Sequence of diffeomorphisms $F_n=f_n\circ\cdots\circ f_2 \circ f_1 $ Given a sequence of diffeomorphisms of class $C^1$ onto $\mathbb{R}^m$ $$f_n:\mathbb{R}^m\longrightarrow \mathbb{R}^m$$ 
Suppose that the sequence of diffeomorphisms $$F_n=f_n\circ\cdots\circ f_2 \circ f_1$$ converges uniformly to application $$F:\mathbb{R}^m\longrightarrow \mathbb{R}^m$$
How prove that $F$ is a diffeomorphism ?
Any hints would be appreciated.
 A: I believe this result is false as stated. Let $m=1$. Let $f_1(x)=\arctan x$, $f_k(x)=x/2$ for all $k\ge 2$. $F_n$ is certainly a diffeo, but the uniform limit is the zero function.
Edit:
I can give an example of a uniformly convergent sequence of diffeomorphisms $f_k\colon \mathbb R\to\mathbb R$ (onto) whose limit is not a diffeomorphism. But I still don't have a counterexample to the problem as stated. Let $f_k(x) = x^3 + (1/k)\arctan x$. Then $f_k\to f$ uniformly, where $f(x)=x^3$.
Next edit:
OK, I believe I have a candidate for a counterexample (with $m=1$) to the problem as stated. Let $f_k(x)=\int_0^x dt/(k+|t|)$. Each $f_k$ is monotone, has range all of $\Bbb R$, and satisfies $f_k'(0)=1/k$. Then $F_n(0)=0$, $F_n'(0)=1/n!$, and I believe that $F_n\to F$ uniformly, where $F'\ge 0$ and $F'(0)=0$, so $F$ is not a diffeomorphism. 
Last (I hope) edit:
Pretty clearly, in my last example, the convergence is not uniform. However, stealing @Daniel Fischer's clever idea (why couldn't I have thought of it?), why not go back to my second paragraph, set $g_1 = f_1$, $g_2=f_2\circ f_1^{-1}$, $\dots$, $g_k = f_k\circ f_{k-1}^{-1}$, $\dots$. Then $F_k = g_k\circ g_{k-1} \circ \dots \circ g_1 = f_k$ converges uniformly to $F(x)=x^3$, which is not a diffeomorphism.
A: It need not be a diffeomorphism.
Consider the homeomorphisms $H_k \colon \mathbb{R} \to \mathbb{R}$, given by
$$
H_k(x) = \begin{cases}
\frac{1}{k}\cdot x &, \lvert x\rvert \leqslant 1\\
\frac1k + \left(2 - \frac1k\right)(x-1) &, 1 \leqslant x \leqslant 2\\
-\frac1k + \left(2 - \frac1k\right)(x+1) &, -2\leqslant x \leqslant -1\\
x &, \lvert x\rvert \geqslant 2
\end{cases}
$$
It is easy to verify that the $H_k$ converge uniformly to
$$
N(x) = \begin{cases}
0 &, \lvert x\rvert \leqslant 1\\
2(x-1) &, 1 \leqslant x \leqslant 2\\
2(x+1) &, -2 \leqslant x \leqslant -1\\
x &, \lvert x\rvert \geqslant 2
\end{cases}$$
($\lvert H_k(x) - N(x)\rvert \leqslant \frac1k$).
Now let $\varphi \in C_c^\infty(\mathbb{R})$ with $\varphi \geqslant 0$, $\varphi(-x) = \varphi(x)$, $\operatorname{supp}\varphi \subset \left[-\frac{1}{4},\frac14\right]$ and $\int_\mathbb{R} \varphi(x)\,dx = 1$.
Let $D_k = \varphi \ast H_k$. Then $D_k$ is a $C^\infty$-diffeomorphism of $\mathbb{R}$ for all $k \in \mathbb{Z}^+$ (it's $C^\infty$ by general properties of convolutions, it's a diffeomorphism since $D_k' = \varphi' \ast H_k = \varphi \ast H_k'$, where $H_k'$ is the almost-everywhere defined derivative of $H_k$. Since $H_k' \geqslant \frac1k$, we have $D_k' \geqslant \frac1k$, so $D_k$ is indeed a diffeomorphism).
The sequence $D_k$ converges uniformly to $\varphi \ast N$ - even all derivatives of the $D_k$ converge uniformly to the corresponding derivative of $\varphi\ast N$ - but $(\varphi\ast N)\big\lvert_{\left[-\frac34,\frac34\right]} \equiv 0$, so $\varphi\ast N$ is not a diffeomorphism.
Now set $f_1 = D_1 = \operatorname{id}$, and for $k > 1$, set $f_k = D_k \circ D_{k-1}^{-1}$. Then $F_n = D_n$ converges uniformly (with all derivatives) to $\varphi\ast N$, which is not a diffeomorphism.
