(3.44) If there is a lift \(\tilde{f}\) then the criterion holds because for any loop \(\gamma\) in \(T\) based at \(t\), \begin{align*} f_*[\gamma]&=(p\circ\tilde{f})_*[\gamma]\\ &=p_*(\tilde{f}_*[\gamma])\in p_*\pi_1(Y,y), \end{align*} as \(\tilde{f}\circ\gamma\) is based at \(y\).

(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.

(6.57) The fact that \(\tilde{f}\) is a lift of \(f\) is clear from the construction: the path \(\widetilde{f\circ\alpha}\) is a lift of \(f\circ\alpha\), so \(p(\tilde{f}(t'))=p(\widetilde{f\circ\alpha}(1))=f(\alpha(1))=f(t')\).

(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 \(F\).