Surjectivity of a function considering $f \circ h=Id_Y$

Question

\begin{align} f: \mathbb{R}^2 &\longrightarrow \mathbb{R} \\ (x,y) & \longmapsto x+y \end{align} Question: Is this function surjective?

It seems clear to me that this function must be surjective, because the point $(x,y) \in \mathbb{R}^2$ maps to the entire codomain $\mathbb{R}$ under the addition given by function. But I wanted to try a different approach:

Question (refined): Is this function surjective, using $\ f\circ h=Id_\mathbb{R}$

Here are my steps. I defined $h$ as the following function: \begin{align}h: \mathbb{R} &\longrightarrow \mathbb{R}^2 \\ x &\longmapsto (x,0) \end{align}

Such that: \begin{align} (f \circ h)(x)=x&=f(h(x)) \\ &=f(x,0)=x \end{align} Are these steps correct, or aren't they even valid?

Note: I am very new to this subject and in my homework assigment I am also allowed to use examples and counter examples, but I am always very eager to expand my knowledge to such sentences as introduced as above.

Namely: if a right inverse exists, such that $f \circ h= Id_Y$, where $h$ is called the right inverse, then the function $f$ is surjective.

Answer 1

Your second argument is fully correct.

One remark about the first one though. You say "this function must be surjective, because the point $(x,y) \in {\mathbb R}^2$ maps to the entire codomain ${\mathbb R}$". The formulation is a bit weird: it's not the point that maps anywhere, but the function. But anyway, this is not really an argument, but just a restatement of what surjectivity means. To really show that the function $f$ maps to the entire codomain, you'd have to take an element of the codomain, say $x$, and find an element of the domain that maps to $x$. Of course, you take $(x,0)$ and then you again have exactly your second argument again.

This way you also see that the first and second way of arguing are fundamentally the same.