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Well, Could any one tell me how to prove this one or any reference?

Let $f$ be a continuos map on $\mathbb{R}^2$, and $S$ be a rectangular region such that as the boundary of $S$ is traversed, the net rotation of the vectors $x-f(x)$ is non zero. Then $f$ has a fixed point in $S$.

I can not understand the statement clearly also. Thank you for help.

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up vote 1 down vote accepted

What you call "net rotation" is usually called winding number of $f(x)-x$. It can be expressed e.g. as an integral (cf. ) that depends continuously on the closed path taken and assumes only integer values, hence is constant as long as the path changes continuously. Therefore, if you let the rectangle shrink to (almost) a point, the winding number of $f(x)-x$ remains non-zero. However, unless the point $x_0$ you shrink towards is a fixpoint of $f$, one can see that the winding number must be small for small rectangles (when $f(x)-x$ is nearly constant, i.e. its changes are governed by some $\varepsilon \ll |f(x_0)-x_0|$).

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could you give me more rigorious mathematical proof? – miosaki Sep 11 '12 at 8:44

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