View source for Open and closed maps

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
{{Begin Theorem}}
Theorem: Given a map <math>f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})</math> that is [[Continuous map|continuous]], [[Surjection|surjective]] and an open or closed map, then the topology {{M|\mathcal{K} }} on {{M|Y}} is the same as the [[Quotient topology]] induced on {{M|Y}} by {{M|f}}
{{Begin Proof}}
'''Note:''' the intuition behind this theorem is fairly obvious, if {{M|f}} is an open map then it takes open sets to open sets, but the quotient topology is precisely those sets in {{M|Y}} whose preimage is open in {{M|X}}


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} }})

To show equality we need only to show {{M|\mathcal{Q}\subseteq\mathcal{K} }}

'''Open case:'''
: Suppose that {{M|f}} is a continuous open 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.
: 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:
: {{M|f(f^{-1}(A))\in\mathcal{K} }} (it is open in {{M|Y}})
:: '''Warning: '''we have not shown that {{M|A\in\mathcal{K} }} yet! 
: 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} }}
: Combining this with {{M|\mathcal{K}\subseteq\mathcal{Q} }} we see:
:: {{M|1=\mathcal{K}=\mathcal{Q} }}


'''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:
:: {{M|1=\mathcal{K}=\mathcal{Q} }}


This completes the proof.
{{End Proof}}
{{End Theorem}}

==See also==
* [[Continuous map]]
* [[Topology]]

==References==
<references/>

{{Definition|Topology}}