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Learn MoreTry this problem from TIFR 2013 problem 21 based on no fixed homeomorphism.

**Question: TIFR 2013 problem 21**

*True/False?*

Every homeomorphism of the 2-sphere to itself has a fixed point.

*Hint:*

\(z= -z\) implies \(z=0\)

**Discussion:**

2-sphere means \( S^2=\left \{(x,y,z)\in\mathbb{R}^3 | x^2+y^2+z^2=1 \right \} \).

i.e, \( S^2=\left \{v\in\mathbb{R}^3 | ||v||=1 \right \} \).

\(||.||\) denotes the usual 2-norm (Euclidean norm).

Let us try \(f:S^2\to S^2\) defined by \(f(v)=-v\) for all \(v\in\mathbb{R}^3\).

The only vector in \(\mathbb{R}^3\) that is fixed by \(f\) is 0, which doesn't lie in \(S^2\).

We hope \(f\) turns out to be a homeomorphism.

\(||f(v)-f(w)||=||-v+w||=||v-w||\). So f is in fact Lipshitz function, so continuous.

\(f(f(v)=v\) for all \(v\in\mathbb{R}^3\). Therefore, \(f\) itself is inverse of \(f\). Which proves that \(f\) is bijective (since, inverse function exists) and homeomorphism (inverse is also continuous).

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