Evenly covered by a continuous map

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Books conflict on whether or not the $V_\alpha$ must be connected.

• By the homeomorphism requirement, the $V_\alpha$ and $U$ have the same number of connected components, that is clear.
• Lee requires the $V_\alpha$ to be connected, that means that $U$ must be connected then. Munkres doesn't. Alec (talk) 21:54, 24 February 2017 (UTC)

Definition

Let $(C,\mathcal{ K })$ and $(X,\mathcal{ J })$ be topological spaces and let $f:C\rightarrow x$ be a continuous map. Let $U\in\mathcal{J}$ be given, so $U$ is an open set of $(X,\mathcal{ J })$, we say that:

• $U$ is evenly covered by $f$ if:
  • $$\exists \{V_\alpha\}_{\alpha\in I}\subseteq\mathcal{K}$$ - there exists an arbitrary collection of open sets - such that:
    1. $$f^{-1}(U)\eq\bigcup_{\alpha\in I}V_\alpha$$ - the union of this family is the entire pre-image of $U$
    2. $\forall\alpha,\ \beta\in I[\alpha\neq\beta\implies V_\alpha\cap V_\beta\eq\emptyset]$ - the $V_\alpha$ are pairwise disjoint
    3. $\forall\alpha\in I\big[\big(V_\alpha\cong_{f\vert_{V_\alpha}^\text{Im}:V_\alpha\rightarrow f(V_\alpha)} U\big)\text{ are }$$\text{homeomorphic}$$\text{ via }f\vert_{V_\alpha}^\text{Im}\big]$ - the restriction onto its image of $f$ for each $V_\alpha$ is a homeomorphism onto $U$
       • Note that this would mean $\forall\alpha\in I[f(V_\alpha)\eq U]$

Derived constraints

TODO: Notes stage, don't trust yet unless you prove

• $U$ and $V_\alpha$ have the same number of connected components - follows by homeomorphism part. Must do some theorem about homeomorphisms and components! But I don't want to say "number"