Hausdorff does not pass to the quotient

Question

Let $Q=[0,1]\times[0,1]\subset \mathbb{R}^2$ and the following relation $$(x,y)\sim (x',y')\iff (x,y)=(x',y'),\; (x',y')=(x,1-y),\;x\neq 0.$$ And $\pi: Q\to Q/\sim := X$ projection on the quotient. $X$ is equipped with quotient topology and $Q$ with induced euclidean topology. Prove that $X$ is $T_1$ but not $T_2$

I've got a couple of question. First thing first I think to prove that $X$ respects the first separation axiom is sufficient to prove that points are closed. So let $[(x,y)]\in X$ and with $\pi^{-1}([(x,y)])=\{(x,y),(x,1-y)\}$ since is the union of two closed sets points are closed. This work if and only if $\pi$ is a continuous map. I have to prove that $Q$ is $T_1$ or since is a subspace of $\mathbb{R}^2$ is sufficient?

Here's another question: is it true that in general the projection on the quotient is always continuous or I can say that because $X$ is equipped with quotient topology?

For the second part I know that the image of Hausdorff space isn't in general Hausdorff ,but I really don't know if I should proceed in general or finding two specific point that could not be separated... my main issues with the second reasoning is that I can't properly understand what are open sets in $X$.

Trying with the first reasoning and picking $U,V \in X$ open sets we got that $\pi^{-1}(U\cap V)$ is open in $Q$ so I think that I have to prove that this intersection isn't empty but I'm a little bit stuck. Thank you for the help!

Answer 1

After working on this for a bit, I think came up with a solution. I'm sharing here for everyone.

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I want to show that $(0,0)$ and $(0,1)$ in $X$ could not be separated. Let $[0,0]$ be the equivalence class of $(0,0)$ - similar notation will be used for $(0,1)$ - and let $U,V$ two open sets in $X$ s.t. $$[0,0] \in U \\ [0,1] \in V$$ Since $\pi$ is continuous $\pi^{-1}(U),\pi^{-1}(V)$ are open in $Q$ and contains $(0,0),(0,1)$. Since we are in $\mathbb{R}^2$ we have two open balls one centered in $(0,0)$ and one in $(0,1)$. Let $\epsilon$ be the minimum radius, If $(x,y)\in B_\epsilon(0,0)$ and $x>0$ this implies that $$\sqrt{x^2+ (1-(1-y))^2} = \sqrt{x^2+y^2} <\epsilon$$ so $(x,1-y) \in B_\epsilon(0,1)$ but $$U \supset \pi\big(B_\epsilon(0,0)\big)\ni \pi(x,y) =[x,y]=[x,1-y]=\pi(x,1-y)\in\pi (B_\epsilon(0,1))\subset V $$ So we can conclude that $U\cap V\neq \emptyset $.

Answer 2

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1. You asked if you have to prove that $Q$ is $T_1$ as it's a subspace of $\mathbb{R}^2$. The answer is no, you don't need to prove this. It's a well-known result that subspaces of $T_1$ spaces are $T_1$, and $\mathbb{R}^2$ with the standard Euclidean topology is $T_1$.

2. Is the projection on the quotient always continuous? Yes, it is, by definition of the quotient topology. The quotient topology is the coarsest topology (i.e., the topology with the fewest open sets) for which the projection map is continuous.

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To prove $T_1$ axiom, we want to show that every singleton set $\{[(x,y)]\}$ is closed in $X$. We have $\pi^{-1}(\{[(x,y)]\}) = \{(x, y),(x, 1-y)\}$ for $x \neq 0$, and $\pi^{-1}(\{[(0,y)]\}) = \{(0, y)\}$. Both of these are closed in $Q$ as they are finite, and the preimage of a closed set under a continuous function (in this case the quotient map $\pi$) is closed. Therefore, $\{[(x,y)]\}$ is closed in $X$ for all $[(x,y)]$, hence $X$ is $T_1$.

However, $X$ is not $T_2$ (Hausdorff). This can be shown by selecting the points $[(x, 1/2)]$ and $[(x, 1/2')]$ where $1/2' = 1 - 1/2$, and for $x \neq 0$. These points are not separated, because any open set of $X$ that contains $[(x, 1/2)]$ also contains $[(x, 1/2')]$, and vice versa. This is due to the specific quotient identification that identifies each point $(x, y)$ with $(x, 1-y)$ for $x \neq 0$. Therefore, $X$ is not Hausdorff, and the proof is complete.