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.

  • 2
    $\begingroup$ @user2369284 American University in Cairo. I don't mind answering questions not related to my post, but I prefer to keep things not related to my post in the chat room/email ( in order to stick to the rules of SE) $\endgroup$ – Amr Dec 29 '13 at 20:24

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.

  • $\begingroup$ I only read your second paragraph and immediately knew what to do. Thanks a lot Daniel. Are you a mathematician, no? $\endgroup$ – Amr Dec 29 '13 at 22:28
  • 1
    $\begingroup$ Not professionally. But passionately ;) $\endgroup$ – Daniel Fischer Dec 29 '13 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.