powers. $\endgroup$– Torsten EkedahlCommented Jul 21, 2010 at 4:02
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