This is an exercise from Fundamental Groups & Covering Spaces by Lima:
If two homeomorphisms $f,g:X\to Y$ are homotopic, then their inverses are also homotopic.
Below is my thought for a solution, but I am not sure about it.
Since $f\simeq g$ then there exists a continuous map $H:X\times I\to Y$ such that $H(x,0)=f(x)$ and $H(x,1)=g(x)$. Moreover, this defines a family of maps $H_t:X\to Y$ where $H_t (x)=H(x,t)$ for all $x\in X$.
To define a homotopy $G:Y\times I\to X$ from $f^{-1}$ to $g^{-1}$, I think it should be something similar to the following: $$G(y,t)=H_t^{-1}(y),$$
for each $y\in Y$, and where $H_t^{-1}$ is the inverse of $H_t$.
Clearly then $G(y,0)=H_0^{-1}(y)=f^{-1}(y)$ and $G(y,1)=H_1^{-1}(y)=g^{-1}(y)$.
What I am not sure about is if each $H_t$ for $0<t<1$ is actually invertible. Since $f$ and $g$ are homeomorphisms, then of course they are invertible, but what about each transitory map in the homotopy?
Best Answer
Assume $f,g:X\rightarrow Y$ are homeomorphisms and are homotopic.
Let $H:X\times I\rightarrow Y$ be a homotopy from $f$ to $g$.
We aim to show that $f^{-1},g^{-1}:Y\rightarrow X$ are homotopic.
The map $G:Y\times I\rightarrow X$, $G(y,t)=f^{-1}(H(g^{-1}(y),1-t)))$ is the desired homotopy