I have the following $3$ definitions of a quotient map of uniform spaces:
A surjective morphism $q:X\to Q$ such that if $q = g\circ f$ where $g, f$ are morphisms and $g$ is a bijection then $g$ is an isomorphism.
A surjective morphism $q:X\to Q$ such that if $f:X\to Y$ is a morphism with $\text{ker}(q)\subseteq \text{ker}(f)$ (where, as usual $\text{ker}(h) = \{(x, y): h(x) = h(y)\}$), then $f = f_0\circ q$ for some morphism $f_0:Q\to Y$.
A polar epimorphism $q:X\to Q$, that is, an epimorphism such that morphisms of the form $h\circ q$ with $h$ a morphism form a polar class, that is, they are a class of all morphisms $c$ with domain $X$ satisfying some equations of the form $c\circ f = c\circ g$.
Those are from the book of Isbell called Uniform Spaces. There is a theorem which says that all polar epimorphisms are surjective, and I understand it. But how can we show that all $3$ definitions are equivalent?
I got a little bit confused with Isbell's phrasing and such, but I think I get it now. The claim that is supposed to be obvious is that $3)$ implies $2)$ and that $2)$ implies $1)$.
Let $q:X\to Q$ be a polar epimorphism and suppose that $f:X\to Y$ is such that $\ker q \subseteq \ker f$. If for some maps $h_1, h_2$ we have $qh_1 = qh_2$ then $fh_1 = fh_2$. It follows that $f$ is in the polar class consisting of left multiples $f_0q$ of $q$, hence $f = f_0q$ for some $f_0$. This proves that $f$ is a quotient in the sense $2)$ since polar epimorphisms are surjective (point $6$ in the book).
To prove $2)$ implies $1)$, let $q = gf$ where $g$ is bijection. Then $\ker q\subseteq \ker f$ so that $f = f_0q$ and since $q$ is surjective, we have $gf_0 = 1_Q$ hence $f_0 = g^{-1}$ and so $g$ is an isomorphism.
The rest of the proof that they are equivalent should be contained in point $7$ of the book.