I want to construct a sort of sequence of right inverses. My question is whether the construction uses the Axiom Of Choice correctly.
Suppose I have a sequence of surjective functions $$ \ldots\stackrel{f_3}{\rightarrow}\mathbb{R}\stackrel{f_2}{\rightarrow}\mathbb{R}\stackrel{f_1}{\rightarrow}\mathbb{R} $$ Assuming the Axiom Of Choice there exists a right inverse function $g_1:\mathbb{R}\rightarrow\mathbb{R}$ such that $g_1(x)\in f_{1}^{-1}(x)$. Therefore we can define, using the Axiom Of Choice again, a function $g_2:\mathbb{R}\rightarrow\mathbb{R}$ such that $g_2(x) \in f_{2}^{-1}(g_1(x))$. Note that $f_1\circ g_1 = \text{id}_\mathbb{R}$ and $f_1\circ f_2 \circ g_2 = \text{id}_\mathbb{R}$.
In general we construct: \begin{align*} g_1(x)&\in f_{1}^{-1}(x) \\ g_n(x) &\in f_{n}^{-1}(g_{n-1})(x) \end{align*} I want to use this construction to state that for each $x_0\in\mathbb{R}$ there exists a sequence of elements $(x_n = g_n(x_0))_{n\in\mathbb{N}}$ such that $f_n(x_n) = x_{n-1}$. However, to create the whole sequence I feel we need to use the Axiom Of Choice countably many times. Is this allowed?
Thank you in advance!
You are not using the axiom of choice infinitely many times. You're using it exactly twice to choose, uniformly, an inverse $g_n$ to $f_n$. Once to prove that each $f_n$ has an inverse, then again to choose the $g_n$'s.
Then you can use recursion, which does not use the axiom of choice at all, to define the function mapping $x\mapsto\langle g_n(x)\mid n\in\Bbb N\rangle$.