Surjective functions are defined as if $x$ is in the domain then $f(x)$ exists (in the codomain/image). Most of the simple proofs I have looked at seem to be showing this by constructing an inverse, e.g., for a function on reals $f(x) = 2x$ is surjective since $f(x/2)=x$. First I thought, wait will that also not always show bijectivity. Then I realised that we might still show for a $g:\{-2,2\}\rightarrow \{4\}$, $g(x)=x^2$ as surjective by $g(\sqrt{x})=x$ but $\sqrt{x}$ is not a function.
But must we always show an inverse relation for a proof of surjection?
Take $f(x)=2x$ for example, we might argue that assume $x \in \mathbb{R}$ and since $2 \in \mathbb{R}$ we know that $2x \in \mathbb{R}$ due to the fact that real numbers are closed under multiplication. Therefore, $f$ is surjective.
Is this a valid argument for showing surjection in at least some cases?
Could you think of a relatively simple function where proving surjection maybe easier this way and harder by showing an inverse relation?
Showing surjectivity means showing that for any element of the codomain of a mapping, in this case your function $f(x)=2x$, there is some element in the domain whose image under the mapping is equal to that element in the codmain. So take such an element in the codomain and attempt to show the existence of an element in the domain that maps to said element.