Let $X=[0,1]\times [0,1]\subset \mathbb{R}$ with the subspace topology. Consider the equivalence relation $\sim$ on $X$ whose non trivial relations are $(0,y)\sim (1,1-y)$ for any $y\in [0,1]$. Let $p:X\rightarrow X^*$ be the quotient map.
I am trying to show that if If $E\subset X$, is defined as $E=[0,1]\times \{ 0,1 \}$, then $P(E)$ is homeomorphic to $S^1$. I am hoping my attempt below makes some sense and is fix-able and if not, I can understand where I am getting confused.
My attempt: $p(E)=[0,1]\times \{ 0,1 \}/\sim$. Then take the mapping \begin{align*} \phi:\;& p(E)\rightarrow S^1\\ &(x,y)\mapsto e^{i\pi(x+y)} \end{align*} We verify that the map is constant on the equivalence class, or that $\phi(0,0)=\phi(1,1)$ and $\phi(0,1)=\phi(1,0)$. The latter is obvious by the symmetric nature of the function. For the former: \begin{align*} &\phi(0,0)=1=e^{2\pi i}=\phi(1,1) \end{align*} But I am a bit hazy about this portion, and the quotient topology in general. I think what I am doing here is defining a map from a space $X$ that factor through $X/\sim$ a la algebra first isomorphism theorem.
In this problem, this would mean I am defining my map $\phi$ which actually goes from $[0,1]\times [0,1]$ and insuring it actually is constant on equivalence classes defined by $\sim$. Is that correct?
Finally, how do I show that this map is actually open and continuous? I am particularly concerned about the point $(1,0)$? If I can actually just work in the product topology on $[0,1]\times [0,1]$, this wouldn't be too bad; the pre-image of any neighborhood around $(1,0)$ will include the end points of the intervals (since both x and y will be near zero) which is fine as those are zero in the subspace topology on $I\times I$ as a subspace of $\mathbb{R}^2$. Is that right?