Define an equivalence relation $\sim$ on $\mathbb{R}$ by $x\sim y\leftrightarrow x-y\in\mathbb Q$. Write $\mathbb R/\mathbb Q$ for the set of equivalence classes under $\sim$. If we assume the axiom of choice, then a choice function on $\mathbb R/\mathbb Q$ would be an injection from $\mathbb R/\mathbb Q$ into $\mathbb{R}$. On the other hand, it is consistent that, without the axiom of choice, no such injection exists (for example see the accepted answer here).
I am curious about the other direction. What can we say about the other direction? With the axiom of choice, as in the accepted answer here, we have an injection from $\mathbb R$ into $\mathbb R/\mathbb Q$. Is AC necessary for this? In other words, without the axiom of choice, can we have an injection from $\mathbb R$ into $\mathbb R/\mathbb Q$?
