Show that $0 \to \mathbb{Q} \to \mathbb{R} \to \mathbb{R}/\mathbb{Q} \to 0$ split?

Question

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.

Answer 1

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 $\mathbb{Q}$.

Answer 2

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.