Bijective continuous non-homeomorphism (TIFR 2013 problem 28)



Let \(f:X\to Y \) be a continuous map between metric spaces. If \(f\) is a bijection, then its inverse is also continuous.


No need to go into weird spaces. Start with familiar spaces, find out counterexamples.


One way to search for counterexamples is finding spaces that can not be homeomorphic but surely has a one way continuous function. For example, start with a not-connected space and have its image connected. Then you can’t go back, because connected sets will go to a connected set only (under continuous functions). Or similar thoughts can be made about compact sets.

To illustrate, I will give one such example.

Consider \( I=[0,1] \). And \(A=[-0.5, 0]\) \(B=(1,2]\). Translation by \(0.5\) takes \(A\) to \([0,0.5]\) and translating \(B\) by \(-0.5\) gives \((0.5,1]\). Therefore, we get \(I\) as the whole image of \(f\) where \(f(x)=x+0.5\), \(x\in A \) and \(f(x)=x-0.5\), \(x\in B\). This \(f\) is a continuous function from \(A \cup B \) to \(I\). This can be checked by sequence criteria, or pasting lemma as well. Note that for pasting lemma we need \(A\) and \(B\) be closed. But they truly are closed because here we are considering the space \(A \cup B \) as a subspace of \(\mathbb{R}\). That means, \(B\) can be written as a closed set in \(\mathbb{R}\) intersection the space \(A \cup B \).

Now that we know \(f\) is continuous, f is bijective is easy to check. And further, by our starting point, we know that inverse function can not be continuous since our domain space \(A \cup B \) is not connected (or compact) while I is.



Leave a Reply

Your email address will not be published. Required fields are marked *