Topological equivalence relations

Question

Consider $(T, \tau)$ a topological space. Now consider $\sim_1,\sim_2$ equivalence relations on T. Let's call $\sim_3= (\sim_1 \vee \sim_2$)

Is it always true that that topological quotient $(T/\sim_3)$ is omeomorphic to $(T/\sim_1)/\sim_2$?

I perfectly know that $\sim_2$ has not sense on $(T/\sim_1)$ but I hope that what I mean is clear.

Answer 1

For $x\in T$ let $[x]$ be the $\sim_1$-equivalence class of $x$. Let $R$ be the relation on $T/\!\sim_1$ defined by $$[x]\mathbin{R}[y]\quad\text{iff}\quad u\sim_2v\text{ for some }u\in[x]\text{ and }v\in[y]\;,$$ and let $\sim_4$ be the transitive closure of $R$; $\sim_4$ is an equivalence relation on $T/\!\sim_1$. I suspect that you’re asking whether $T/(\sim_1\lor\sim_2)$ is homeomorphic to $(T/\!\sim_1)/\!\sim_4$; at least, that’s the most reasonable interpretation of $(T/\!\sim_1)/\!\sim_2$ that I can think of, and if you mean anything else, the answer is likely to be no, they’re not homeomorphic.

Fix $x,y\in T$; then $x\sim_3y$ iff there is a chain $$x=x_0\sim_1 x_1\sim_2 x_2\sim_1\ldots\sim_2x_{2n}=y\,.$$ We also have $[x]\sim_4[y]$ iff there is a chain $$[x]=[x_0]\mathbin{R}[x_1]\mathbin{R}\ldots\mathbin{R}[x_n]=[y]\,.$$

Suppose that $[x]\sim_4[y]$, with $[x]=[x_0]\mathbin{R}[x_1]\mathbin{R}\ldots\mathbin{R}[x_n]=[y]$. Then there are $u_k\in T$ for $k=0,\dots,n-1$ and $v_k\in T$ for $k=1,\dots,n$ such that $x_k\sim_1u_k\sim_2v_{k+1}\sim_1x_{k+1}$ for $k=0,\dots,n-1$. It follows that

$$\begin{align*} x&=x_0\sim_1u_0\sim_2v_1\sim_1u_1\sim_2v_2\sim_1u_2\sim_2\ldots\\ &\ldots\sim_2v_{n-1}\sim_1u_{n-1}\sim_2v_n\sim_1x_n\sim_2x_n=y \end{align*}$$

and hence that $x\sim_3y$.

Conversely, suppose that $x\sim_3y$, with $x=x_0\sim_1 x_1\sim_2 x_2\sim_1\ldots\sim_2x_{2n}=y$. Then

$$[x]=[x_0]\mathbin{R}[x_2]\mathbin{R}[x_4]\mathbin{R}\ldots\mathbin{R}[x_{2n}]=[y]\;,$$

so $[x]\sim_4[y]$.

In short, $x\sim_3y$ iff $[x]\sim_4[y]$. From here you should be able to show that $T/\!\sim_3\cong(T/\!\sim_1)/\!\sim_4$.

Answer 2

Let $f\colon X\to Y$ and $g\colon Y \to Z$ be quotient maps.

Then $g \circ f$ is a quotient map:

It is continuous because the composition of continuous functions is continuous.

Let $U\subseteq Z$.

$$(g\circ f)^{-1}[U] = f^{-1}[g^{-1}[U]],$$ so if $(g\circ f)[U]$ is open, so is $g^{-1}[U]$, and this so is $U$.

What, then, is the equivalence relation induced by $g\circ f$?

$p \sim_{g\circ f} q$ iff $g(f(p))=g(f(q))$

To get much further, I think you'll need to get more specific about those relations you handwaved.

Answer 3

Think there is a problem with your question.

Unless $\sim_1 \vee \sim_2$ means something like "the equivalence relation generated by $\sim_1$ and $\sim_2$", literally the "union" $\sim_3$ of two equivalence relations needs not to be an equivalence relation. Namely, transitivity might fail because

$$ x \sim_1 y \qquad \text{and} \qquad y \sim_2 z \qquad \text{doesn't imply} \qquad x \sim_1 z \qquad \text{or} \qquad x \sim_2 z \ . $$

For this reason, $\sim_2$ does not necessarily induce a well-defined equivalence relation on the quotient space $T/\! \!\sim_1$. If you try to define $\sim_2$ on classes $\widetilde{x}, \widetilde{y} \in T/\! \!\sim_1 $ as it would probably be your first guess,

$$ \widetilde{x} \sim_2 \widetilde{y} \qquad \Longleftrightarrow \qquad x \sim_2 y $$

you have to face the problem that, if you take different representatives of classes $ x'\in\widetilde{x}$ and $ y'\in \widetilde{y} $, having

$$ x \sim_1 x' \ , \quad y \sim_1 y' \qquad \text{and} \qquad x \sim_2 y $$

doesn't imply that you would obtain

$$ x' \sim_2 y' \ . $$

That is, $\sim_2$ might not be well-defined on $\sim_1$-classes. So you cannot speak of the quotient $(T/\!\!\sim_1)/\!\!\sim_2$.

Or am I missing something?