Quotients and subspaces that intersect each equivalence class

Question

My question is not really a question, but a proof checking and a clarification. Suppose that $X$ is a topological space, and let $\sim$ an equivalence relation on this space. Suppose that there exist a subspace $Y$ of $X$ that intersect each equivalence class. I want to prove that the two spaces $X/\sim$ And $Y/\sim$ are homeomorphic. I think this is quite useful in a lot of situations, but I have not found this result in my books, so I have thought to ask, just to be sure. I have thought about a proof: we denote with $\pi_X:X\to X\sim$ and with $\pi_Y:Y\to Y/\sim$ the two projection maps, and with $i:Y\to X$ the canonical topological embedding. We consider now the following diagram: $\require{AMScd}$ \begin{CD} Y @>{\pi_Y}>> Y/\sim\\ @V{\pi_X\circ i}VV\\ X/\sim \end{CD} Now, if the maps $\pi_Y$ and $\pi_X\circ i$ are two identifications, since they are constant on the fibres of the other, the universal property of identifications gives us the existence of two maps $f:X/\sim \to Y/\sim$ and $g:Y/\sim\to X/\sim$ continuous that make the diagram commute, and we have that $g=f^{-1}:$ $$\pi_Y=f\circ(\pi_X\circ i)=f\circ g\circ \pi_Y \Rightarrow f\circ g=id_{Y/\sim},$$ and similarly $f\circ g=id_{X/\sim}.$ If all the reasoning is right, what are the most simple conditions on $Y$ to have that $\pi_X\circ i$ is an identification?

Answer 1

$\pi_{X}\circ i$ may not be a quotient map( what you call an "identification"). For example, consider the spaces $$X=[0,1]\cup[2,3],\ Y=\{q|\ q\text{ is a rational in } [0,1]\}\cup \{r|\ r \text{ is an irrational in }[2,3]\},$$ and let $x\sim y$ if $|x-y|=0\text{ or }2$. Clearly $Y$ intersects every equivalence class. The set $A=\{[x]|x\in \mathbb{Q}\cap X\}$ is not open in the space $X/\sim$, for $\pi_{X}^{-1}(A)=X\cap\mathbb{Q}$ is not open in $A$. But $(\pi_{X}\circ i)^{-1}(A)=[0,1]\cap\mathbb{Q}$ is open in $Y$. Thus $\pi_{X}\circ i$ is not a quotient map.

Here's where you made a mistake: the line where you said "$\pi_{X}\circ i$ is the composition of an identification, $\pi_{X}$, and a homeomorphism, hence it is an identification." $i$ is in general not a homeomorphism unless $X=Y$. (If $Y\neq X$, $i$ is not surjective.)