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I'm doing some basic topology and analysis in the book "Tensor Analysis on Manifolds" by Bishop and Goldberg, and a homeomorphism is defined as a bijection, $f: X \to Y$, such that $f$ and $f^{-1}$ are both continuous.

This made me wonder whether there exists a continuous bijection whose inverse is not continuous. So far, I haven't been able to find such a function.

ted
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2 Answers2

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Consider the interval $[0,2\pi)$ and map it to the unit circle. So, $f(t) = (\cos t, \sin t)$.

Now, clearly this is a continuous bijection since closer points on the interval are mapped to closer points in the circle.

On the other hand, the inverse is not continuous. If you consider the point $(1,0)$ on the circle, its inverse contains points near both $0$ and $2\pi$.

ThePortakal
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Yes, the identity function from $\mathbb R$ with discrete topology to $\mathbb R$ with the usual topology.

Kavi Rama Murthy
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  • Okay, next question. Can such an example exist if we restrict consideration to a continuous bijection from (possibly a subset of) $\mathbb{R}^n$ to (possibly a subset of) $\mathbb{R}^n$, both with the standard Euclidean topology? And what if we restrict the domain and range to be path-connected subsets of $\mathbb{R}^n$? – ted Dec 02 '19 at 07:59
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    @ted In that case it is known that $f^{-1}$ is also continuous, but the result is quiet non-trivial. It is easy in the case of $\mathbb R$. – Kavi Rama Murthy Dec 02 '19 at 08:01