TITLE: Is every $f\colon\mathbb{R\to R}$ can be written as the pointwise difference of two injective functions?
QUESTION [6 upvotes]: Let $\mathscr {F}=\{f\in\mathbb{R}^\mathbb R\colon f \ \text{is injective}\}$ by $\mathbb {R}^\mathbb{R}$ all functions from $\mathbb R$ to $\mathbb R.$ My question Is the following true?
$$
\mathscr F-\mathscr F=\mathbb R^\mathbb R
$$
My guess, it is correct. Does anyone has a counterexample? any idea will be appreciated

REPLY [10 votes]: Certainly assuming the axiom of choice we can do this. Here's an outline of how to do this (filling in the gaps is a good exercise):
Fixing $f:\mathbb{R}\rightarrow \mathbb{R}$, enumerate the reals as $(r_\alpha)_{\alpha<\mathfrak{c}}$. Now recursively for each $\alpha<\mathfrak{c}$ pick reals $b_\alpha, c_\alpha$ which have not already been used such that $c_\alpha-b_\alpha=f(r_\alpha)$ (these exist since we've only used $<\mathfrak{c}$ reals so far).
Now let $g(r_\alpha)=c_\alpha$ and $h(r_\alpha)=b_\alpha$.

In the absence of choice, the situation is less clear - I suspect in that case the answer is no, that is, that it is consistent with ZF (= set theory without choice) that there is a function $f:\mathbb{R}\rightarrow\mathbb{R}$ which is not the difference of two injective functions. However, such consistency proofs are rather difficult, and I don't see one immediately.
("Paging Karagila, Asaf Karagila ...")