If $q: X\to Y$ is a quotient map, does there necessarily exist some equivalence relation ~ such s.t. $Y\cong X/$~

Solution 1:

Yes, this is classical: if $f: X \to Y$ is a quotient map (in the sense that $U \subseteq Y$ open iff $f^{-1}[U]$ open in $X$). Then define $$\forall x,x' \in X: x \sim x' \iff f(x)=f(x')$$

It's trivial to check this is an equivalence relation on $X$, the equivalence relation induced by $f$ (I've seen $R_f$ used for this as well). I'll denote the class of $x$ by $[x]$, and the map $x \to [x]$ by $q: X \to X{/}\sim$.

Then the map $$h: (X{/}\sim) \to f[X]; h([x]) = f(x)$$

is well-defined and a homeomorphism between $X{/}\sim$ and $f[X]$, where of course $X{/}\sim$ has the quotient topology w.r.t. $q$, as expected, and $f[X]$ the subspace topology w.r.t. $Y$.

Sketch of proof: that $h$ is a bijection is set-theoretically obvious; that $h$ is continuous follows from $h^{-1}[U \cap f[X]] = q[f^{-1}[U]]$, which is open in $X{/}\sim$ (when $U \subseteq Y$ is), as the $q$-image of a $\sim$-saturated open set etc. and if $O \subseteq X{/}\sim$ is open, $h[O]=f[q^{-1}[O]]$ which is open in $Y$ (and $f[X]$) by the quotient-property of $f$.

In short, if $f$ is quotient and surjective (aka onto), indeed $Y=f[X]$ can be seen as an "equivalence-relation-quotient" of $X$ as well. This "validates" the name quotient map for maps like $f$, one could say.