Natural map from $\mathbb{X}$ to $\mathbb{X}/\mathbb{M}$, that is not closed I know that by considering projection $q : \mathbb{R}^2 \to \mathbb{R}$, $(x, y) \to x$, and the closed subset 
$$G = \left\{(x, y) : y \ge \frac 1 x, x > 0\right\}$$
will prove that $q$ is not a closed map.  
But I'm having some difficulty in understanding why we can consider the projection map $q$ as a natural map to the quotient space.  
The definition I have learnt for the natural map is
$Q:\mathbb{X}\rightarrow\mathbb{X}/\mathbb{M}$ such that
$Q(x)=x+\mathbb{M}$.
So why is this the same as the projection

The term 'projection' for the natural quotient map $X\to X/M$ is rather illustrative.
However, the projection $q:\Bbb R^2\to \Bbb R$ can also be viewed as a quotient map, namely take $X=\Bbb R^2 $ (as a vector space), and its subspace $M=\{(0,y):y\in\Bbb R\}$. 
Then, identifying the quotient $\Bbb R^2 /M$ with $\Bbb R$ (by $x\mapsto (x,0)+M$), we get $q$ as the quotient map.
Note that this also generalizes to arbitrary projection $p:X\to Y$ where $X$ is a vector space and $Y$ is a subspace:
Then we have $X/\ker p\cong Y$, and along this isomorphism, the quotient map $X\to Y$ is just $p$. 
