# 8.05 Galois correspondence, 1

## Video

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

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**8.04 Deck group**. - Next video:
**8.06 Galois correspondence, 2**. - Index of all lectures.

## Notes

*(0.00)* Given a space \(X\) with basepoint \(x\in X\), a covering
space \(p\colon Y\to X\) and a point \(y\in p^{-1}(x)\) we get a
subgroup \(p_*\pi_1(Y,y)\subset\pi_1(X,x)\). Which subgroups of
\(\pi_1(X,x)\) arise this way?

*(0.38)* Let \(X\) be a path-connected, locally path-connected
topological space. Suppose that there is a simply-connected covering
space \(u\colon\tilde{X}\to X\). Then for any subgroup
\(H\subset\pi_1(X,x)\) there is a covering space \(p\colon Y\to X\)
and a point \(y\in Y\) such that \(p_*\pi_1(Y,y)=H\).

*(1.32)* Observe that \(Deck(\tilde{X},u)=\pi_1(X,x)\). This group
acts on \(\tilde{X}\). Our strategy is as follows:

- We will first prove that this action is properly discontinuous.
- Then we will form the quotient \(Y=\tilde{X}/H\). Since \(\tilde{X}\) is simply-connected and since the action is properly discontinuous, we deduce that \(\pi_1(Y,[\tilde{x}]_H)=H\) (where \(\tilde{x}\in\tilde{X}\) is a basepoint and \([\tilde{x}]_H\in Y\) denotes its orbit under \(H\)).
- Finally, we need to check that the projection map \(p\colon Y\to X\) is a covering map; the projection \(p\) is the map defined as follows: a point \(y\in\tilde{X}\) defines an equivalence class \([y]_H\in\tilde{X}/H\) and an equivalence class \([y]_X\in\tilde{X}/Deck(X,x)=X\); we define \(p([y]_H)=[y]_X\).

*(4.44)* Once we have proved these facts, we obtain a covering space
\(p\colon Y\to X\) with fundamental group isomorphic to \(H\). It is
an exercise to think about why \(p_*\pi_1(Y,[\tilde{x}]_H)\) is
precisely the subgroup \(H\subset\pi_1(X,x)\), rather than just
*some* subgroup of \(\pi_1(X,x)\) isomorphic to \(H\).

*(6.04)* First, we will prove that the deck group of any covering
space \(p\colon Y\to X\) acts properly discontinuously. Pick a point
\(y\in Y\) and set \(x=p(y)\). To prove that the action of
\(Deck(Y,p)\) is properly discontinuous, we need to find a
neighbourhood \(V\) of \(y\) such that \(F(V)\cap V\) is empty for
any \(F\in Deck(Y,p\) not equal to the identity. To that end, pick
an elementary neighbourhood \(U\) containing \(x\) and let \(V\) be
the elementary sheet in \(Y\) over \(U\) containing \(y\).

*(8.12)* Take \(g\in Deck(Y,p)\). We want \(Vg\cap V=\emptyset\)
(writing our action on the right). Because \(g\) is a covering
transformation, \(Vg\) is another elementary sheet for \(p\) living
over \(U\). We have two possibilities: either \(Vg=V\) or else
\(Vg\cap V=\emptyset\). If \(Vg=V\) then \(yg=y\) (as \(y\) is the
unique preimage for \(y\) under \(p\) in \(V\)), so \(g\) fixes a
point and hence agrees with the identity (by the uniqueness theorem
for covering spaces). Therefore \(Vg\cap V=\emptyset\) unless
\(g=id_Y\).

*(11.02)* Finally, we show that if a group \(G\) acts properly
discontinuously on a space \(Z\) and \(H\subset G\) is a subgroup
then the quotient map \(p\colon Z/H\to Z/G\) is a covering map. (To
relate this to point (3) mentioned earlier, take \(Z=\tilde{X}\),
\(G=\pi_1(X,x)\) so \(X=Z/G\).)

*(11.50)* We will write \([z]_H\in Z/H\) and \([z]_G\in Z/G\) for
the respective equivalence classes of a point \(z\in Z\). The map
\(p\colon Z/H\to Z/G\) is \(p([z]_H)=[z]_G\). We need to check that
\(p\) is well-defined and continuous. If \([z]_H=[z']_H\) then there
exists \(h\in H\) such that \(z=z'h\). Since \(H\subset G\) this
implies that \([z]_G=[z']_G\). There is a quotient map \(q_G\colon
Z\to Z/G\) which is continuous by definition of the quotient
topology on \(Z/G\). There is also a quotient map \(q_H\colon Z\to
Z/H\), which is again continuous. By the video on continuous maps
out of quotient spaces, \(p\colon Z/H\to Z/G\) is the unique map we
need to get \(q_G=p\circ q_H\) and is therefore continuous.

*(14.53)* To prove that \(p\) is a covering map, we use the fact
that \(q_G\colon Z\to Z/G\) is a covering map. We have elementary
neighbourhoods \(U\subset Z/G\) and local inverses for \(q_G\)
defined over \(U\). If we compose these local inverses with the map
\(q_H\), we get local inverses to \(p\) defined on the same
elementary neighbourhoods. This shows that \(p\) is a covering map.

## Pre-class questions

- We've used the following fact a few times without really explaining why, so you should try to justify it. Why are different elementary sheets over the same elementary neighbourhood disjoint?

## Navigation

- Previous video:
**8.04 Deck group**. - Next video:
**8.06 Galois correspondence, 2**. - Index of all lectures.