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

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**8.02 Covering transformations**. - Index of all lectures.

*(0.00)* In this section, we will see a very general result which
tells us when we can lift a map to a covering space. This generalises
path- and homotopy-lifting.

*(4.27)* The converse is harder: if the criterion holds then we need
to construct \(\tilde{f}\). We require \(\tilde{f}(t)=y\), which is
a start, but we need to extend this to a map \(\tilde{f}\colon T\to
Y\). Given another point \(t'\in T\), pick a path \(\alpha\) from
\(t\) to \(t'\) (we can do this because \(T\) is
path-connected). Restrict \(f\) to the path \(\alpha\). This yields
a path \(f\circ\alpha\) in \(X\), so path-lifting gives us a path
\(\widetilde{f\circ\alpha}\colon[0,1]\to Y\) with initial condition
\(y\). We will define \(\tilde{f}(t')=\widetilde{f\circ\alpha}(1)\).

*(6.22)* We need to check:

- that this is a lift of \(f\),
- that it is the unique lift of \(f\) satisfying the condition \(\tilde{f}(t)=y\),
- that it is well-defined,
- that it is continuous.

*(7.40)* The uniqueness of \(\tilde{f}\) follows from the
uniqueness lemma for lifts which we proved in an earlier section.

*(8.01)* To see that \(\tilde{f}\) is well-defined, pick two
different paths \(\alpha_1\) and \(\alpha_2\) from \(t\) to
\(t'\). Let \(\tilde{f}_1(t')=\widetilde{f\circ\alpha_1}(1)\) and
\(\tilde{f}_2(t')=\widetilde{f\circ\alpha_2}(1)\). We need to check
that \(\tilde{f}_1(t')=\tilde{f}_2(t')\).

Note that \(f(\alpha_1(1))=f(\alpha_2(2))=f(t')\). The loop \(\gamma:=\alpha_2^{-1}\cdot\alpha_1\) is a loop in \(T\) based at \(t\) and its image under \(f\) is a loop in \(X\) based at \(x\). By assumption, \(f_*[\gamma]\in p_*\pi_1(Y,y)\), which means that there is a loop \(\delta\) in \(Y\) based at \(y\) such that \(p\circ\delta=f\circ\gamma\). Now I claim that \(\tilde{f}_1(t')=\delta(1/2)=\tilde{f}_2(t')\).

*(11.12)* To see that \(\tilde{f}_1(t')=\delta(1/2)\), note
that \(\delta|_{[0,1/2]}\) is a lift of \(f\circ\alpha_1\) starting
at \(y\), so
\(\delta(1/2)=\widetilde{f\circ\alpha_1}(1)=\tilde{f}_1(t')\). Similarly,
\(\delta^{-1}|_[0,1/2]\) is a lift of \(f\circ\alpha_2\) starting at
\(y\). Therefore \(\delta^{-1}(1/2)=\widetilde{f\circ\alpha_2}\),
but \(\delta^{-1}(1/2)=\delta(1/2)\) so
\(\tilde{f}_2(t')=\widetilde{f\circ\alpha_2}(1)=\delta(1/2)\).

*(12.22)* Finally, we need to show that \(\tilde{f}\) is
continuous. Take an open set \(V\subset Y\); we want to show that
\(\tilde{f}^{-1}(V)\) is open. In fact, it's sufficient to check
this for a \(V\) running over a base of the topology of \(Y\). We
will take the base comprising the elementary sheets over elementary
neighbourhoods of the covering map.

*(14.50)* Pick a point \(t'\in T\) in \(\tilde{f}^{-1}(V)\). The
point \(f(t')\) then lives in an elementary neighbourhood \(U\subset
X\). We need to find an open set \(W\subset T\) containing \(t'\)
such that \(\tilde{f}(W)\subset V\). Take \(f^{-1}(U)\subset T\);
this is open because \(f\) is continuous and let \(W\subset
f^{-1}(U)\) be a path-connected open subset of \(f^{-1}(U)\)
containing \(t'\); I can do this because \(T\) is locally
path-connected (every open neighbourhood of a point \(t'\) contains
a path-connected open neighbourhood of \(t'\)).

*(17.08)* To show that such a \(W\) satisfies \(\tilde{f}(W)\subset
V\), pick \(t''\in W\) and a path \(\alpha\) in \(W\) from \(t'\) to
\(t''\). To define \(\tilde{f}(t'')\), we simply take
\(\widetilde{f\circ\alpha}(1)\), where \(\widetilde{f\circ\alpha}\)
is the unique lift of \(f\circ\alpha\) starting at
\(\tilde{f}(t')\). This lift is given by \(q\circ f\circ\alpha\),
where \(q\colon U\to V\) is the local inverse for the covering map
over the elementary neighbourhood \(U\). This implies that
\(\tilde{f}(t'')=q(f(\alpha(1)))\in V\), so \(\tilde{f}^{-1}(V)\)
contains \(W\). This proves that \(\tilde{f}\) is continuous.

*(19.32)* The video then recaps the proof, because it is quite
complicated.

Consider the covering map \(p\colon\mathbf{R}^2\to T^2\) coming from
the properly discontinuous action of \(\mathbf{Z}^2\) on
\(\mathbf{R}^2\). Since \(\pi_1(S^2,t)=\{1\}\) for any basepoint
\(t\in S^2\), any continuous map \(F\colon S^2\to T^2\) satisfies
\(F_*\pi_1(S^2,t)=\{1\}\subset\pi_1(T^2,F(t))\). For any \(y\in
p^{-1}(F(t))\), we have \(p_*\pi_1(\mathbf{R}^2,y)=\{1\}\), so
\[F_*\pi_1(S^1,t)\subset p_*\pi_1(\mathbf{R}^2,y),\] and the map
\(F\) lifts to a map \(\tilde{F}\colon S^2\to\mathbf{R}^2\). This
lift is nullhomotopic because \(\mathbf{R}^2\) is contractible. Let
\(H\) be such a nullhomotopy. Then \(p\circ H\) is a nullhomotopy of
\(F\).

The set of based homotopy classes of maps from \(S^2\) into another
space \(X\) is called \(\pi_2(X)\); this theorem shows that
\(\pi_2(T^2)\) is trivial. More generally, the

- Can you think of another space with trivial \(\pi_2(X)\)?
- I claim that any covering space of a CW complex is a CW complex. How
would I construct the cells of this covering space?

- Next video:
**8.02 Covering transformations**. - Index of all lectures.