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# No fixed point Homeomorphism (TIFR 2013 problem 21)

Question:

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).

August 16, 2017