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TIFR 2013 problem 21 | No fixed point Homeomorphism

Try 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).