The question is as follows,

Assume $\displaystyle q:X \rightarrow Y$ is onto. Define a relation of $\displaystyle X$ given by $\displaystyle x_1 {\sim}_q x_2$ iff $\displaystyle q(x_1) = q(x_2)$.

1) Show that every equivalence class with respect to $\displaystyle {\sim}_q$ is of the form $\displaystyle q^{-1}(\{y\})$ for some element $\displaystyle y \in Y$.

So, the equivalence classes of $\displaystyle x$ are given by

$\displaystyle [x] = \{x^{\prime} \in X : q(x) = q(x^{\prime})\}$

which Ipossiblycould write as

$\displaystyle [x] = \{x^{\prime} \in X : x = q^{-1}(q(x^{\prime}))\}$

where, $\displaystyle q(x^{\prime}) \in Y$. But, how can this inverse exist in the first place; he defined the map to be onto, and not one-to-one! Am I thinking about this the wrong way?

2) Show that the there is a one-to-one correspondence between $\displaystyle Y$ and the set of equivalence classes of $\displaystyle {\sim}_q$My argument goes a little like this; we have our set $\displaystyle Y$, and denote our set $\displaystyle {\sim}_q$ by $\displaystyle \{[x_1], [x_2], \ldots, [x_n]\}$. Then, we have that each of these elements are unique. If this is the case, then each equivalence on the set $\displaystyle X$ corresponds to a unique element on the set $\displaystyle Y$. This satisfies injectivity.

I'm not sure if this is even a correct argument, or if I'm on the right tracks, and how would this all be formalised?

3) Prove that for every equivalence relation $\displaystyle \sim$ on $\displaystyle X$, one can find a $\displaystyle Y$ and an onto map $\displaystyle q:X \rightarrow Y$ such that $\displaystyle \sim = \sim$

I have not got a clue what on earth this is even asking! Any help on this would be much appreciated.

Thank you all in advance,

HTale.