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Background: (From the definition of Homeomorphism in Topology, by Munkres )

Let $X$ and $Y$ represent topological spaces and $$f \colon X \longrightarrow Y$$ be a bijective function. Then if $f$ and $f^{-1}$ are continuous functions, $f$ is a homeomorphism.


(1) Would it suffice to say that if $f$ is a bijective, continuous function then $f$ is a homeomorphism?

(1.b) If not, when does it ever happen that $f$ is a continuous, bijective function and $f^{-1}$ is not also a continuous function?

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Related: – Jonas Meyer Jan 20 '13 at 7:18
up vote 4 down vote accepted

The answer to question $1$ is no. Let $X$ be a set and $\tau$, $\tau'$ two topologies on $X$ with $\tau'$ strictly coarser than $\tau$. Then $\operatorname{id} : (X, \tau) \to (X, \tau')$ is a continuous bijection, but its inverse is not continuous.

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I think @JonasMeyer is correct, but I see what you mean. Thank you. – providence Jan 20 '13 at 7:26
This is a nice class of examples. It arises for example when considering the norm and weak topologies on a Banach space. – Jonas Meyer Jan 20 '13 at 7:31
  1. It's not sufficient. There are counterexamples.
  2. If the map fails to be open, for example: $\theta\mapsto e^{\theta i}$ for $\theta\in[0,2\pi)$. The domain is not compact and the range is, so it is not a homeomorphism. But it is easy to check that this is a continuous bijection.
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