 TIFR 2013 Problem 28 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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## Problem:True/False?

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

## Hint:

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

## Discussion:

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.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuity, Bijective, Continuity of inverse function
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert