Difference between revisions of "Open and closed maps"

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(Created page with "Due to the parallel definitions and other similarities there is little point to having separate pages. ==Definition== ===Open map=== A <math>f:(X,\mathcal{J})\rightarrow (Y,\...")
 
m (→‎Importance with respect to the quotient topology)
 
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{{Todo|Come back after writing about how the quotient topology is the strongest topology - rather than doing that proof here}}
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Using [[Quotient topology#Theorems|the first theorem]], automatically we see that {{M|K}} is weaker than the quotient topology (that is {{M|\mathcal{K}\subseteq\mathcal{Q} }})
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To show equality we need only to show {{M|\mathcal{Q}\subseteq\mathcal{K} }}
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'''Open case:'''
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: Suppose that {{M|f}} is a continuous open map
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: Let {{M|A\in\mathcal{Q} }} be given
 +
: we need to show this {{M|\implies A\in\mathcal{K} }} then we use the [[Implies and subset relation]] to get what we need.
 +
: This means {{M|f^{-1}(A)\in\mathcal{J} }} but as {{M|f}} is an open mapping {{M|\forall U\in\mathcal{J}[f(U)\in\mathcal{K}]}} so:
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: {{M|f(f^{-1}(A))\in\mathcal{K} }} (it is open in {{M|Y}})
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:: '''Warning: '''we have not shown that {{M|A\in\mathcal{K} }} yet!
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: Because {{M|f}} is [[Surjection|surjective]] we know {{M|1=f(f^{-1}(A))=A}} (as for all things in {{M|A}} there must be ''something'' that maps to them, by surjectivity)
 +
: So we conclude {{M|A\in \mathcal{K} }}
 +
 
 +
: So {{M|A\in\mathcal{Q}\implies A\in\mathcal{K} }} or {{M|\mathcal{Q}\subseteq\mathcal{K} }}
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: Combining this with {{M|\mathcal{K}\subseteq\mathcal{Q} }} we see:
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:: {{M|1=\mathcal{K}=\mathcal{Q} }}
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 +
 
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'''Closed case:'''
 +
: Suppose {{M|f}} is a continuous closed map
 +
: Let {{M|A\in\mathcal{Q} }} be given
 +
: we need to show this {{M|\implies A\in\mathcal{K} }} then we use the [[Implies and subset relation]] to get what we need.
 +
: Obviously, because {{M|f}} is continuous {{M|f^{-1}(A)\in\mathcal{J} }} - that is to say {{M|f^{-1}(A)}} is open in {{M|X}}
 +
: That means that {{M|X-f^{-1}(A)}} is closed in {{M|X}} - just by definition of an [[Open set|open set]]
 +
: Since {{M|f}} is a closed map, {{M|f(X-f^{-1}(A))}} is also closed
 +
: BUT! {{M|1=f(X-f^{-1}(A))=Y-A}} by [[Surjection|surjectivity]] (as the set {{M|X}} take away ALL the things that map to points in {{M|A}} will give you all of {{M|Y}} less the points in {{M|A}})
 +
: As {{M|f(X-f^{-1}(A))}} is closed and equal to {{M|Y-A}}, {{M|Y-A}} is closed
 +
: this means {{M|Y-(Y-A)}} is open, but {{M|1=Y-(Y-A)=A}} so {{M|A}} is open.
 +
: That means {{M|A\in\mathcal{K} }}
 +
:
 +
: We have shown {{M|A\in\mathcal{Q}\implies A\in\mathcal{K} }} and again by the [[Implies and subset relation]] we see {{M|\mathcal{Q}\subseteq\mathcal{K} }}
 +
: Combining this with {{M|\mathcal{K}\subseteq\mathcal{Q} }} we see:
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:: {{M|1=\mathcal{K}=\mathcal{Q} }}
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 +
 
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This completes the proof.
 
{{End Proof}}
 
{{End Proof}}
 
{{End Theorem}}
 
{{End Theorem}}

Latest revision as of 08:00, 8 April 2015

Due to the parallel definitions and other similarities there is little point to having separate pages.

Definition

Open map

A [math]f:(X,\mathcal{J})\rightarrow (Y,\mathcal{K})[/math] (which need not be continuous) is said to be an open map if:

  • The image of an open set is open (that is [math]\forall U\in\mathcal{J}[f(U)\in\mathcal{K}][/math])

Closed map

A [math]f:(X,\mathcal{J})\rightarrow (Y,\mathcal{K})[/math] (which need not be continuous) is said to be a closed map if:

  • The image of a closed set is closed


TODO: References - it'd look better


Importance with respect to the quotient topology

The primary use of recognising an open/closed map comes from recognising a Quotient topology, as the following theorem shows

Theorem: Given a map [math]f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})[/math] that is continuous, surjective and an open or closed map, then the topology [ilmath]\mathcal{K} [/ilmath] on [ilmath]Y[/ilmath] is the same as the Quotient topology induced on [ilmath]Y[/ilmath] by [ilmath]f[/ilmath]

Note: the intuition behind this theorem is fairly obvious, if [ilmath]f[/ilmath] is an open map then it takes open sets to open sets, but the quotient topology is precisely those sets in [ilmath]Y[/ilmath] whose preimage is open in [ilmath]X[/ilmath]


Using the first theorem, automatically we see that [ilmath]K[/ilmath] is weaker than the quotient topology (that is [ilmath]\mathcal{K}\subseteq\mathcal{Q} [/ilmath])

To show equality we need only to show [ilmath]\mathcal{Q}\subseteq\mathcal{K} [/ilmath]

Open case:

Suppose that [ilmath]f[/ilmath] is a continuous open map
Let [ilmath]A\in\mathcal{Q} [/ilmath] be given
we need to show this [ilmath]\implies A\in\mathcal{K} [/ilmath] then we use the Implies and subset relation to get what we need.
This means [ilmath]f^{-1}(A)\in\mathcal{J} [/ilmath] but as [ilmath]f[/ilmath] is an open mapping [ilmath]\forall U\in\mathcal{J}[f(U)\in\mathcal{K}][/ilmath] so:
[ilmath]f(f^{-1}(A))\in\mathcal{K} [/ilmath] (it is open in [ilmath]Y[/ilmath])
Warning: we have not shown that [ilmath]A\in\mathcal{K} [/ilmath] yet!
Because [ilmath]f[/ilmath] is surjective we know [ilmath]f(f^{-1}(A))=A[/ilmath] (as for all things in [ilmath]A[/ilmath] there must be something that maps to them, by surjectivity)
So we conclude [ilmath]A\in \mathcal{K} [/ilmath]
So [ilmath]A\in\mathcal{Q}\implies A\in\mathcal{K} [/ilmath] or [ilmath]\mathcal{Q}\subseteq\mathcal{K} [/ilmath]
Combining this with [ilmath]\mathcal{K}\subseteq\mathcal{Q} [/ilmath] we see:
[ilmath]\mathcal{K}=\mathcal{Q}[/ilmath]


Closed case:

Suppose [ilmath]f[/ilmath] is a continuous closed map
Let [ilmath]A\in\mathcal{Q} [/ilmath] be given
we need to show this [ilmath]\implies A\in\mathcal{K} [/ilmath] then we use the Implies and subset relation to get what we need.
Obviously, because [ilmath]f[/ilmath] is continuous [ilmath]f^{-1}(A)\in\mathcal{J} [/ilmath] - that is to say [ilmath]f^{-1}(A)[/ilmath] is open in [ilmath]X[/ilmath]
That means that [ilmath]X-f^{-1}(A)[/ilmath] is closed in [ilmath]X[/ilmath] - just by definition of an open set
Since [ilmath]f[/ilmath] is a closed map, [ilmath]f(X-f^{-1}(A))[/ilmath] is also closed
BUT! [ilmath]f(X-f^{-1}(A))=Y-A[/ilmath] by surjectivity (as the set [ilmath]X[/ilmath] take away ALL the things that map to points in [ilmath]A[/ilmath] will give you all of [ilmath]Y[/ilmath] less the points in [ilmath]A[/ilmath])
As [ilmath]f(X-f^{-1}(A))[/ilmath] is closed and equal to [ilmath]Y-A[/ilmath], [ilmath]Y-A[/ilmath] is closed
this means [ilmath]Y-(Y-A)[/ilmath] is open, but [ilmath]Y-(Y-A)=A[/ilmath] so [ilmath]A[/ilmath] is open.
That means [ilmath]A\in\mathcal{K} [/ilmath]
We have shown [ilmath]A\in\mathcal{Q}\implies A\in\mathcal{K} [/ilmath] and again by the Implies and subset relation we see [ilmath]\mathcal{Q}\subseteq\mathcal{K} [/ilmath]
Combining this with [ilmath]\mathcal{K}\subseteq\mathcal{Q} [/ilmath] we see:
[ilmath]\mathcal{K}=\mathcal{Q}[/ilmath]


This completes the proof.


See also

References

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