Let $X$ be a Hausdorff paracompact space, and let $\pi : Y \to X$ be a covering space (i.e. locally on $X$ it looks like the projection $F \times U \to U$ where $F$ is discrete). Is it true that $Y$ is paracompact as well?
Thanks
Let $X$ be a Hausdorff paracompact space, and let $\pi : Y \to X$ be a covering space (i.e. locally on $X$ it looks like the projection $F \times U \to U$ where $F$ is discrete). Is it true that $Y$ is paracompact as well?
Thanks
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It is hard to obtain a rigorous proof from the condition “it looks like”, but I guess that the space $Y$ is paracompact because it has a locally finite closed cover of sets homeomorphic to closed subsets of $X$. If the latter hold then the space $Y$ is paracompact because a closed subspace of a Hausdorff paracompact space is paracompact and by [Eng, Theorem 5.1.34], if a topological space $Y$ has a locally finite closed cover consisting of Hausdorff paracompact subspaces, then $Y$ is itself Hausdorff paracompact.
Theorem 5.1.34 follows from invariance of paracompactness under perfect mappings, along with Theorems 5.1.30 and 3.7.22.
I recall that in [Eng] all paracompact spaces are assumed to be Hausdorff.
Theorem 5.1.30. The sum $\bigoplus_{s\in S} X_s$ is paracompact if and only if all spaces Xs are paracompact.
Proof. If the sum $\bigoplus_{s\in S} X_s$ is paracompact, then all the $X_s$'s are paracompact by virtue of the last corollary.
Conversely, if all the $X_s$'s are paracompact, then for every open cover $\mathcal V = \{V_t\}_{t\in T}$ of the sum $\bigoplus_{s\in S} X_s$ the family $\bigcup_{s\in S}\mathcal A_s$ – where $\mathcal A_s $ is a locally finite open refinement of the open cover $\{X_s\cap V_t\}_{t\in T}$ of the subspace $X_s$ – is an open locally finite cover of the sum $\bigoplus_{s\in S} X_s$ that refines $\mathcal V$.
Theorem 3.7.22. Let $\mathcal P$ be a (finitely) additive topological property which is an invariant of perfect mappings. If a space $X$ can be represented as the union of a locally finite (finite) family $\{X_s\}_{s\in S}$ of closed subspaces each of which is a Hausdorff space and has the property $\mathcal P$, then $X$ also has the property $\mathcal P$.
Proof. The mapping $\nabla_{s\in S} i_x: \bigoplus_{s\in S} X_s \to X$ is perfect. $\square$
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.