# 8.05 Galois correspondence, 1

## Video

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

## 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:

1. We will first prove that this action is properly discontinuous.
2. 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 $$H$$).
3. 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

1. 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?