# 8.05 Galois correspondence, 1

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

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