(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.