The following question I have tried to solve and have also tried to find reference for. A reference would be appreciated.
Suppose that $f_i, g_i :\mathbb{R^n} \to \mathbb{R^n}$ are continuous functions such that $f_i \circ g_i =g_i \circ f_i = \text{id}_\mathbb{R^n}$ for each $i \in \mathbb{N}$. Let $\Omega_{\{f_i\}}$ be the set of points for which $\{f_i\}_i$ converges pointwise to some function, say $f$. Similarly, let $\Omega_{\{g_i\}}$ be the set of points for which $\{g_i\}_i$ converges pointwise to some function, say $g$. Then show that $$f \circ g = \text{id}_{\Omega_{\{g_i\}}}$$ and $$g \circ f = \text{id}_{\Omega_{\{f_i\}}}$$ specifically that such compositions make sense (i.e. that $\Omega_{\{f_i\}} \subseteq g(\Omega_{\{g_i\}})$ and $\Omega_{\{g_i\}} \subseteq f(\Omega_{\{f_i\}})$.
Counterexample on $\mathbb R^1:$ For $m=1,2,\dots,$ define
$$f_m(x)= \begin{cases}x,& x\le 0\\x^m,& 0<x<1\\ x,& x\ge 1\\\end{cases}$$
Then each $f_m$ is a homeomorphism of $\mathbb R$ onto $\mathbb R.$ Set $g_m= (f_m)^{-1},$ i.e.,
$$g_m(x)= \begin{cases}x,& x\le 0\\x^{1/m},& 0<x<1\\ x,& x\ge 1\\\end{cases}$$
Check that $f_m$ converges pointwise everywhere to the function $f$ that is $0$ on $(0,1)$ and is the identity elsewhere. Similarly, $g_m$ converges pointwise everywhere to the function $g$ that is $1$ on $(0,1)$ and is the identity elsewhere. We have $f\circ g=1$ on $(0,1),$ and $g\circ f=0$ on $(0,1),$ showing the desired conclusion does not hold.