# 8.01 Lifting criterion

## Video

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

## Notes

### Lifting criterion

(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.

(0.23) Let $$p\colon Y\to X$$ be a covering space and let $$T$$ be a path-connected, locally path-connected space. Suppose there is a continuous map $$f\colon T\to X$$. Given a point $$t\in T$$ (with $$f(t)=x$$) and a point $$y\in p^{-1}(x)$$, there exists a unique continuous lift $$\tilde{f}\colon T\to Y$$ with $$\tilde{f}(t)=y$$ if and only if $f_*\pi_1(T,t)\subset p_*\pi_1(Y,y).$

(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.

### Applications

(21.10) Any continuous map $$F\colon S^2\to T^2$$ is nullhomotopic.

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 higher homotopy groups of a space $$X$$ are the groups $$\pi_n(X)$$ of based homotopy classes of maps $$S^n\to X$$. The same reasoning shows that $$\pi_m(X)$$ is trivial for all $$m\geq 2$$ whenever $$X$$ admits a contractible covering space.

## Pre-class questions

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