Natural map from $\mathbb{X}$ to $\mathbb{X}/\mathbb{M}$, that is not closed

Question

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

Answer 1

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 $p$.

Answer 2

The set of equivalence classes of $\sim,$ where $(x,y)\sim (x',y')\iff x=x',$ is $\Bbb R^2/\Bbb M=\{\{x\}\times \Bbb R: x\in \Bbb R\}.$

The quotient topology on $\Bbb R^2/\Bbb M$ is the $\subset$-largest topology such that $q((x,y))=\{x\}\times \Bbb R$ is continuous.

The product topology on $\Bbb R^2$ is the $\subset$-smallest topology on $\Bbb R^2$ such that both $p_1((x,y)=x$ and $p_2((x,y))=y$ are continuous.

So the map $\psi(\{x\}\times \Bbb R)=(x,0)$ is a homeomorphism to the sub-space $\Bbb R\times \{0\}$ (a sub-space of $\Bbb R^2$).

The maps $q$ and $p_1^*((x,y))=(x,0)$ are not the same functions, but can be called "the same for topological purposes", as $\psi$ is a homeomorphism and $\psi\circ q=p_1^*.$