We define the maps $f_+$ and $f_-$ from $\mathbb{Q}[X]/(X^2 -2) \mathbb{Q}[X]$ to $\mathbb{Q} + \mathbb{Q} \cdot \sqrt2$ in the following manner: For any residue class $g+ (X^2-2) \mathbb{Q}[X]$, we write:
$$ f_+(g+ (X^2-2) \mathbb{Q}[X]) = g \left( \sqrt2 \right)$$
$$ f_-(g+ (X^2-2) \mathbb{Q}[X]) = g \left(- (\sqrt2) \right)$$
I want to show that these maps are well-defined. Because the degree of our polynomial $X^2-1$ is two, every representative in the residue class will have a degree less than this, upon division. We thus expect polynomials in the domain of the form $aX+b$ with $a,b \in \mathbb{Q}$; the elements in the image will have the form $p+ q \sqrt2$, with $p,q \in \mathbb{Q}$.
I have a hard time figuring out what this map precisely does to elements in the domain. If I would understand this, it might be easier to prove it is well-defined ( it does not matter which representative we pick).
The tip in the comments answered my question. The function is well-defined because any equivalent elements $g$ and $g'$ have images $g(\pm \sqrt{2})$ and $ g'(\pm \sqrt{2})$.