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

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**8.04 Deck group**. - Index of all lectures.

*(2.15)* If we take \(\pi_1(S^1,1)=\mathbf{Z}\) and push it
forward along \(p_n\) then we get the subgroup
\[(p_n)_*\pi_1(S^1,1)=n\mathbf{Z}\subset\pi_1(S^1,1)=\mathbf{Z}.\]
Note that, in this example,
\[Deck(S^1,p_n)\cong\mathbf{Z}/(n\mathbf{Z})=\pi_1(X,x)/p_*\pi_1(Y,y)\]
where \(p\colon Y\to X\) is the covering space.

*(5.40)* Is this a general phenomenon? No: the subgroup
\(p_*\pi_1(Y,y)\subset\pi_1(X,x)\) might not be a normal subgroup, in
which case the quotient \(\pi_1(X,x)/p_*\pi_1(Y,y)\) doesn't even make
sense. However, in the case where \(p_*\pi_1(Y,y)\) is a normal
subgroup of \(\pi_1(X,x)\), it will turn out to be true that
\(Deck(Y,p)=\pi_1(X,x)/p_*\pi_1(Y,y)\). We single out these covers
with a definition:

*(9.56)* Moreover, \(F\in Deck(Y,p)\) is determined uniquely by its
value on a particular point \(y\in p^{-1}(x)\): indeed, any covering
transformation is determined uniquely by its value at a
point. Therefore, if the action of \(Deck(Y,p)\) is transitive on
\(p^{-1}(x)\), then \(Deck(Y,p)\) is as big as it could possibly be:
if it were any bigger, two of its elements would necessarily agree at
some point \(y\in p^{-1}(x)\), and hence would be equal.

*(12.29)* In our earlier examples, the deck group acts transitively.
For \(p_n\colon S^1\to S^1\), the \(n\)th roots of unity act
transitively by rotation on \(p_n^{-1}(1)\), which it itself the set
of \(n\)th roots of unity. For \(p\colon\mathbf{R}\to S^1\), the group
\(\mathbf{Z}\) acts transitively by translation on
\(p^{-1}(1)=\mathbf{Z}\).

*(14.03)* Below is a covering space of the figure 8 which is not
normal.

*(19.34)* By assumption, \(Deck(Y,p)\) acts transitively, so
there exists a deck transformation \(F\colon Y\to Y\) such that
\(F(y)=\sigma_\beta(y)\) and deck transformations are
homeomorphisms, so
\[(F_*\colon\pi_1(Y,y)\to\pi_1(Y,\sigma_\beta(y))\] is an
isomorphism. We also know that \(p\circ F=p\), so \(p_*\circ
F_*=p_*\). Putting all this together gives:
\begin{align*}
\beta p_*\pi_1(Y,y)\beta^{-1}&= p_*\pi_1(Y,\sigma_\beta(y))\\
&=p_*(F_*(\pi_1(Y,y)))\\
&=p_*\pi_1(Y,y).
\end{align*}
Since this holds for all \(\beta\in\pi_1(X,x)\), this shows that
\(p_*\pi_1(Y,y)\subset\pi_1(X,x)\) is normal.

*(21.50)* Conversely, suppose that \(p_*\pi_1(Y,y)\) is a
normal subgroup. Let \(y'\) be another point in \(p^{-1}(x)\) and
let \(\tilde{\beta}\) be a path in \(Y\) from \(y\) to \(y'\). Then
\(y'=\sigma_\beta(y)\) and \[p_*\pi_1(Y,y')=\beta
p_*\pi_1(Y,y)\beta^{-1}.\] Since \(p_*\pi_1(Y,y)\) is normal, we get
\(p_*\pi_1(Y,y')=p_*\pi_1(Y,y)\), so the existence theorem for
covering isomorphisms implies that there exists a covering
transformation \(F\colon Y\to Y\) such that \(F(y)=y'\). Therefore
\(Deck(Y,p)\) acts transitively on \(p^{-1}(x)\).

- Consider the non-normal covering space example. What is the subgroup
\(p_*\pi_1(Y,y)\subset\pi_1(X,x)\)? Can you see it is not normal
without appealing to the lemma we used?

- Previous video:
**8.02 Covering transformations**. - Next video:
**8.04 Deck group**. - Index of all lectures.