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

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## Example

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

- [ilmath]f[/ilmath] is clearly continuous
- [ilmath]f[/ilmath] is also easily seen to be bijective

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

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

## Proof

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