View source for Example:A bijective and continuous map that is not a homeomorphism

{{Stub page|grade=C|msg=Maybe add others? Tidy the page up a bit!}}
==Example==
Let {{M|X:\eq[0,1)\subset\mathbb{R} }} with the [[subspace]] topology and consider: {{M|f:\mathbb{R}\rightarrow\underbrace{\mathbb{S}^1}_{\subseteq\mathbb{C} } }} by {{M|f:r\mapsto e^{2\pi j x} }}, then:
* {{M|f}} is clearly continuous
* {{M|f}} is also easily seen to be bijective
But {{M|f}} is not a [[homeomorphism]]{{rITTMJML}}, that is specifically: {{M|f^{-1}:\mathbb{S}^1\rightarrow [0,1)}} is not [[continuous]].

Clearly this is equivalent to {{M|f}} (not) being an [[open map]] as {{M|(f^{-1})^{-1}(U)\eq f(U)}}, which we require to be open for all {{M|U}} open in {{M|[0,1)}}
==Proof==
{{Requires proof|grade=A|msg=Would be good to do [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 23:04, 22 February 2017 (UTC)}}
==References==
<references/>
{{Example Of|Topology}}