Subject: Composites of covering maps Date: Tue, 3 Aug 2004 14:42:22 +0100 From: "Ronald Brown" To: "Don Davis" reply to r.brown@bangor.ac.uk With regard to the question from Sang-Eon Han, the following is exercise 1.9 of Chapter 9 of the 1968 and 1988 editions of my book on Topology. \label{Ex:9.1:9} Let $p : \tilde{X} \to X$ be a connected covering map which is non-trivial (i.e., it is not a homeomorphism). Let $Y = X \times X \times X \times \dots$ be the countable product of $X$ with itself, and let $\tilde{X}^{n}$ be the $n$-fold product of $\tilde{X}$ with itself. Let $\tilde{Y}_{n} = \tilde{X}^{n} \times Y$. Define $p_{n} : \tilde{Y}_{n} \to Y$ by $(\tilde{x}_{1}, \dots, \tilde{x}_{n}, x_{1}, x_{2}, \dots) \mapsto (p \tilde{x}_{1}, \dots, p \tilde{x}_{n}, x_{1}, x_{2}, \dots)$. Prove that $p_{n}$ is a covering map. Let $$\tilde{Z} = \coprod_{n\geqslant 1} \tilde{Y}_{n}$$ and let $Z$ be the countably infinite topological sum of $Y$ with itself. Let $q = \coprod p_{n} : \tilde{Z} \to Z$ and let $r : Z \to Y$ be the obvious projection. Prove that $q$ and $r$ are covering maps but that the composite $r q$ is not a covering map. If my memory is right, the example came from the book by Hilton and Wylie. Exercise 3 of 9.5 studies this further, using a condition on $Y$. Ronnie Brown ----- Original Message ----- > > Subject: question > Date: Tue, 3 Aug 2004 10:51:32 +0900 > From: "???" > > I would like to know the literature(books or paper) somewhere with > relation to the following well-known fact : > > Let p: X \to Y and q: Y \to Z be covering maps. > We know that if q^{-1}(z) is finite for each z in Z, then the > composition map of the maps p and q is also a covering map and > further, > if q^{-1}(z) is not finite the statement aboe fails. > > Sincerely yours, > > Sang-Eon Han >