The answer is yes. Every function on the reals is the sum of two injective functions, and this can be done in a highly effective manner, constructing the two functions $g,h$ from $f$ without any need for the axiom of choice or transfinite recursion.

**Theorem.** (ZF)

- Every function on the reals is the sum of two injective functions: for every $f:\newcommand\R{\mathbb{R}}\R\to\R$ there are injective $g,h:\R\to\R$ with $f=g+h$.
- Furthermore, this is possible with functions $g$ and $h$ that are arithmetically definable from $f$, and indeed, the digits of $g(x)$ and $h(x)$ are uniformly computable from oracles specifying the digits of $x$ and $f(x)$.
- In particular, every Borel function $f$ is the sum of two injective Borel functions $g$ and $h$.

The proof will rely on the following key lemma.

**Key lemma.** There is a pairing function on the reals $x,y\mapsto\langle x,y\rangle$ such that $y-\langle x,y\rangle$ also is a pairing function. That is, for reals $x,y$ we can define a real number $\langle x,y\rangle$ for which:

- From the value of $\langle x,y\rangle$ we can uniformly recover both $x$ and $y$.
- From the value of the difference $y-\langle x,y\rangle$ we can uniformly recover both $x$ and $y$.

**Proof.** Given $x$ and $y$, let $z$ be a binary sequence that encodes both $x$ and $y$ in some sensible concrete manner. We shall first specify the even digits of $\langle x,y\rangle$ with $0$s and $1$s in such a way so as to list out $z$ on those digits, so that the $k$th digit of $z$ is the same as the $2k$th digit of $\langle x,y\rangle$. This will ensure that $\langle x,y\rangle$ is a pairing function, regardless of how we define the odd digits. Next, we specify the odd digits of $\langle x,y\rangle$ in such a way that the parity pattern of the odd digits of the difference $y-\langle x,y\rangle$ is again the binary pattern of $z$. So from the value of $y-\langle x,y\rangle$ we can recover $z$ and hence both $x$ and $y$. $\Box$

**Proof of theorem.** Let's now prove the theorem. Note that the pairing function $\langle x,y\rangle$ in the key lemma is computable in the relevant sense from oracles for $x$ and $y$. Let
$$g(x)=\langle x,f(x)\rangle\qquad$$
and
$$h(x)=f(x)-\langle x,f(x)\rangle.$$
The function $g$ is injective, since from $\langle x,f(x)\rangle$ we can recover $x$. The function $h$ is injective, since the key lemma ensures that from $f(x)-\langle x,f(x)\rangle$ we can recover $x$. And clearly $f=g+h$, so we have achieved $f$ as the sum of two injective functions.

The proof is effective, since the functions $g$ and $h$ are computable from $f$. $\Box$

If we were working on Baire space $\mathbb{N}^{\mathbb{N}}$ instead of $\R$, that is, with infinite sequences of natural numbers $x:\mathbb{N}\to\mathbb{N}$ instead of real numbers, it is a slightly more natural context, since working digit-by-digit is more natural in Baire space. In particular, in that context the functions $g$ and $h$ arising in the proof would be continuous, if $f$ is, since in order to know the first $k$ values of $g$ or $h$ it would suffice to know the first $k$ values of $x$ and $f(x)$. On the real numbers, however, this reasoning does not quite transfer to $\R$, because the non-unique representations cause certain boundary issues — interleaving digits is not a continuous process since $1=0.999\bar 9$. And so my functions $g$ and $h$ are not continuous, even when $f$ is. ~~But I think it might be possible to patch this up somehow to achieve the continuous case as a special case of the fully general case as I have.~~ (Update: Will Sawin explains how the answer of Joonas Ilmavirta shows that the continuous case will be impossible.)

Let me finally observe that once we know the two-injective-summand case is true, then we easily get three injective summands or any number by simple scaling. For example, from $f=g+h$ we can write $f=g+\frac12 h+\frac12 h$ and so forth.

**Constructive logic?** My answer has a computable nature, but I am unsure whether the proof can be made completely constructive, that is, in constructive logic. I would welcome answers by those who could say something about that. (Update: Will Sawin explains in his comment below why we should expect no constructive proof.)