If A is compact and $A/{\sim}$ is Hausdorff, then $X/{\sim}$ and $A/{\sim}$ are homeomorphic.

Question

Let $X$ be a topological space and let $\sim$ be an equivalence relation on $X$.
Call a subset $A \subset X$ $\mathit{full}$ if every equivalence class intersects $A$.
If A is compact and $A/{\sim}$ is Hausdorff, then $X/{\sim}$ and $A/{\sim}$ are homeomorphic.

I came across this Lemma on the web, and I can't see why it is true.

The basic idea of the proof is to use the well-known fact: A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

So the proof goes as follows: First, it shows that there exists a continuous bijection $f$ from $A/{\sim}$ to $X/{\sim}$.

And it is easy to show that $A/{\sim}$ is compact.

But I couldn't show that $X/{\sim}$ is Hausdorff.

Why it is Hausdorff? and why the assumption that $A/{\sim}$ is Hausdorff is needed?

Answer 1

The result is false without some restriction on the equivalence relation. Let $X=[0,1]$ with the usual topology, let $A=\{0,1\}$, and let $\sim$ be the equivalence relation on $X$ whose equivalence classes are $\{0\}$ and $(0,1]$. Then $A/\!\sim$ is homeomorphic to $A$, the discrete two-point space, but $X/\!\sim$ is homeomorphic to the Sierpiński space, which is not even $T_1$.

Answer 2

This is just my first impression only as I do not have time to work out the details or check the validity of my claims. I assume that $A$ is full, otherwise the claim is obviously false (for, let $A$ be a singleton). You may think about the commutative diagram (Sorry, I don't know how to draw but I can describe...):

$i:A\rightarrow X$, the usual inclusion map $i(x)=x$

$q_{1}:X\rightarrow X/\sim$, the canonical quotient map, $q_{1}(x)=\{y\in X\mid x\sim y\}$

$q_{2}:$$A\rightarrow A/\sim$, the canonical quotient map, $q_{2}(x)=\{y\in A\mid x\sim y\}$

$j:A/\sim\rightarrow X/\sim$, the map defined by $j(\alpha)=\beta$ such that $\alpha\subseteq\beta$. (That is, given $\alpha\in A/\sim$, choose $x\in\alpha$. Then define $\beta=q_{1}(x)$.

Note that $q_{1}\circ i=j\circ q_{2}$.

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To show that $X/\sim$ is Hausdorff:

Let $\beta_{1},\beta_{2}\in X/\sim$ and $\beta_{1}\neq\beta_{2}$. Since $A$ is full, there exist $x_{1},x_{2}\in A$ such that $x_{1}\in\beta_{1}$ and $x_{2}\in\beta_{2}$. Check that $q_{2}(x_{1})\neq q_{2}(x_{2})$. Since $A/\sim$ is Hausdorff, there exist disjoint open sets $U,V$ such that $q_{2}(x_{2})\in U$ and $q_{2}(x_{2})\in V$. Think about $U, V$ and the commutative diagram...