How to show that this sequence split?
I'm trying to construct a map $\phi: \mathbb{R}/\mathbb{Q} \to \mathbb{R}$ by $\overline{r} + \mathbb{Q} \mapsto r$. Let the quotient map be $\pi$ and $\pi \circ \phi(\overline{r}) = \pi(r) = \overline{r}$ is the identity on $\mathbb{R}/\mathbb{Q}$. Is this correct?
Is there a theoretical way of showing this? I want to use some projective module or divisible or injective module theory but it is not clear what ring we are considering.
You can consider the sequence in the category of modules over $\mathbb{Q}$, or over $\mathbb{Z}$.
In both cases the sequence is split, because $\mathbb{Q}$ is injective in both cases. However, writing explicitly the retraction is not possible, because it requires the axiom of choice, at least a weak form of it.
Since $\mathbb{R}/\mathbb{Q}$ is torsion free as an abelian group, it’s actually immaterial what ring you consider among $\mathbb{Z}$ or $\mathbb{Q}$. Defining the retraction needs a basis of $\mathbb{R}$ over $\mathbb{Q}$.
First, as @Qurultay says, $\phi$ is not well-defined. You cannot, given just an equivalence class, say what its representative is. Now, given the Axiom of Choice, one can say that a choice function exists, but then the definition is circular; $\phi(S) = r$, where r is the representative of S chosen by the choice function, but then you might as well say "Let $\phi$ be a choice function."
If $\phi$ is defined as being a choice function, and $\pi$ is defined as being a function that take an element to the equivalence class it's in, then by definition $\phi$ chooses an element of $\overline{r}$, thus the equivalence class that $\phi (\overline r)$ is in is $\overline r$, thus $\pi (\phi(\overline r)) = \overline r$. I'm not sure what else needs to be proven.
