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I am trying a new approach to an already-solved problem, but I need help to see if I'm on point. Munkres Chapter 53, question 6 [abridged] asks, given a covering map $p: E \to B$:

Show that "if $B$ is completely regular, then so is $E$."

Now, first and foremost, I have solved this problem, but I did it via explicit construction of a function $f$ that 'forces' $E$ to be completely regular, given that $B$ is. My question is regarding another possible approach to this proof:

By Munkres's Topology 2nd Ed. Theorem 34.3, a space is completely regular if and only if it is homeomorphic to $[0,1]^{\Omega}$ for some index $\Omega$. Because $B$ is completely regular, we know it is homeomorphic to such an arbitrary product of the closed interval, say for the purposes of this argument that it's product index is $J$. Call this homeomorphism $h$, i.e. $h: B \to [0,1]^J$. Let $E = \bigcup_{\alpha \in K} E_{\alpha}$ be the indexed disjoint union of 'slices' homeomorphic to $B$ under the restriction of $p$ to each slice. But since we know that each 'slice' of $E$ is mapped homeomorphically onto $B$, i.e. for each $\alpha \in K, \space p|_{\alpha} : E_{\alpha} \to B$ is a homeomorphism, and since a composition of homeomorphisms is itself a homeomorphism, each 'slice' $E_{\alpha}$ is homeomorphic to the same closed unit interval product, i.e. let $g_{\alpha} := h \circ p|_{\alpha}: E_{\alpha} \to [0,1]^J$ is a homeomorphism. This shows that each 'slice' is completely regular. Yet we cannot extend $g_{\alpha}$ to a homeomorphism $g: E \to [0,1]^J$ because if we 'generalize' the functions $p|_{\alpha}$, we get the function $p$ (our covering map), which is not necessarily a homeomorphism.

Since we cannot extend $g_{\alpha}$ implicitly, all we need to show then is that a disjoint union of completely regular spaces is completely regular. If so, then the hypothesis is fulfilled, as $E = \bigcup_{\alpha \in K} E_{\alpha}$. It seems like this proposition (almost trivially) answers itself, because every subset of $E$ is either entirely contained in a completely regular slice of some $E_{\alpha}$, hence it is completely regular, or it is the arbitrary union of disjoint completely regular spaces, i.e. by the arguments above, we already know $U \cap E_{\beta}$ is completely regular for any $\beta \in K$. But since each such subspace is completely regular, and the union of these subspaces are the entire set in question, if we can 'build' such a $U$ arbitrarily as the composition of completely regular spaces, does complete regularity of $U$ hold?

Embedded questions notwithstanding, am I on the right track? Does this suffice for a proof? I'm worried that the 'disjoint union' does not preserve complete regularity for an arbitrary $E$ (i.e. I'm stupidly overlooking the obvious...)

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  • $\begingroup$ Completely regular means it is homeomorphic to a subspace of the product, not necessarily the whole product. $\endgroup$ – Matt Samuel May 23 '15 at 1:45
  • $\begingroup$ Yes, thank you for the clarification. This discrepancy would not, however, change the 'meat and potatoes' of the first part of the proof. For simply take $h: B \to U \subset [0,1]^J$, then the requisite $g_{\alpha}$ would still be a homeomorphism from $E_{\alpha}$ to $U \subset [0,1]^J$, i.e. each 'slice' would still be completely regular. $\endgroup$ – Alex May 23 '15 at 11:53
  • $\begingroup$ Every space is a union of points which are completely regular. $\endgroup$ – Matt Samuel May 23 '15 at 12:45
  • $\begingroup$ I don't follow. Can you explain? Points in themselves are not completely regular, and if your claim is true, then every topological space has a completely regular subset?! This can't be true; take any Hausdorff space that isn't regular or a T-1 space that isn't Hausdorff, etc... Again, what I'm missing in the above proof is a rigorous demonstration that $E_{\alpha}$ is completely regular $\forall \alpha \in K$ $\implies$ $E$ is completely regular. $\endgroup$ – Alex May 23 '15 at 13:28
  • $\begingroup$ Single points are completely regular, trivially. The point is that this arbitrary union idea without regard to the topology won't work. It would be much easier to use the continuous real valued function definition $\endgroup$ – Matt Samuel May 23 '15 at 13:32

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