Q:

Question about a proof in GreenBerg-Harper algebraic topology

I am currently self-studying Greenberg-Harper algebraic topology. In the proof of the covering homotopy theorem, the book makes the following claim without justification:
Let $p:E\rightarrow X$ be a covering map. $Y$ be any topological space. Let $F:Y\times \mathbb{I}\rightarrow X$ be a continuous function. For any $y\in Y$, there exists an open set $N_y$ containing $y$ and real numbers $0=t_0\leq t_1\leq t_2\leq...\leq t_n=1$ such that for every $i\in \{0,2,...,n-1\}$, $F[N_y\times [t_i,t_{i+1}]]\subseteq U$ for some evenly covered open subset of $X$.
Question:  Why does this hold ?
Thank you a lot.

A:

Fix $y \in Y$. For every $s \in [0,1]$, choose an evenly covered neighbourhood $U_{y,s}$ of $F(y,s)$. By continuity, there is a neighbourhood $V_{y,s}$ of $y$ and a connected open neighbourhood $W_s$ of $s$ such that $F(V_{y,s}\times W_s) \subset U_{y,s}$.
$[0,1]$ is compact, so there are finitely many $s_1,\dotsc,s_n$ such that $[0,1] \subset \bigcup\limits_{k=1}^n W_{s_k}$. Let $N_y = \bigcap\limits_{k=1}^n V_{y,s_k}$. Now choose $t_0 = 0$, $t_i \in W_{s_i} \cap W_{s_{i+1}}$, and $t_n = 1$.
It's a typical compactness argument. Every point $(y,s)$ has a neighbourhood $O_{s}$ that is mapped into an evenly covered open set, the union of these $O_s$ is a neighbourhood of $\{y\}\times[0,1]$, hence there is a neighbourhood $N_y$ of $y$ such that $N_y\times [0,1] \subset \bigcup O_s$. Here you then also need to cut the interval $[0,1]$ into short enough pieces such that every piece is mapped into an evenly covered open set.

