Let $A$ be the set obtained by removing the set $$Y:=\{(0, y)\ |\ y\neq 0\}$$ from $\mathbb{R}^2$. Clearly, the set $Y$ is the $y$-axis minus the origin. Let $$\tau_1: \text{subspace topology on A induced from the usual topology of}\ \mathbb{R}^2.$$ Let $$\tau_2:\ \text{quotient topology on A due to the following surjective map}\ \Phi ,$$ where $$\Phi:\mathbb{R}^2\to A$$ defined by $$\Phi(x, y)= (x, y)\ \text{for}\ x\neq 0, \ \text{and}\ \Phi(0, y)= (0, 0)\ \text{for all}\ y.$$ Let $\tau_2$ be the resulting quotient topology on $A$.
I need to show that $\tau_1$ and $\tau_2$ are comparable.
My thought: I want to show that either $\tau_1\subseteq \tau_2$ or $\tau_2\subseteq \tau_1$. I know that, by definition, a set $V$ is open in $\tau_1$ if $V=W\cap A$, where $W$ is open in $\mathbb{R}^2$. Further, by definition, $\tau_2$ is the smallest topology such that $\Phi$ is continuous. How to use these to show the result. Thanks in advance for the help.
These are not the same topologies.
Let $U\subset A$ the open ball of radius one around the origin. This is the open ball with the non origin points of the $y$ axis missing. It can be expressed as the intersection of $U$ and $B_1(0,0)=\{x:\lvert x\rvert\lt 1\}$ so $U$ is open under the subspace topology.
But $\Phi^{-1}(U)=B_1(0,0)\cup Y$ which is not open in $\mathbb R^2$ so $U$ is not open in the quotient topology.