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I am trying to understand the Burau representation of the braid group, and I am stuck on a particular assertion regarding lifts of homeomorphisms. This braid group/burau context may or may not be necessary: (paraphrasing)

Let $D_n$ be the $n$-times punctured disk. We have a homomorphism $\phi: D_n \to \mathbb{Z}$ sending a curve to its winding number. Let $p:\tilde{D}_n \to D_n$ be the covering corresponding to the kernel of $\phi$. Let $h$ be some homeomorphism of $D_n$ fixing the boundary pointwise (a braid). For any loop $\gamma$, we know $\phi(h\gamma) = \phi \gamma$, so $h$ lifts to a homeomorphism $\tilde{h}$ of $\tilde{D}_n$.

I (think I) understand everything but the last statement. I am interpreting the $\phi(h\gamma) = \phi \gamma$ condition to be satisfying the lifting criterion---it is saying in particular that $(h \circ p)_\ast (\pi_1(\tilde{D}_n)) \subseteq p_\ast (\pi_1(\tilde{D}_n))$. I believe that one can obtain the lift $\tilde{h}: \tilde{D}_n \to \tilde{D}_n$, but how do I know it is a homomorphism? If you can prove the inverse lifts, you can use this answer, but I don't know how to do this.

Edit: After typing this out and reading it, I see that the $\phi(h\gamma) = \phi \gamma$ condition also guarantees the inverse lifts. However, think my question may still be interesting in the general setting.

Say I have a cover $p:\tilde{X} \to X$ and a homeomorphism $h:X \to X$, with the property that $(h \circ p)_\ast (\pi_1(\tilde{X},\tilde{x}_0)) \subseteq p_\ast (\pi_1(\tilde{X},\tilde{x}_0))$, so that $h \circ p$ lifts to a map $\tilde{h}:\tilde{X} \to \tilde{X}$. Must $\tilde{h}$ be a homeomorphism?

My suspicion is that the answer is no, since we are not guaranteed that $f^{-1}$ lifts, but it is not clear to me whether this fact is necessary.

Noah Caplinger
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Indeed, $\tilde{h}$ need not be an isomorphism in general. Writing $G=\pi_1(X,x_0)$ and $H=\pi_1(\tilde{X},\tilde{x}_0)$, we may identify $H$ as a subgroup of $G$ via $p$, and our assumption is that $h_*(H)\subseteq H$. From the commutative square $$\require{AMScd} \begin{CD} \tilde{X} @>{\tilde{h}}>> \tilde{X}\\ @V{p}VV @VV{p}V \\ X @>{h}>> X \end{CD}$$ we see that $\tilde{h}_*:H\to H$ is just the restriction of $h_*:G\to G$. So, if $h_*$ maps $H$ to a proper subgroup of itself, $\tilde{h}_*$ will not be surjective, so $\tilde{h}$ is not a homeomorphism.

For an explicit example, you could take $X=B\mathbb{Q}$ and $h:X\to X$ the homeomorphism induced by the multiplication by $2$ map $\mathbb{Q}\to\mathbb{Q}$. Then $h_*$ maps the subgroup $\mathbb{Z}\subset\mathbb{Q}$ to itself, so $h$ lifts to the associated covering space, but the lift is not a homeomorphism since multiplication by $2$ is not surjective on $\mathbb{Z}$.

Eric Wofsey
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  • Thanks for your response. What is the space $B\mathbb{Q}$? – Noah Caplinger Dec 18 '20 at 04:05
  • The classifying space of the group $\mathbb{Q}$. Explicitly, you can construct it as the geometric realization of the nerve of the group $\mathbb{Q}$ considered as a 1-object category. – Eric Wofsey Dec 18 '20 at 04:37
  • All that matters about it here is that it is a space with given fundamental group (namely, $\mathbb{Q}$) which is functorial in the group, so any automorphism of $\mathbb{Q}$ induces a homeomorphism on the space. – Eric Wofsey Dec 18 '20 at 04:38