existence of a unique lift of homotopy (Hatcher's Algebraic Topology)

Question

I am currently reading Hatcher's algebraic topology, and I am confused about his proof. I would appreciate a little help.

Let $p$ be a covering map of $\widetilde{X}$ onto $X$. Given a map $F: Y \times I \rightarrow X$ and a map $\widetilde{F}: Y \times \{0\} \rightarrow \widetilde{X}$ lifting $F|Y \times \{0\}$, then there is a unique map $\widetilde{F} : Y \times I \rightarrow \widetilde{X}$ lifting $F$ and restricting to the given $\widetilde{F}$ on $Y \times \{0\}$.

Proof: Pick $y_0 \in Y$ and for each $t \in I$, there exists a product neighborhood $N_t \times (a_t,b_t)$ such that $F(N_t \times (a_t,b_t))$ is contained in an evenly covered neighborhood of $F(y_0, t)$. By compactness of $\{y_0\} \times I$, finitely many products $N_t \times (a_t,b_t)$ cover $\{y_0\}\times I$. This implies that we can choose a single neighborhood $N$ of $y_0$ and a partition $0 = t_0<t_1<\cdots<t_m = 1$ of $I$ so that for each $i$, $F(N \times [t_i,t_{i+1}])$ is contained in an evenly covered neighborhood $U_i$.

Everything is okay until the bolded next. It can happen that $(a_i, b_i) \cap (a_{i+1}, b_{i+1}) = \phi$ (anything in $[b_t, a_{t+1}]$ is covered by other open intervals in the cover), and there's no guarantee that $F(N \times [t_i, t_{i+1}]) \subset U_i$ for some evenly covered $U_i$. What kind of choice do I have to make for the finite subcover so that $F(N \times [t_i, t_{i+1}]) \subset U_i$?

Thank you very much in advance.

Answer 1

Note this first: you have an open cover $\{y_0\}\times I$ if and only if you have a collection of product spaces of the type $A_i\times B_i,$ where $\{A_i\}$ is an open cover of $\{y_0\}$ and $\{B_i\}$ is an open cover of $I$.

So, we have two collections here as well - $\{N_t\}$ and $T=\{(a_t,b_t)\}.$ Let $N=\bigcap_tN_t;$ this set is clearly open as the collection is finite and contains $\{y_0\}.$ Since $T$ covers $I,$ there is some set, say $(a_0,b_0)$ which contains $0$ and also for any set $A\in T$ there is another set $B\in T$ such that $A\cap B\neq \emptyset$ (if there was, then a limit point of $A$ would not be contained in the union of all sets of $T$, and $T$ would not cover $I$).

Suppose $(a_1,b_1)$ is the set such that $(a_1,b_1)\cap(a_0,b_0)\neq\emptyset.$ Let $t_0=0$ and $t_1\in (a_1,b_0);$ then we have that $F(N\times[t_0,t_1])\subseteq U_\alpha$ for some evenly covered set $U_\alpha.$ Now, inductively for a finite number of steps, you can find the partition you were seeking after, which eventually ends at $t_m=1.$

Answer 2

First observe that Hatcher is a little imprecise. He should correctly say that there exists a product neighborhood $N_r \times J_r$ of $(y_0,r)$ such that

1. $J_r = (a_r,b_r)$ with $0 < a_r < b_r < 1$ for $0 < r < 1$

2. $J_0 = [0,b_0)$ with $0 < b_0 < 1$

3. $J_1 = (a_1,1]$ with $0 < a_1 < 1$

4. $F(N_r \times J_r)$ is contained in an evenly covered neighborhood of $F(y_0, r)$.

It seems that Hatcher does not want to argue with Lebesgue numbers. If he would do that, he could choose a Lebesgue number for the open cover $J_r$, $r \in I$, of $I$ which would immediately show that we can take $t_i = i/n$ for a sufficiently large $n$. Then $[t_i,t_{i+1}] \subset J_{r_i}$ for suitable $r_i \in I$. Taking $N = \bigcap N_{r_i}$, we see that $F(N \times [t_i,t_{i+1}])$ is contained in an evenly covered neighborhood of $F(y_0, r_i)$.

Instead he argues that finitely many $N_{r_i} \times J_{r_i}$, $i =1,\ldots,m$, cover $\{y_0\} \times I$. That is correct, but I would suggest a little modification. For $0 < r < 1$ choose smaller open intervals $J'_r = (a'_r,b'_r)$ containing $r$ such that $J''_r = [a'_r,b'_r] \subset (a_r,b_r)$. Similarly let $J'_0 = [0,b'_0)$ such that $J''_0 = [0,b'_0] \subset [0,b_0)$ and $J'_1 = (a'_1,1]$ such that $J''_1 = [a'_1,1] \subset (a_1,1]$.

Then finitely many $N_{r_i} \times J'_{r_i}$, $i =1,\ldots,m$, cover $\{y_0\} \times I$. W.l.o.g. we may assume that $r_1 < r_2 < \ldots < r_m$. Note that we must have $r_1 = 0, r_m = 1$. Taking $N = \bigcap N_{r_i}$, we see that $F(N \times J''_{r_i})$ is contained in an evenly covered neighborhood of $F(y_0, r_i)$.

Let $T$ denote the finite set of all endpoints of the closed intervals $J''_{r_i}$. Clearly $0,1 \in T$. Write $T = \{t_0,\ldots,t_n\}$ with $0 = t_0 < t_1 < \ldots < t_n = 1$. Each $[t_j,t_{j+1}]$ is contained in some $J''_{r_i}$: We have $t_1 \le b'_0$ because $b'_0 \in T$. Thus $[t_0,t_1] = [0,t_1] \subset J_0 = J''_{r_1}$. Next consider $t_j$ with $0 < j < n$. Note that then $t_j < 1$. We have $t_j \in J'_{r_i}$ for some $i$. The point $t_j$ cannot be the right endpoint $b'_{r_i}$ of $J''_{r_i}$. This shows that $t_{j+1} \le b'_{r_i}$, thus $[t_j,t_{j+1}] \subset J''_{r_i}$.

This completes the proof of Hatcher's theorem.