Brouwer fixed-point theorem: Difference between revisions

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Proof outline: punctuation
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== Proof outline ==
 
A full proof of the theorem would be too long to reproduce here, but the following paragraph outlines a proof omitting the difficult part. It is hoped that this will at least give some idea why the theorem might be expected to be true. Note that the boundary of ''D''<sup>&nbsp;''n''</sup> is ''S''<sup>&nbsp;''n''<tt>-</tt>1</sup>, the (''n''<tt>-</tt>1)-[[sphere]] .
 
Suppose ''f''&nbsp;:&nbsp;''D''<sup>&nbsp;''n''</sup>&nbsp;<tt>-></tt>&nbsp;''D''<sup>&nbsp;''n''</sup> is a continuous function that has no fixed point. The idea is to show that this leads to a contradiction. For each ''x'' in ''D''<sup>&nbsp;''n''</sup>, consider the straight line that passes through ''f''(''x'') and ''x''. There is only one such line, because ''f''(''x'') &ne; ''x''. Following this line from ''f''(''x'') through ''x'' leads to a point on ''S''<sup>&nbsp;''n''<tt>-</tt>1</sup>. Call this point ''g''(''x''). This gives us a continuous function ''g''&nbsp;:&nbsp;''D''<sup>&nbsp;''n''</sup>&nbsp;<tt>-></tt>&nbsp;''S''<sup>&nbsp;''n''<tt>-</tt>1</sup>. This is a special type of continuous function known as a retraction: every point of the [[codomain]] (in this case ''S''<sup>&nbsp;''n''<tt>-</tt>1</sup>) is a fixed point of the function.