Difference between revisions of "Notes:Quotient topology"
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* {{M|1=x\sim y\iff q(x)=q(y)}} and that's where this comes from | * {{M|1=x\sim y\iff q(x)=q(y)}} and that's where this comes from | ||
===Passing to the quotient=== | ===Passing to the quotient=== | ||
| − | This is very similar to [[quotient (function)|the quotient of a function]].<br/> | + | <div style="float:right;margin:0.1em;"> |
| − | *Let X and Z be [[topological space|topological spaces]] | + | {| class="wikitable" border="1" |
| + | |- | ||
| + | |<center><span style="font-size:1.2em;"><m>\xymatrix{ | ||
| + | X \ar[d]_q \ar[dr]^f & \\ | ||
| + | Y \ar@{.>}[r]_{\bar{f}} & Z | ||
| + | }</m></span></center> | ||
| + | |- | ||
| + | ! Passing to the quotient | ||
| + | |} | ||
| + | </div>This is very similar to [[quotient (function)|the quotient of a function]].<br/> | ||
| + | * Let X and Z be [[topological space|topological spaces]], | ||
| + | * let {{M|q:X\rightarrow Y}} be a quotient map, | ||
| + | * let {{M|f:X\rightarrow Z}} be ''any'' continuous mapping such that {{M|1=q(x)=q(y)\implies f(x)=f(y)}} | ||
| + | Then | ||
| + | * There exists a unique continuous map, {{M|\bar{f}:Y\rightarrow Z}} such that {{M|1=f=\bar{f}\circ q}} | ||
==Munkres== | ==Munkres== | ||
Revision as of 15:52, 21 April 2016
Note to readers: the page quotient topology as it stands right now (Monday, 8/Dec/2025 at 22:44) is an embarrassment to me. However before I can clean it up I must unify it. I've been using it for almost 2 years now though I promise! Gosh this is embarrassing.
According to John M. Lee
Let [ilmath]\sim[/ilmath] denote an equivalence relation, let [ilmath](X,\mathcal{J})[/ilmath] be a topological space. We get a map, [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] that takes [ilmath]\pi:x\rightarrow[x][/ilmath]
- The quotient topology on [ilmath]\frac{X}{\sim} [/ilmath] is the finest such that [ilmath]\pi[/ilmath] is continuous
Let [ilmath]\mathcal{K} [/ilmath] denote a topology on [ilmath]\frac{X}{\sim} [/ilmath], then we may define [ilmath]\mathcal{K} [/ilmath] as:
- [ilmath]\mathcal{K}:=\{U\in\mathcal{P}(\frac{X}{\sim})\ \vert\ \pi^{-1}(U)\in\mathcal{J} \}[/ilmath], that is:
- [ilmath]U\in\mathcal{P}(\frac{X}{\sim})[/ilmath] is open if [ilmath]\pi^{-1}(U)[/ilmath] is open in [ilmath]X[/ilmath] - we get "only if" by going the other way. I must make a page about how definitions are "iff"s
Note: more than one book is very clear on "[ilmath]U\in\mathcal{P}(\frac{X}{\sim})[/ilmath] is open in [ilmath]\frac{X}{\sim} [/ilmath] if and only if [ilmath]\pi^{-1}(U)\in\mathcal{J} [/ilmath], not sure why they stress it so.
Quotient map
A map between two topological spaces [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath] is a quotient map if:
- It is surjective
- The topology on [ilmath]Y[/ilmath] ([ilmath]\mathcal{K} [/ilmath]) is the quotient topology that'd be induced on [ilmath]Y[/ilmath] by the map [ilmath]q[/ilmath]
Lee actually defines the quotient topology using maps first, then constructs the equiv relation version, but we can can define an equivalence relation as follows:
- [ilmath]x\sim y\iff q(x)=q(y)[/ilmath] and that's where this comes from
Passing to the quotient
| Passing to the quotient |
|---|
- Let X and Z be topological spaces,
- let [ilmath]q:X\rightarrow Y[/ilmath] be a quotient map,
- let [ilmath]f:X\rightarrow Z[/ilmath] be any continuous mapping such that [ilmath]q(x)=q(y)\implies f(x)=f(y)[/ilmath]
Then
- There exists a unique continuous map, [ilmath]\bar{f}:Y\rightarrow Z[/ilmath] such that [ilmath]f=\bar{f}\circ q[/ilmath]
