# Let $A$ be the subspace of $ℝ×ℝ$ s.t. $x≥0$ or $y=0$. Let $q:A→ℝ$ be obtained by restricting $π_1$. Show that $q$ is a quotient map.

Let $\pi_1:ℝ×ℝ→ℝ$ be a projection on the first coordinate. Let $A$ be the subspace of $ℝ×ℝ$ consisting of all points $(x,y)$ for which $x≥0$ or $y=0$ (or both). Let $q:A→ℝ$ be obtained by restricting $π_1$. Show that $q$ is a quotient map.

Clearly $q$ is continuous and surjective. So now I want to prove that $q$ is open or closed map, and then I'm done. But I don't see how I could prove that.

I also tried proving that for a subset $U$ of $\mathbb{R}$ that $\pi_1^{-1}(U)$ being open implies that $U$ is open, but that didn't work for me as well.

Any hints :) ?

Sometimes it's easier to verify the quotient map property with open sets, and sometimes using closed sets seems more appropriate. I think if you're already dealing with closed sets like $\Bbb R_{\ge0}×\Bbb R$ and $\Bbb R\times\{0\}$ (whose union is $A$), then using a closed preimage seems wiser.
So let $C$ be a subset of $\Bbb R$ whose preimage $q^{-1}[C]=\pi_1^{-1}[C]\cap A$ is closed in $A$. Let $$C_-=C\cap\Bbb R_{\le0},\qquad C_+=C\cap\Bbb R_{\ge0}$$ Then $\pi_1^{-1}[C_+]=q^{-1}[C]\cap(\Bbb R_{\ge0}×\Bbb R)$ is closed in $\Bbb R^2$, so $C_+$ is closed in $\Bbb R$.
Now $C_-$ is homeomorphic, via the projection, to $$C_-×\{0\}=q^{-1}[C_-]=q^{-1}[C]\cap(\Bbb R_{\le0}×\{0\})$$ which is closed in $A$, thus closed in $\Bbb R×\{0\}$. So $C_-$ is closed in $\Bbb R$. It follows that $C$ is closed.