Notes:Universal property of the quotient topology

Problem

I really want to do this using equivalence relations rather than a mapping, I also need to work out exactly which theorem is which.

Theorems

John M. Lee

Characteristic property of the quotient topology

[ilmath]\xymatrix{ X \ar[d]_q \ar@{->}[dr]^{f\circ q} & \\ Y \ar[r]_f & Z}[/ilmath]
Diagram

• Suppose [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] are topological spaces
• Suppose [ilmath]q:X\rightarrow Y[/ilmath] is a quotient map, then:
• For any topological space [ilmath](Z,\mathcal{ L })[/ilmath] and any map [ilmath]f:X\rightarrow Z[/ilmath]
• [ilmath]f[/ilmath] is continuous [ilmath]\iff[/ilmath] [ilmath]f\circ q[/ilmath] is continuous

Passing to the quotient

[ilmath]\xymatrix{ X \ar[d]_q \ar[dr]^{f} & \\ Y \ar@{.>}[r]_{\bar{f} } & Z}[/ilmath]
Diagram

• Suppose [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] are topological spaces
• Suppose [ilmath]q:X\rightarrow Y[/ilmath] is a quotient map, then:
• Suppose [ilmath](Z,\mathcal{ L })[/ilmath] is a topological space and [ilmath]f:X\rightarrow Z[/ilmath] is any map that is constant on the fibres of [ilmath]q[/ilmath]
• Then there exists a unique continuous map [ilmath]\bar{f}:Y\rightarrow Z[/ilmath] such that [ilmath]f=\bar{f}\circ q[/ilmath]

Munkres

Theorem 22.2

This is just "passing to the quotient" but stated slightly differently.

Theory

Lee's characteristic property does say "you can stick any topology you like, if there's a continuous map from [ilmath]\frac{X}{\sim} [/ilmath] to it, then there's a continuous map from the parent to it" which is "universal"

Passing to the quotient uses that and is something else. Very close concept, but still distinct.