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{{Stub page|grade=A*|msg=Doing some work while I've got a bit of time, majority written right before bed, so needs checking. This is also "the theorem" of connectedness}}
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==Statement==
Let {{Top.|X|J}} and {{Top.|Y|K}} be [[topological spaces]] and let {{M|f:X\rightarrow Y}} be a ''[[continuous]]'' [[map]]. Then, for any {{M|A\in\mathcal{P}(X)}}, we have{{rITTBM}}:
* If {{M|A}} is a {{link|connected subset|topology}} of {{Top.|X|J}} then {{M|f(A)}} is connected subset in {{Top.|Y|K}}
==Proof==
We do this proof by [[contrapositive]], that is noting that: {{M|1=(A\implies B)\iff((\neg B)\implies(\neg A))}}, as such we will show:
* if {{M|f(A)}} is {{link|disconnected|topology}} then {{M|A}} is disconnected
Let us begin:
*Suppose {{M|f(A)}} is disconnected, and write {{M|(f(A),\mathcal{K}_{f(A)})}} for the [[topological subspace]] on {{M|f(A)}} of {{Top.|Y|K}}.
** Then there exists {{M|U,V\in\mathcal{K}_{f(A)} }} such that {{M|U}} and {{M|V}} [[disconnected (topology)|disconnect]] {{M|f(A)}}<ref group="Note">Recall {{M|U}} and {{M|V}} are said to disconnect {{M|f(A)}} if:
# they are [[disjoint]],
# they are both [[non-empty]], and
# {{M|1=U\cup V=f(A)}}</ref>
*** Notice that {{M|f^{-1}(U)}} and {{M|f^{-1}(V)}} are [[disjoint]]
**** PROOF HERE
*** Notice also that {{M|A\subseteq f^{-1}(U)\cup f^{-1}(V)}} (''we may or may not have: {{M|1=A=f^{-1}(U)\cup f^{-1}(V)}}, as there might exist elements outside of {{M|A}} that map into {{M|f(A)}} notice'')
**** PROOF HERE
{{Caution|We do not know whether or not {{M|f^{-1}(U)}} or {{M|f^{-1}(V)}} are open in {{M|X}} (and they may form a strict superset of {{M|A}} so cannot be open in {{M|A}})}}
{{Requires proof|grade=A|msg=Working on it}}
===OLD WORKINGS===
* OLD
** WORK
*** {{M|f^{-1}(U)}} and {{M|f^{-1}(V)}} are disjoint by {{M|f}} being a function and their union contains {{M|A}} (but could be bigger than it, as we might not have {{M|1=f^{-1}(f(A))=A}} of course!)
*** We apply the right-hand part of:
**** [[a subset of a topological space is disconnected if and only if it can be covered by two non-empty-in-the-subset and disjoint-in-the-subset sets that are open in the space itself]]
This completes the proof
==Notes==
<references group="Note"/>
==References==
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