# Lifting of a continuous map through a covering map

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## Contents

## Definition

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space, let [ilmath](E,\mathcal{ H })[/ilmath] be a covering space of [ilmath]X[/ilmath], with covering map [ilmath]p:E\rightarrow X[/ilmath]. Then^{[1]}^{[2]}:

- if we're given a continuous map [ilmath]f:Y\rightarrow X[/ilmath] for an arbitrary topological space [ilmath](Y,\mathcal{ K })[/ilmath] such that:
- there exists a continuous map, [ilmath]\varphi:Y\rightarrow E[/ilmath], such that [ilmath]p\circ\varphi\eq f[/ilmath] (the diagram on the right commutes)
- then [ilmath]\varphi[/ilmath] is called a
*lifting*of [ilmath]f[/ilmath] (through [ilmath]p[/ilmath])

- then [ilmath]\varphi[/ilmath] is called a

- there exists a continuous map, [ilmath]\varphi:Y\rightarrow E[/ilmath], such that [ilmath]p\circ\varphi\eq f[/ilmath] (the diagram on the right commutes)

**Caveat:**I am not sure if we require [ilmath]Y[/ilmath] be a connected topological space or not^{[Note 1]} - however if we do then the unique lifting property applies.

## See next

## Notes

- ↑ Author notes for future use:
- Lee requires that the disjoint elements of an even covering which are homeomorphic to the open neighbourhood be themselves connected topological (sub)spaces
- Lee then requires the covering space be connected and locally path-connected
- He does not require [ilmath]Y[/ilmath] to be connected. But for the unique lifting property he does require [ilmath]Y[/ilmath] to be connected

- Gamelin and Greene do not require anything be connected until they reach lifts (here) where [ilmath]Y[/ilmath] must be connected.

- Lee requires that the disjoint elements of an even covering which are homeomorphic to the open neighbourhood be themselves connected topological (sub)spaces