# TIFR 2013 Problem 28 Solution -Bijective continuous non-homeomorphism

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

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