Homotopy of composition function implies homotopy of individual functions

Question

If $f$, $g$ are homotopic homeomorphisms and $f\circ f^{-1} \sim g\circ g^{-1}$ then is $f^{-1}$ homotopic to $g^{-1}$ ? I know the works if the individual functions are homotopic but I'm not sure about the backwards direction?

Thank you.

Answer 1

Certainly. Note that $f$, $g$ need not even be homeomorphisms. All we require is that they are homotopic homotopy equivalences. As pointed out in the comments, the condition that $g\circ g^{-1}\simeq f\circ f^{-1}$ is vacuous, since both of these composites are homotopic to $id_Y$ (assuming $f,g:X\rightarrow Y$). In fact we have

$$f^{-1}\circ g\simeq f^{-1}\circ f\simeq id_X,$$ so it follows that

$$f^{-1}\simeq f^{-1}\circ (g\circ g^{-1})=(f^{-1}\circ g)\circ g^{-1}\simeq id_X\circ g^{-1}=g^{-1}.$$