8.02 Covering transformations


Below the video you will find accompanying notes and some pre-class questions.


Definition, existence and uniqueness

  • (0.12) Given two covering spaces \(p_1\colon Y_1\to X\) and \(p_2\colon Y_2\to X\) of the same space \(X\), a covering transformation \(F\colon(Y_1,p_1)\to (Y_2,p_2)\) is a continuous map \(F\colon Y_1\to Y_2\) such that \(p_1=p_2\circ F\) [NOTE: I got this formula back-to-front in the video].
  • A covering transformation is called a covering isomorphism if \(F\) is also a homeomorphism.
  • A deck transformation or covering automorphism of a covering space \((Y,p)\) is a covering isomorphism \((Y,p)\to (Y,p)\).
  • The deck group \(Deck(Y,p)\) is the group of all deck transformations of \((Y,p)\).

(3.32) Let \(p_1\colon Y_1\to X\) and \(p_2\colon Y_2\to X\) be covering spaces, let \(x\in X\) and let \(y_1\in p_1^{-1}(x)\) and \(y_2\in p_2^{-1}(x)\). Then there exists a covering transformation \(F\colon (Y_1,p_1)\to (Y_2,p_2)\) such that \(F(y_1)=y_2\) if and only if \[(p_1)_*\pi_1(Y_1,y_1)\subset (p_2)_*\pi_1(Y_2,y_2)\] as subgroups of \(\pi_1(X,x)\). Moreover, if such a covering transformation exists then it is unique (uniquely determined by the condition \(F(y_1)=y_2)\)).

(6.38) There is a covering isomorphism \(F\colon (Y_1,p_1)\to (Y_2,p_2)\) if and only if \((p_1)_*\pi_1(Y_1,y_1)=(p_2)_*\pi_1(Y_2,y_2)\) as subgroups of \(\pi_1(X,x)\).

(7.40) The theorem follows immediately from the lifting criterion applied to the following situation. Take \(T=Y_1\) and \(f=p_1\colon Y_1\to X\). A covering transformation \(F\colon Y_1\to Y_2\) is precisely a lift of \(p_1\) to a map \(Y_1\to Y_2\). We see that the statement of the theorem is just the lifting criterion applied to this siutation.


(9.44) Let \(X=S^1\) and take \(Y_1=S^1\), \(Y_2=S^1\) with covering maps \(p_1(z)=z^m\) and \(p_2(z)=z^n\). When is there a covering transformation \(F\colon Y_1\to Y_2\)? The pushforward \((p_1)_*\pi_1(Y_1,1)\) is the subgroup \(m\mathbf{Z}\subset\mathbf{Z}\). The pushforward \((p_2)_*\pi_1(Y_2,1)\) is the subgroup \(n\mathbf{Z}\subset\mathbf{Z}\). Therefore the criterion from the theorem holds if and only if \(n\) divides \(m\).

(11.54) For example, if \(m=6\) and \(n=2\) then we can take \(F(z)=z^3\) and we get \(p_1(z)=z^6=(z^3)^2=F(p_1(z))\) so \(F\) is a covering transformation. We could also take \(F(z)=-z^3\) because \((-z^3)^2=z^6\) too. This exhausts the possible covering transformations \(Y_1\to Y_2\) because a covering transformation is determined by its value at a single point, \(F(1)\), and we need \(F(1)=\pm 1\) because \(1=p_1(1)=p_2(F(1))=(F(1))^2\).

(14.47) Note that \(z\mapsto z^3\) is again a covering map. We will now see that this is a general feature.

Covering transformations are covering maps

(15.00) Assume that \(p_1\colon Y_1\to X\) and \(p_2\colon Y_2\to X\) are covering spaces and that \(Y_2\) is path-connected. A covering transformation \(F\colon Y_1\to Y_2\) is a covering map.

(16.04) Each \(x\in X\) has an elementary neighbourhood \(U_1\) for \(p_1\) and \(U_2\) for \(p_2\), so \(U:=U_1\cap U_2\) is an elementary neighbourhood for both simultaneously.

(17.55) \(F\) is surjective. To see this, pick a point \(y_2\in Y_2\). We want to find a point \(z\in Y_1\) such that \(F(z)=y_2\). Pick a point \(y_1\in Y_1\) and a path \(\alpha\) in \(Y_2\) from \(F(y_1)\) to \(y_2\). This projects along \(p_2\) to give a path \(p_2\circ\alpha\) from \(p_1(y_1)\) to \(p_2(y_2)\). Path-lifting for \(p_1\) yields a path \(\widetilde{p_2\circ\alpha}\) in \(Y_1\) from \(y_1\) to some point \(z\). Applying \(F\) to this gives a path \(F\circ\widetilde{p_2\circ\alpha}\) in \(Y_2\) from \(y_1\) to \(F(z)\).

(20.15) We have \(p_2\circ(F\circ\widetilde{p_2\circ\alpha})=p_1\circ\widetilde{p_2\circ\alpha}=p_2\circ\alpha\), so \(F\circ\widetilde{p_2\circ\alpha}\) and \(\alpha\) are both paths lifting \(p_2\circ\alpha\) along \(p_2\) satisfying the initial condition \(F(\widetilde{p_2\circ\alpha}(0))=\alpha(0)=y_2\). Therefore they agree by uniqueness of lifts, so their endpoints, \(F(z)\) and \(y_2\), agree, so \(F(z)=y_2\) and we see that \(F\) is surjective.

(22.32) \(F\) is a covering map. Fix a point \(y_2\in Y_2\). We want to find an elementary neighbourhood around \(y_2\) together with a local inverse for \(F\). Let \(x=p_2(y_2)\) and let \(x\in U\subset X\) be an elementary neighbourhood simultaneously for \(p_1\) and \(p_2\). For each \(y_1\in F^{-1}(y_2)\) (which is non-empty because \(F\) is surjective) we have an elementary sheet \(y_1\in V\subset Y_1\) for \(p_1\) living over \(U\) and a local inverse \(q\colon U\to V\) with \(q(x)=y_1\). Let \(y_2\in W\subset Y_2\) be an elementary sheet for \(p_2\) over \(U\). The map \(q\circ (p_2)|_W\colon W\to V\) is a local inverse for \(F\). If we do this for all our elementary sheets, we deduce that \(F\) is a covering map.

Pre-class questions

  1. In the proof that covering transformations are covering maps, why is \(q\circ (p_2)|_W\) a local inverse for \(F\)?
  2. Because I was using the lifting criterion, I should have added an assumption about my spaces in the first theorem. What should I have said?


CC-BY-SA, Jonny Evans 2017