Proving that a particular restriction of a projection is a quotient map

I was hoping somebody could help me with the following problem:

Let $\pi: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be the
projection onto the first coordinate, and let $p=\pi|_X$, where
$X=(\mathbb{R}_{\geq0} \times \mathbb{R}) \cup (
\mathbb{R} \times \{0\})$ (so $X$ should be the x-axis union everything to the
right of, and including, the y-axis…right?). Show that $p$ is a
quotient map, but $p$ is not an open map or a closed map.

Using a property of the subspace topology (namely that we can restrict the codomain of a continuous function and still retain continuity) $p$ must be continuous. Surjectivity is also apparent. However, I’m not sure how to prove that $p$ is a quotient map, and that it is neither open nor closed. I feel like examining neighborhoods of the origin might give a clue towards a solution, but I’ve tried a handful of examples of such neighborhoods and gotten nowhere. Any help is appreciated, thanks!

Solutions Collecting From Web of "Proving that a particular restriction of a projection is a quotient map"

Hint: (1) $\pi$ is a quotient (as it is open), hence $\pi|_X$ is also.

(2) $\{x \in \mathbb R^2 \mid x_1 \ge 0, x_1x_2 = 1\}$ is closed in $X$.

(3) $\{x \in X \mid \left|x-(0,2)\right| < 1 \}$ is open in $X$.

You can use the following lemma:

Let $q:X\to Y$ be a quotient map, $A$ a subspace of $X$, and $q’=q|_A:A\to q(A)$ the restriction. Then $q’$ is a quotient map if and only if each closed and $q’$-saturated subset of $A$ is the intersection of $A$ with a closed and $q$-saturated subset of $X$. The same is true if “closed” is replaced by “open”.

We actually only need the “if”-direction. Assume that $C$ is closed and saturated in $X$. Being closed and saturated means that it is of the form $D\times\{0\}\cup E\times\Bbb R$ for closed $D\subseteq(-\infty,0]$ and $E\subseteq[0,\infty)$. But then its $\pi$-saturation is $(D\cup E)\times\Bbb R$ which is again closed.