(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 $F$.