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

Stub grade: C

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

Maybe add others? Tidy the page up a bit!

Example

Let $X:\eq[0,1)\subset\mathbb{R}$ with the subspace topology and consider: $f:\mathbb{R}\rightarrow\underbrace{\mathbb{S}^1}_{\subseteq\mathbb{C} }$ by $f:r\mapsto e^{2\pi j x}$, then:

• $f$ is clearly continuous
• $f$ is also easily seen to be bijective

But $f$ is not a homeomorphism, that is specifically: $f^{-1}:\mathbb{S}^1\rightarrow [0,1)$ is not continuous.

Clearly this is equivalent to $f$ (not) being an open map as $(f^{-1})^{-1}(U)\eq f(U)$, which we require to be open for all $U$ open in $[0,1)$

Proof

Grade: A

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:

Would be good to do Alec (talk) 23:04, 22 February 2017 (UTC)