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Let $f,g:\mathbb{Z}\to \{0,1\}$ be maps such that the preimages of both 0 and 1 under both $f$ and $g$ are infinite, i.e. $$ |f^{-1}(0)|=|f^{-1}(1)|=|g^{-1}(0)|=|g^{-1}(1)|=\infty. $$

In other words, if we think of $f$ and $g$ as infinite sequences of $0$'s and $1$'s, then both of these sequences contain infinitely many $0$'s and $1$'s.

Question: Does there exist a bijection $\phi:\mathbb{Z}\to \mathbb{Z}$ such that $f\circ \phi =g$?

For simple examples of $f$ and $g$ I can construct such a map $\phi$ explicitly. Does $\phi$ exist in general, and if so, how can this be shown? Does the existence of $\phi$ follow from the axiom of choice?

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Yes, for given two functions $\varphi_0:g^{-1}\{0\}\to f^{-1}\{0\}$ and $\varphi_1:g^{-1}\{1\}\to f^{-1}\{1\}$. Since $\Bbb Z=g^{-1}\{0\}\cup g^{-1}\{1\}$ as disjoint union, there exists one and only one function $\varphi:\Bbb Z\to\Bbb Z$ extending both $\varphi_0$ and $\varphi_1$ and we have $f\circ\varphi=g$.

The condition \begin{align} &g^{-1}\{0\}\neq\varnothing\implies f^{-1}\{0\}\neq\varnothing& &g^{-1}\{1\}\neq\varnothing\implies f^{-1}\{1\}\neq\varnothing \end{align} is necessary and sufficient for the existence of such map $\varphi$.

Finally, the axiom of choice is not required, for if we pick $a\in f^{-1}\{0\}$ and $b\in f^{-1}\{1\}$, then we can take $$\varphi(x)= \begin{cases} a&g(x)=0\\ b&g(x)=1 \end{cases}$$

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