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Here is a simple proof that every function is the sum of two surjective functions.

Pick any onto functions $g:[0, \infty) \to \mathbb R$ and $h: (-\infty, 0) \to \mathbb R$. Define $$ g(x)= f(x)-h(x) \qquad \forall x < 0\\ h(x)= f(x)-g(x) \qquad \forall x \geq 0\\ $$

Note: It is actually possible to show any function $f$ can be written as $f=g+h$, where $g,h$ are surjective and satisfy the Intermediate Value Property. This can be done via the standard argument that each function on $\mathbb R$ can be written as the sum of two functions with the IVP.

Here is a simple proof that every function is the sum of two surjective functions.

Pick any onto functions $g:[0, \infty) \to \mathbb R$ and $h: (-\infty, 0) \to \mathbb R$. Extend them to onto functions $\mathbb R\to\mathbb R$ as follows: $$ g(x)= f(x)-h(x) \qquad \forall x < 0\\ h(x)= f(x)-g(x) \qquad \forall x \geq 0\\ $$

Note: It is actually possible to show any function $f$ can be written as $f=g+h$, where $g,h$ are surjective and satisfy the Intermediate Value Property. This can be done via the standard argument that each function on $\mathbb R$ can be written as the sum of two functions with the IVP.

Here is a simple proof that every function is the sum of two surjective functions.

Let $f: \mathbb R \to \mathbb R$ be any function.

Let $O,E$ be the set of real numbers whose first digit after period is odd or even, respectively, that is $$ O=\{ m.a_1a_2 \ldots a_n \ldots | a_1 \mbox{ is odd } \} \\ E=\{ m.a_1a_2 \ldots a_n \ldots | a_1 \mbox{ is even } \} \\ $$

Then, the functions $h:O \to \mathbb R, g : E \to \mathbb R$ $$ h( m.a_1a_2 \ldots a_n \ldots)= m.a_2 \ldots a_n \ldots \\ g( m.a_1a_2 \ldots a_n \ldots)= m.a_2 \ldots a_n \ldots \\ $$ which erase the first digit after the dot are onto.

Extend them via $$ f=g+h \,. $$

Simplification I realised that the argument is much more complicated than it needs be:

Pick any onto functions $g:[0, \infty) \to \mathbb R$ and $h: (-\infty, 0) \to \mathbb R$. Define $$ g(x)= f(x)-h(x) \qquad \forall x < 0\\ h(x)= f(x)-g(x) \qquad \forall x \geq 0\\ $$

Note: A more complicated version of this proof shows that any function $f$ can be written as $f=g+h$, where $g,h$ are surjective and satisfy the Intermediate Value Property.

Here is a simple proof that every function is the sum of two surjective functions.

Pick any onto functions $g:[0, \infty) \to \mathbb R$ and $h: (-\infty, 0) \to \mathbb R$. Define $$ g(x)= f(x)-h(x) \qquad \forall x < 0\\ h(x)= f(x)-g(x) \qquad \forall x \geq 0\\ $$

Note: It is actually possible to show any function $f$ can be written as $f=g+h$, where $g,h$ are surjective and satisfy the Intermediate Value Property. This can be done via the standard argument that each function on $\mathbb R$ can be written as the sum of two functions with the IVP.

Here is a simple proof that every function is the sum of two surjective functions.

Let $f: \mathbb R \to \mathbb R$ be any function.

Let $O,E$ be the set of real numbers whose first digit after period is odd or even, respectively, that is $$ O=\{ m.a_1a_2 \ldots a_n \ldots | a_1 \mbox{ is odd } \} \\ E=\{ m.a_1a_2 \ldots a_n \ldots | a_1 \mbox{ is even } \} \\ $$

Then, the functions $h:O \to \mathbb R, g : E \to \mathbb R$ $$ h( m.a_1a_2 \ldots a_n \ldots)= m.a_2 \ldots a_n \ldots \\ g( m.a_1a_2 \ldots a_n \ldots)= m.a_2 \ldots a_n \ldots \\ $$ which erase the first digit after the dot are onto.

Extend them via $$ f=g+h \,. $$

Note: A more complicated version of this proof shows that any function $f$ can be written as $f=g+h$, where $g,h$ are surjective and satisfy the Intermediate Value Property.

Here is a simple proof that every function is the sum of two surjective functions.

Let $f: \mathbb R \to \mathbb R$ be any function.

Let $O,E$ be the set of real numbers whose first digit after period is odd or even, respectively, that is $$ O=\{ m.a_1a_2 \ldots a_n \ldots | a_1 \mbox{ is odd } \} \\ E=\{ m.a_1a_2 \ldots a_n \ldots | a_1 \mbox{ is even } \} \\ $$

Then, the functions $h:O \to \mathbb R, g : E \to \mathbb R$ $$ h( m.a_1a_2 \ldots a_n \ldots)= m.a_2 \ldots a_n \ldots \\ g( m.a_1a_2 \ldots a_n \ldots)= m.a_2 \ldots a_n \ldots \\ $$ which erase the first digit after the dot are onto.

Extend them via $$ f=g+h \,. $$

Simplification I realised that the argument is much more complicated than it needs be:

Pick any onto functions $g:[0, \infty) \to \mathbb R$ and $h: (-\infty, 0) \to \mathbb R$. Define $$ g(x)= f(x)-h(x) \qquad \forall x < 0\\ h(x)= f(x)-g(x) \qquad \forall x \geq 0\\ $$

Note: A more complicated version of this proof shows that any function $f$ can be written as $f=g+h$, where $g,h$ are surjective and satisfy the Intermediate Value Property.