2

Let

  • $(E,\tau)$ be a topological space
  • $\sim$ be an equivalence relation on $E$
  • $[x]$ denote the equivalence class of $x$ with respect to $\sim$ for $x\in E$
  • $E_\sim$ denote the quotient space of $E$ by $\sim$ and $$\pi_\sim:E\to E_\sim\;,\;\;\;x\mapsto[x]$$

Note that $$\tau_\sim:=\left\{U\subseteq E_\sim:\bigcup U\in\tau\right\}$$ is the final topology on $E_\sim$ with respect to $\pi_\sim$.

Are we able to show that $$\mathcal B(E_\sim)\stackrel{(1)}=\underbrace{\left\{B\subseteq E_\sim:\bigcup B\in\mathcal B(E)\right\}}_{=:\:\mathcal B_1}\stackrel{(2)}=\underbrace{\left\{B\subseteq E_\sim:\pi_\sim^{-1}(B)\in\mathcal B(E)\right\}}_{=:\:\mathcal B_2}?$$

Clearly, since $\pi_\sim$ is $(\tau,\tau_\sim)$-continuous, it is Borel measurable and hence $\mathcal B(E_\sim)\subseteq\mathcal B_2$.

0xbadf00d
  • 13,422

1 Answers1

1

In general, without additional conditions imposed on $\sim$, the answer can be negative both for (1) and (2). Indeed, let $(E,\tau)$ be the set of reals endowed with the standard topology and a unique non-trivial equivalence class of $\sim$ is $\Bbb Q$. Then $[0]\in\mathcal B_1\cap \mathcal B_2$. On the other hand, it is easy to check that a family $$\mathcal B’=\{B\in\mathcal B(E_\sim): [0]\in B\mbox{ iff }\pi^{-1}_\sim(B)\mbox{ is comeager}\}$$
is a $\sigma$-algebra on $E_\sim$ containing all its open subsets. (Recall that a subset of $E$ is comeager if it is a complement to a countable union of nowhere dense subsets of $E$). Thus $\mathcal B(E)\subset\mathcal B’(E)$. But $\pi^{-1}_\sim ([0])=\Bbb Q$ is not comeager, a contradiction.

Alex Ravsky
  • 90,434