# 8.02 Covering transformations

## Video

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

- Previous video:
**8.01 Lifting criterion**. - Next video:
**8.03 Normal covering spaces**. - Index of all lectures.

## 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

- In the proof that covering transformations are covering maps, why is \(q\circ (p_2)|_W\) a local inverse for \(F\)?
- 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?

## Navigation

- Previous video:
**8.01 Lifting criterion**. - Next video:
**8.03 Normal covering spaces**. - Index of all lectures.