How to prove that in this context epimorphisms are 'surjective' In article AN ELEMENTARY THEORY OF THE CATEGORY OF SETS of William Lawvere I met a proposition left to reader
(poor me) and I hope someone can help me with it. It wouldn't surprise me
if it is not difficult and I am just struggling with a blind
spot. The author describes a category having certain properties. Let me also say that composition $A\stackrel{f}{\rightarrow}B\stackrel{g}{\rightarrow}C$
is denoted as $fg$ (so not $gf$ as most of us are used to). I will mention below what according to my view could be relevant when it comes to my question.


*

*All finite limits and colimits exist. There is a unique initial object
$0$ and a unique terminal object $1$. The notation $x\in A$ is
used for $x:1\rightarrow A$. Practicized terminology: $x$ is an element of $A$.

*Functor $A\times-$ has a right adjoint.

*Object $1$ is a generator (I would call it a separator). This comes to $f,g:A\rightarrow B$ are equal if $xf=xg$ for all arrows $x:1\rightarrow A$.

*If some arrow $x:1\rightarrow A$ exists and $f:A\rightarrow B$ then some $g:B\rightarrow A$ exists with $fgf=f$ (axiom of choice).

*If $A\neq0$ then some arrow $x:1\rightarrow A$ exists.
On page $18$ it is remarked that $f:A\rightarrow B$ is an epimorphism (left-cancellable here) iff: $$\forall y\in B\exists x\in A\left[xf=y\right]$$
It looks like: $f$ is epi iff it is 'surjective' in a special meaning. I managed to prove that the condition is sufficient (making use of
the fact that $1$ serves as separator) but am stuck on the necessity.
My question can also be formulated as:

Let it be that there is an arrow $y:1\rightarrow B$ such that there
  are no arrows $x:1\rightarrow A$ with $xf=y$. Based on this
  construct (or prove the existence of) two distinct arrows $g,h:B\rightarrow C$ with $fg=fh$ (proving that $f$ is not epi).

I hope that there are persons here that are familiar with the article. 
Thanks in advance.
 A: Suppose $f: A\to B$ is an epimorphism. If $A\neq 0$, then $A$ has a point by assumption, and hence the axiom of choice guarantees the existence of $g: B\to A$ such that $A\xrightarrow{f} B\xrightarrow{g} A\xrightarrow{f} B = A\xrightarrow{f} B$. Since $f$ is epi, we may cancel it from the left, hence $B\xrightarrow{g} A\xrightarrow{f} B = \text{id}_B$. In particular, any point of $B$ lifts to a point of $A$ by composition with $g$.
Suppose $A=0$. If $B=0$, then $f=\text{id}_0$ and the statement is clear. Suppose $B\neq 0$. Since $A=0$ is initial and $0\to B$ is an epimorphism, any two morphisms $\alpha,\beta: B\to C$ are equal. In particular, for some choice of point $x: 1\to B$ (which exists because $0\neq B$) we must have $B\to 1\xrightarrow{x} B=\text{id}_B$. As $1\xrightarrow{x} B\to 1=\text{id}_1$ holds, too, we conclude $B\cong 1$, hence the morphism $f$ equals the canonical morphism $0\to 1$. Then $0\to X$ is an epimorphism for any $X$ since we have the factorization $(0\to 1) = (0\to X\to 1)$, so all $0\neq X$ are terminal, and hence the category is just $\{0<1\}$ up to equivalence. In this case, $0\to 1$ is an epimorphism which does not lift all points. However, this example is after all excluded by Axiom 8 (which however comes after the Exercise).
