# Example of topological spaces with continuous bijections that are not homotopy equivalent

In one of the books on algebraic topology (I don't remember exactly which one) there was an exercise to build an example of two topological spaces having two continuous bijections between them which are not homotopy equivalent. To be honest, this exercise confuses me a little because, as I understand, each pair of homeomorphic spaces is homotopy equivalent by construction. On the other hand, the existence of bijective continuous mapping between spaces automatically provides their homeomorphism (correct me if I'm wrong). Thus, in this logic, if we have two continuous bijections between topological spaces this will inevitably lead to the homotopy equivalence between them. I guess, however, that there is a simple counterexample related to the discrete topologies which breaks such a reasoning (see, for example, this: Is a bijective homotopy equivalence with bijective homotopy inverse a homeomorphism?), but I have certain difficulties in discovering it. Is there any suggestion?

• The two continuous bijections will not be inverse to each other. So they could fail to be homeomorphisms. – GEdgar Jun 24 '15 at 13:36
• Are you asking that the two continuous bijections not be homotopy equivalent, or that the two spaces not be homotopy equivalent? Both are possible, but they'll probably require different examples. – Jack Lee Jun 24 '15 at 13:40
• Initial exercise asks about spaces, not mappings. – Alfred Rutkowski Jun 24 '15 at 13:44
• @GEdgar, thanks. So, it only remains to build such spaces that any continuous bijection from one to another has discontinuous inverse mapping. – Alfred Rutkowski Jun 24 '15 at 13:52
• @AlfredRutkowski More is needed if you want the spaces to have different homotopy types. – Peter Franek Jun 24 '15 at 13:56

The interval $[0,1)$ admits a continuous bijection to the circle (use $\exp(i 2\pi t)$ as your map). But the two spaces are not homotopy equivalent since the circle is not simply connected.