# 8.02 Covering transformations

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

# Notes

## 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.

## Example

(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?