# Evenly covered by a continuous map

From Maths

(Redirected from Evenly covers (topology))

**Stub grade: A****

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

find another source! I have encountered many variations

**Provisional page grade: A***

This page is provisional

This page is provisional and the information it contains may change before this notice is removed (in a backwards incompatible way). This usually means the content is from one source and that source isn't the most formal, or there are many other forms floating around. It is on a to-do list for being expanded.The message provided is:

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^{[1]}^{[2]}:- [math]\exists \{V_\alpha\}_{\alpha\in I}\subseteq\mathcal{K} [/math] - there exists an arbitrary collection of open sets - such that:
- [math]f^{-1}(U)\eq\bigcup_{\alpha\in I}V_\alpha[/math] - the union of this family is the entire pre-image of [ilmath]U[/ilmath]
^{[Note 1]} - [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
- [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]

- [math]f^{-1}(U)\eq\bigcup_{\alpha\in I}V_\alpha[/math] - the union of this family is the entire pre-image of [ilmath]U[/ilmath]

- [math]\exists \{V_\alpha\}_{\alpha\in I}\subseteq\mathcal{K} [/math] - there exists an arbitrary collection of open sets - such that:

### 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"

## Notes

- ↑ Note that both sides of the [ilmath]\eq[/ilmath] are open: