Lifting of a continuous map through a covering map

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:

Very important page, make sure that it is done with the utmost generality!

Definition

$\xymatrix{ & & E \ar[d]_p \\ Y \ar[rr]^f \ar@{-->}[rru]^\varphi & & X }$
If $\varphi$ exists such that all maps are continuous and the diagram commutes then $\varphi$ is a lifting of $f$ through $p$

Let $(X,\mathcal{ J })$ be a topological space, let $(E,\mathcal{ H })$ be a covering space of $X$, with covering map $p:E\rightarrow X$. Then:

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

Caveat:I am not sure if we require $Y$ be a connected topological space or not - however if we do then the unique lifting property applies.

See next

• Unique lifting property
• Path lifting property
• Homotopy lifting property
• Monodromy theorem