If you have a surjective function from the unit square, [0,1] X [0,1] onto a bigger square, such as [0,3] X [0,3], will there always be a fixed point under any injection of the unit square into the big square (i.e. will there always be [itex]x[/itex] s.t. [itex]f(x)=i(x)[/itex], where [itex]i[/itex] is an injection?)(adsbygoogle = window.adsbygoogle || []).push({});

It seems to me that there will, because the same idea that holds in proving that a continuous function from a disk to a disk seems to hold here, although somewhat different.

Include the unit square into the bigger square (however you want), and then, assuming there exist no fixed points, draw a line starting at [itex]f(x)[/itex] (of the image of the unit square) through [itex]x[/itex] onto the boundary of the smaller included square. Then, for all points, [itex]x[/itex], not in the image of the smaller square, let [itex]F(x)[/itex]=[itex]x[/itex]. Then let [itex]F(x)[/itex] be defined on the entire large square by the two conditions I just outlined. I *think* this is continuous by the pasting lemma, but I'm not completely sure. Then, if it is, we've defined a retraction of the large square onto the square annulus. So, this implies that the homomorphism induced by inclusion of the square annulus in the square is injective, but we know this cannot be the case, as the square has trivial fundamental group and the square annulus does not.

Does this make sense? My main concerns are the continuity of the function defined above and whether or not any injection of the smaller square into the bigger square will allow me to create this "square-annulus type" object.

If not, are there other conditions I can impose to make it true. Would this be true if I simply let the injection be the inclusion mapping?

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# Fixed Point Theorems

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