# Evenly covered by a continuous map

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Books conflict on whether or not the [ilmath]V_\alpha[/ilmath] must be connected.
• By the homeomorphism requirement, the [ilmath]V_\alpha[/ilmath] and [ilmath]U[/ilmath] have the same number of connected components, that is clear.
• Lee requires the [ilmath]V_\alpha[/ilmath] to be connected, that means that [ilmath]U[/ilmath] must be connected then. Munkres doesn't. Alec (talk) 21:54, 24 February 2017 (UTC)

## Definition

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

• [ilmath]U[/ilmath] is evenly covered by [ilmath]f[/ilmath] 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 [ilmath]U[/ilmath][Note 1]
2. [ilmath]\forall\alpha,\ \beta\in I[\alpha\neq\beta\implies V_\alpha\cap V_\beta\eq\emptyset][/ilmath] - the [ilmath]V_\alpha[/ilmath] are pairwise disjoint
3. [ilmath]\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 } [/ilmath][ilmath]\text{homeomorphic} [/ilmath][ilmath]\text{ via }f\vert_{V_\alpha}^\text{Im}\big][/ilmath] - the restriction onto its image of [ilmath]f[/ilmath] for each [ilmath]V_\alpha[/ilmath] is a homeomorphism onto [ilmath]U[/ilmath]
• Note that this would mean [ilmath]\forall\alpha\in I[f(V_\alpha)\eq U][/ilmath]

### Derived constraints

TODO: Notes stage, don't trust yet unless you prove
• [ilmath]U[/ilmath] and [ilmath]V_\alpha[/ilmath] 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"