Let
- $(X,\mathcal X)$ be a measurable space
- $\sim$ be an equivalence relation on $X$
- $X/\sim$ denote the quotient space of $X$ by $\sim$
Is there a $\sigma$-algebra $\mathcal X/\sim$ on $X/\sim$ such that $\mathcal X$ is "canonically" embedded into $\mathcal X$?
If necessary, assume that $(Y,\mathcal Y)$ is another measurable space such that $\sim$ is the kernel of a $(\mathcal X,\mathcal Y)$-measurable function $f:X\to Y$. Maybe we can define an equivalence relation (denoted by the same symbol $\sim$) on $\mathcal F$ via $$A\sim A':\Leftrightarrow\exists B\in\mathcal G:A,A'\subseteq f^{-1}(B)\tag1\;\;\;\text{for }A,A'\in\mathcal F$$ in that case. However, I'm mostly interested in the general case.
Any function $f:X\to Y$ induces a $\sigma$-algebra $\mathcal Y$ on $Y$ characterized by:$$B\in \mathcal Y\iff f^{-1}(B)\in\mathcal X$$
If $\sim$ is an equivalence relation on $X$ then we have the natural function $\nu:X\to X/\sim$ that is prescribed by:$$x\mapsto\text{ equivalence class represented by }x$$
Applying the procedure on function $\nu$ we find the $\sigma$-algebra $\mathcal Y$ that is characterized by:$$B\in\mathcal Y\iff\cup B\in\mathcal X$$So that $\mathcal Y=\{\cup B\mid B\in\mathcal X\}$.
This is the first (and for me only) $\sigma$-algebra that comes to mind in this situation. However, I am not sure whether this is really what you are after.