Hatcher Exercise 2.2.2


Theorem: If is a continuous map, then there is an with or .


Proof: It suffices to show that at least one of and must have a fixed point. If has a fixed point, then clearly it satisifes . If has a fixed point, then it satisfies .

To that end, suppose towards contradiction that and both have no fixed points. Since has no fixed points, we have . At the same time, since has no fixed points, we have as well, but:

and we have a contradiction. So at least one of and must have a fixed point.

Corollary: Every map has a fixed point.


Proof: is a cover of , and by the lifting lemma lifts to a map . If is the canonical cover, then we can lift the whole map to a map , as in the diagram:

Then, there is an with or . Then, , since . Thus, has a fixed point .

We end by demonstrating a map from that does not have a fixed point. First consider a linear transformation without eigenvectors in . A perfectly good example is the map defined by the matrix:

This map has eigenvalues .

We identify with the set of lines through the origin in (i.e. one dimensional subspaces), and show that induces a well-defined map. Indeed, has full rank so it certainly sends one-dimensional subspaces to other one-dimensional subspaces, so this is a well defined map . Since it has no real eigenvalues, none of these one-dimensional subspaces are mapped onto themselves, so this map has no fixed points.