Let $f, g : X \rightarrow Y$ be homeomorphisms between $X$ and $Y$. If $f$ is homotopic to $g$, then $f^{-1}$ is also homotopic to $g^{-1}.$
$\textbf{Proof}$ Let $H : X \times [0,1] \rightarrow Y$ be the homotopy between $f$ and $g$. I would like to construct a homotopy $G : Y \times [0,1] \rightarrow X.$ I am not sure but I guess it has to do with $H$. Since $f^{-1}: Y \rightarrow X$, I try something like $G(s,t) = H(g \circ f^{-1}(s),t)$ which maps from $Y \times [0,1] \rightarrow X.$ It clearly does not work. I also try $f^{-1}, g^{-1}, f \circ g^{-1}$, and try to compose some homotopy together, but it does not work. I try to think about it, but not sure where to begin.
Any suggestion ?
If you look at domain and codomain of $G$ then you will quickly find out that you don't have much choice:
$$G(s,t)=g^{-1}\bigg(H\big(f^{-1}(s),t\big)\bigg)$$
Now if $H(s,0)=f(s)$ and $H(s,1)=g(s)$ then
$$G(s,0)=g^{-1}(f(f^{-1}(s)))=g^{-1}(s)$$ $$G(s,1)=g^{-1}(g(f^{-1}(s)))=f^{-1}(s)$$