Let $H\lhd G$ and $\psi:G\to G/H,x\mapsto xH$ be a function. Show it is surjective

Question

Let $H\lhd G$ and $\psi:G \to G / H$, $x \mapsto xH$ be a function. I am struggling to show that the map is surjective.

Is it just so simple to say let $y:=xH \in G / H$ than it is trivial that we get $x \in G$ s.t. $\psi(x)=xH$ and hence it is surjective? Many thanks for your help!

Answer 1

Yes, your argument is correct.

A very minor gripe: you don't "get" $x$ such that $\psi(x)=xH$. You actually start with one, and as you say, it trivially satisfies $\psi(x)=xH$.

Answer 2

Take any element $y \in G/H$. By definition of $G/H$, there exists $x \in G$ such that $y=xH$. Then by definition of $\psi$, this can be rewritten as $y=\psi(x)$. So every element of $G/H$ is in the image of $\psi$, i.e. $\psi$ is surjective.