Continuity of inverse for the uniform convergence Let $X$ be a compact metric space, and denote by $G$ the group of all homeomorphisms of $X$ endowed with the topology of uniform convergence. Is it true that the inverse mapping $h \mapsto h^{-1}$ is continuous for this topology ?
 A: Yes, it is continuous. To see this, assume $f_{n}\to f$ uniformly.
If $f_{n}^{-1}\to f^{-1}$ uniformly would not hold, there would be
some $\varepsilon>0$ and a sequence $\left(x_{n}\right)_{n\in\mathbb{N}}$
in $X$ such that
$$
d\left(f_{n}^{-1}\left(x_{n}\right),f^{-1}\left(x_{n}\right)\right)\geq\varepsilon
$$
for all $n\in\mathbb{N}$. As $X$ is compact and metric, there is
a convergent subsequence $x_{n_{k}}\to x$ for some $x\in X$. Let
us write $\left(x_{n}\right)_{n}$ again for this subsequence for
simplicity.
Then
\begin{eqnarray*}
d\left(f\left(f_{n}^{-1}\left(x_{n}\right)\right),x_{n}\right) & = & d\left(f\left(f_{n}^{-1}\left(x_{n}\right)\right),f_{n}\left(f_{n}^{-1}\left(x_{n}\right)\right)\right)\\
 & \leq & d_{{\rm sup}}\left(f,f_{n}\right)\to0
\end{eqnarray*}
and hence
$$
d\left(f\left(f_{n}^{-1}\left(x_{n}\right)\right),x\right)\leq d\left(f\left(f_{n}^{-1}\left(x_{n}\right)\right),x_{n}\right)+d\left(x_{n},x\right)\to0.
$$
But (using continuity of $f^{-1}$), this implies $f_{n}^{-1}\left(x_{n}\right)\to f^{-1}\left(x\right)$
and hence
$$
\varepsilon\leq d\left(f_{n}^{-1}\left(x_{n}\right),f^{-1}\left(x_{n}\right)\right)\leq d\left(f_{n}^{-1}\left(x_{n}\right),f^{-1}\left(x\right)\right)+d\left(f^{-1}\left(x\right),f^{-1}\left(x_{n}\right)\right)\to0,
$$
a contradiction.
