Is an infinite composition of bijections always a bijection?

Main Question

Suppose I have a sequence of real valued functions $f_1:X_0\rightarrow X_1,...,f_n:X_{n-1} \rightarrow X_n,...,$ and I then, with $\circ$ denoting function composition, define

$$g_n : X_0 \rightarrow X_n\;\text{by}\;g_n(x_0) = (f_n \circ ... \circ f_1)(x_0).$$

For the sake of having a notation available in case things get complicated, I declare

$$\bigcirc_{i=1}^n f_i := (f_n \circ ... \circ f_1),$$

and also

$$\left[\bigcirc_{i=1}^n f_i := (f_n \circ ... \circ f_1)\right] (x_0) := (f_n \circ ... \circ f_1)(x_0).$$

For whatever subset $X \subseteq X_0$ it is such that $g:X \rightarrow \mathbb{R}$ can be defined by $g(x) = \lim_{n\rightarrow \infty} g_n(x),$ is $g$ guaranteed to be a bijection?

Side Question

Does anyone have knowledge (and references) for people working with applying a non-constant sequence of maps as we do here? If the sequence $f_i$ is equal to some particular $f$ for all $i \in [1..],$ then $\bigcirc_{i=1}^n$ is just written $f^{i},$ the common notation for function iteration. It seems less common to apply a sequence of maps, but I've been able to do some cool stuff with this idea.