# Brouwer's Fixed Point Theorem

by Ed Quanta
Tags: brouwer, fixed, point, theorem
 P: 297 I have read the following : "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.
 Sci Advisor HW Helper P: 9,398 Notice the word retraction in the statement, given a map f with no fixed points one can construct a retraction by projection along the rays x to f(x) to boundary. **f is not a retraction**, and it is not important what f does to the boundary, or any other point. Rereading your post, I also see that you think lack of existence of retraction implies no fixed point, which is not the argument at all instead the argument is based on no fixed point implies existence of retraction, # since there are no retractions of the alleged type..

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