a counterexample of covering projection

Question

Let $X=S^1\times S^1\times\cdots$ be a countable product of 1-spheres and for $n\geq 1$ let $\tilde{X_n}=R^n\times S^1\times S^1\times\cdots$.Define $p_n:\tilde{X_n}\rightarrow X$ by $$p_n(t_1,\cdots,t_n,z_1,z_2,\cdots)=(e^{t_1},\cdots,e^{t_n},z_1,z_2,\cdots)$$ Let $\tilde{X}=\vee \tilde{X_n}$ and define $p:\tilde{X}\rightarrow X$ so that $p\mid \tilde{X_n}=p_n$.The components of $\tilde{X}$ are the spaces $\tilde{X_n}$ and the map $p \mid \tilde{X_n}=p_n:\tilde{X_n}\rightarrow X$ is a covering projection.However,$p$ is not a covering projection,because no open subset of $X$ is evenly covered by $p$.Can someone interpret it in more details?

Answer 1

Let $U\subseteq X$ be a basic open set. Recall that in the product topology, the basic open sets are of the form $$U = U_1\times U_2\times U_3\times\ldots \times U_N\times S^1\times S^1 \ldots$$ where each $U_i\subseteq S^1$ is open.

We will show that $U$ is not evenly covered, meaning $p^{-1}(U)$ is not a disjoint union of sets each mapped homeomorphically onto $U$. Note that since every open set is built out of basic ones, and because shrinking and evenly covered open set keeps it evenly covered, this will show that no open subset of $U$ is evenly covered, so $p$ is not a covering map.

Given $U$ as above, by shrinking each $U_i$, we may assume each $U_i$ is connected. Now, assume for a contradiction that $p^{-1}(U) = \coprod_\alpha V_\alpha$ for which $p|_{V_\alpha}$ is a homeomorphism onto $U$.

Since $U$ is connected, each $V_\alpha$ is connected, so each $V_\alpha$ is a subset of $X_n = \mathbb{R}^n\times S^1\ldots\times S^1$ for some $n$. Let's focus on $n = N+1$ and consider only a particular $V_\alpha$s contained in $X_{N+1}$. We must have $p|_{V_\alpha} = p_{N+1}|_{V_\alpha}$ a homeomorphism onto $U$. Let $V\subseteq \mathbb{R}$ be the projection of $V_\alpha$ onto the $N+1$st coordinate. Then $p_{N+1}|_V:V\subseteq \mathbb{R}\rightarrow S^1$ must be a homeomorphism, which is clearly impossible.

This contradiction implies $U$ is not evenly covered, so $p$ is not a covering map.