6
$\begingroup$

This question arose in a first-year graduate topology course. But not a homework!

Recall, given a covering map $p:\tilde{X}\rightarrow X$ of path connected spaces. A deck transformation should be defined as a homeomorphism $f:\tilde{X}\rightarrow\tilde{X}$ satisfying $p(f(x))=p(x)$.

The teacher made a mistake in the class. He only required $f$ to be a continuous map. So I am wondering why continuous is not enough (Since no textbook use it as definition, there should be a counterexample I think). I showed $f$ must be surjective and open, but cannot make it further.

Thank you!

$\endgroup$
8
  • $\begingroup$ Presumably your teacher said that $f$ must be continuous and bijective? $\endgroup$
    – Yemon Choi
    Commented Nov 5, 2011 at 2:58
  • $\begingroup$ Well the point of a deck transformation is to form a group under composition and have it permute the elements of each fiber of the covering space. A collection of continuous functions won't do this. $\endgroup$ Commented Nov 5, 2011 at 5:21
  • $\begingroup$ Actually covering space theory is rigid so it will do in this case, see my answer. $\endgroup$ Commented Nov 5, 2011 at 5:31
  • $\begingroup$ Yes, there is a counterexample where $f$ is continuous but not a homeomorphism ( see reference in my comment to Benjamin) $\endgroup$ Commented Nov 5, 2011 at 7:59
  • 1
    $\begingroup$ In defense of my silly answer, let me say what made me answer too quickly late at night. I was thinking of the fact that in topos style Galois theory each endomorphism is an automorphism. But SGA style Galois theory looks at only finite coverings. In this case, one is looking at finite G-sets and so all endomorphisms are automorphisms. @Torsten, you correctly recovered Georges' comments. $\endgroup$ Commented Nov 5, 2011 at 16:56

1 Answer 1

7
$\begingroup$

Assume given a group $G$, a subgroup $H$ and a $g\in G$ such that $gHg^{-1}$ is properly contained in $H$. Let now $Y$ be a topological space with an action of $G$ such that $Y\rightarrow Y/G=:X$ is a covering map (such $Y$'s are plentiful). Let $\tilde X:=Y/H$ so that $p\colon\tilde X\rightarrow X$ is another covering map. Now, $g\cdot\colon Y\rightarrow Y$ maps $H$-orbits to $H$-orbits and hence induces a map $f\colon\tilde X\rightarrow\tilde X$ which fulfils $p\circ f=p$. It is not injective however as the fibres look like cosets $H/gHg^{-1}$ which by assumption contains more than one element. (Using some covering space theory it is easy to see that possibly - I don't think this possibility exists - excluding some very strange examples any example must appear in this way.)

It remains to show that such $(G,H,g)$ exist. One can actually start with any $H$ and an injective non-surjective endomorphism of it but it is easier to give a concrete example (the general construction is very similar). Hence we let $H':=\mathbb Z[1/2]$, the group of rational numbers whose denominators are powers of $2$. We have an action of $\mathbb Z$ on $H'$ where $1\in\mathbb Z$ acts by multiplication by $2$. We then let $G$ be the semi-direct product of $H'$ and $\mathbb Z$, $H:=\mathbb Z\subseteq H'$ and $g=(1,0)$. Note that this group is finitely generated (by $(1,0)$ and $(0,1)$) but it seems not to be finitely presented. However, it is easy enough to modify it to be finitely presented: Just take the group generated by $g$ and $h$ and relation $ghg^{-1}=h^2$ with $H$ generated by $h$. It maps to the semi-direct product which shows that $gHg^{-1}$ is indeed properly contained in $H$. Hence we can get examples where $X$ is a finite CW-complex.

Addendum: I just saw Benjamin's answer when I posted this. He makes use of the claim that any $G$-endomorphism of a transitive $G$-set is an automorphism. However, this claim seems to be false and under my assumptions we get an example of an endomorphism of $G/H$ that is not bijective. Curiously enough my first reaction was the same as Benjamin's. Maybe both of us encountered such statements when learning of finite groups where of course it is true.

Addendum 1: In a comment to Benjamin's answer (that may disappear as Benjamin has deleted his answer) Georges Elencwajg makes a reference to page 179 of Lima's Fundamental Groups and Covering Spaces where a counterexample to bijectivity is given where $X$ is the figure eight with a $2$-cell adjoined. Let me further add that my last example with the group generated by $g$ and $h$ is most likely to be the same example. We can indeed construct a CW-complex with my example as its fundamental group by starting with the figure eight and then adding the relation $ghg^{-1}=h^2$ by adjoining the appropriate 2-cell.

$\endgroup$
1
  • 1
    $\begingroup$ You are correct. I shouldn't answer questions at 1am Ben $\endgroup$ Commented Nov 5, 2011 at 7:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .