Consider the surjective map $f : \mathbb{R}^2 → H^+$ given by $f(x, y) = (x, |y|)$, where $H^+ := \{(x, y) ∈ \mathbb{R}^2| y ≥ 0\}$ is the closed upper half space. Show that the quotient topology induced on $H^+$ by the map $f$ agrees with the subspace topology on H+, where we are assuming that $R^2$ has been given the standard topology.
Here is my attempt:
Define $T_q$ to be the quotient topology induced by $f$ and $T_s$ to be the subspace topology on $H_+$. Then for $V \in T_q$, there is a $U \in T_{st}$ such that $U = f^{-1}(V)$. Define $W := H_+ \cap U$. Then $W \in T_s$ as $U$ is open in $T_{st}$. We claim $V = W$, which will mean that in general that $T_q \subseteq T_s$.
Let $(x, y) \in V$. Since $f((x, y)) = (x, y)$, we have that $(x, y) \in U$. But $(x, y) \in V \subseteq H_+$, meaning $(x, y) \in H_+ \cap U = W$. Therefore, $V \subseteq W$. Conversely, let $(x, y) \in W$. We then have that $y \geq 0$ and $f((x, y)) \in V$. But $f((x, y)) = (x, y)$ since $y \geq 0$, meaning $(x, y) \in V$, and moreover, $W \subseteq V$. Overall we get that $V = W \implies T_q \subseteq T_s$.
I am not sure how to prove that $T_s \subseteq T_q$
Take an open $W\subset H^+$ in the subspace topology. Then you want to show that $f^{-1}(W)=W\cup \{(x,y): (x,-y)\in W\}\subset \mathbb{R}^2$ is open.
You may want to consider two cases: $(x,y)$ on the $x$-axis or not.