If $f_n \to f$ uniformly, does $f_n^{-1} \to f^{-1}$? Let $f_n \to f$ uniformly locally on $\mathbb{R}$. All these functions are from $\mathbb{R}$ to $\mathbb{R}$, are bijective and increasing and continuous. Furthermore $f_n \in C^\infty$. Does it follow that $f_n^{-1} \to f^{-1}$ uniformly locally?
(We have that $f_n$ and $f$ pass through the origin, so maybe we can have local convergence outside $\{0\}$).
Or do I need more assumptions?
 A: First, we will consider the case $f(x)=x$ for all $x$.
Consider an interval $[a,b]$, and a positive real $\epsilon$. Since $(f_n(a-1))_n$ converges to $a-1$ there is an integer $n_0$ such that $f_n(a-1)\leq a$ for $n\geq n_0$. Similarly there is an integer $n_1$ such that $f_n(b+1)\geq b$ for $n\geq n_1$. Now, the sequence $(f_n)_n$ converges uniformly on $[a-1,b+1]$, so there is an integer $n_2$ such that
$\sup_{[a-1,b+1]}|f_n(x)-x|<\epsilon$, for $n\geq n_2$. Let $N=\max(n_0,n_1,n_2)$. and consider $n\geq N$, then
$$\eqalign{
\sup_{x\in[a,b]}|f_n^{-1}(x)-x|&\leq \sup_{x\in[f_n(a-1),f_n(b+1)]}|f_n^{-1}(x)-x|\cr
&=\sup_{t\in[ a-1 , b+1 ]}|f_n^{-1}(f_n(t))-f_n(t)|\cr
&=\sup_{t\in[ a-1 , b+1 ]}|f_n(t)-t|\leq \epsilon
}
$$
This proves that $(f_n^{-1})_n$ converges uniformly to $f$ on $[a,b]$, and the desired conclusion follows in this case.
The general case follows by applying the previous case to $(f_n\circ f^{-1})_n$. the details are not difficult, using the fact that the image under $f$ or $f^{-1}$ of a compact interval is a compact interval.
Remark. If we omit the hypothesis that $f^{-1}$ exists, then the other conditions do not guarantee its existence. Indeed, 
for $n>0$ there is a $C^\infty$ function of compact support $g_n$ that satisfy the following conditions:


*

*$\forall\,x\in\mathbb{R}, \quad g_n(x)\in[0,1]$.

*$\forall\,x\in[-1,1], \quad g_n(x)=1-\frac{1}{n}$.

*$\forall\,x\notin(-1-\frac{1}{n},1+\frac{1}{n}), \quad g_n(x)=0$.


Then we define $f_n(x)=x-\int_0^xg_n(t)dt$  for $x\in\mathbb{R}$. This sequence satisfies the proposed conditions and converges uniformly on $\mathbb{R}$ to a function that is constant on $[-1,1]$.
A: I'll assume that the family $f^{-1}_n$ is equicontinuous. I don't know it this assumption can be avoided.
Let $\varepsilon>0$. Let be $y\in\Bbb R$. Let $x=f^{-1}(y)$, $y_n=f_n(x)$. Use the equicontinuity to claim that there exists some $\delta>0$ such that for every $z\in(y-\delta,y+\delta)$ and for all $n\in\Bbb N$ we have $|f_n^{-1}(y)-f_n^{-1}(z)|<\varepsilon$. We know that there exists some $n_0\in \Bbb N$ such that $|f_n(x)-f(x)|=|y_n-y|<\delta$ for $n\geq n_0$. Then
$$\begin{align}
|f_n^{-1}(y)-f^{-1}(y)|&=|f_n^{-1}(y)-x|\\
&=|f_n^{-1}(y)-f_n^{-1}(y_n)|\\
&<\varepsilon
\end{align}$$
