I have read the following :(adsbygoogle = window.adsbygoogle || []).push({});

"The usual proof of Brouwer's fixed-point theorem makes use of some machinery from simplical homology theory. First we establish that there does not exist a "retraction" of an n-cell onto its boundary, which is to say, there is no continuous mapping from an n-cell to its boundary such that every point on the boundary maps to itself. Given this, Brouwer's fixed-point theorem follows easily, because if x and f(x) are everywhere distinct in the n-cell, we can map each point x unambiguously to a point on the boundary by simply projecting along the ray from f(x) through x to the boundary, as illustrated below for a disk.

This mapping is continuous, and every point on the boundary maps to itself, so the mapping is a "retraction", which contradicts the fact that no retraction of an n-cell exists."

Can someone explain this to me more clearly? I am not able to follow this argument. I don't see why assuming there is no continuous mapping from an n-cell to its boundary such that every point on the boundary maps to itself would imply that x and f(x) must be everywhere distinct in the n-cell.

And also even if x and f(x) are everywhere distinct, why would this mean that every point on the boundary must map to itself? Does the boundary have to remain the same after a continuous mapping? Sorry if I am being unclear and totally missing the point.

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# Brouwer's Fixed Point Theorem

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