TITLE: Joy of Cats Proposition 8.16(4)
QUESTION [1 upvotes]: I am trying to prove Proposition 8.16(4) of Joy of Cats. Before I go into the question itself, let me describe the relevant definitions,

Definition 1.
Let $(\bf{A},U)$ be a concrete category over $\bf{X}$ (i.e., $U:\bf{A}\to\bf{X}$ is faithful).

A structured arrow with domain $\bf{X}$ is a pair $(f,A)$ consisting of an $\bf{A}$-object $A$
  and an $\bf{X}$-morphism $X\overset{f}{\to}U(A)$. 
A structured arrow $(f,A)$ is said to be generating provided that for any pair of
  $\bf{A}$-morphisms $r, s : A\to B$ the equality $U(r)\circ f = U(s)\circ f$ implies that $r = s$.
A generating arrow $(f,A)$ is called extremally generating provided that each $\bf{A}$-monomorphism $m : A' \to A$, through which $f$ factors (i.e., $f = U(m) \circ g$ for some $\bf{X}$-morphism $g$), is an $\bf{A}$-isomorphism.

Definition 2. Let $\bf{A}$ be a category. An $\bf{A}$-epimorphism $f$ is said to be a *extremal epimorphism if whenever $f=m\circ g$ for some $\mathbf{A}$-morphisms $m,g$ where $m$ is a $\mathbf{A}$-monomorphism, we have $m$ is a $\mathbf{A}$-isomorphism.

As I have mentioned earlier, I am trying to prove Proposition 8.16(4) of Joy of Cats which says, 

Let $(\mathbf{A},\mathscr{U})$ be a concrete category over $\bf{X}$. Let $(f,A)$ be an structured arrow with domain $\bf{X}$. If $X \overset{f}{\to} U(A)$ is an $\mathbf{X}$-extremal epimorphism and if $U$ preserves monomorphisms then $(f,A)$ is extremally generating.

Here is my progress so far,

To prove that $(f,A)$ is extremally generating first observe that since $f$ is an $\mathbf{X}$-epimorphism, it follows by 8.16(3) that $(f,A)$ is -generating. Now let $B \overset{m}{\to} A$ be an $\mathbf{A}$-monomorphism such that $f=U(m)\circ_{\mathbf{X}}g$ for some $\mathbf{X}$-morphism $g$. We need to prove that $m$ is an $\mathbf{A}$-isomorphism. Since $U$ preserves monomorphisms and since $m$ is an $\mathbf{A}$-monomorphism, it follows that $U(m)$ is an $\mathbf{X}$-monomorphism. Furthermore since $f$ is an $\mathbf{X}$-extremal epimorphism and since $f=U(m)\circ_{\mathbf{X}}g$ it follows that $U(m)$ is a $\mathbf{X}$-isomorphism.

But from here I can't proceed any further.
Any hint(s) is(are) appreciated.

REPLY [3 votes]: The proposition, as written, is true if and only if the forgetful functor $U$ is conservative. Indeed, if there exists $m\colon B\to A$ such that $U(m)$ is an iso but $m$ is not an iso, then $m$ is a monomorphism in $\mathbf{A}$ since $U$ reflects monomorphisms, and $f=U(m)$ is an extremal epimorphism, but it is not extremally generating since it factors through $U(m)$, even though $m$ is not an isomorphism. Conversely, if $U$ is conservative then you can finish your proof easily.
Note that surjective functions in the category of sets coincide with extremal epimorphisms (even without the axiom of choice), and the forgetful functor $\mathbf{Top}\to \mathbf{Set}$ preserves monomorphisms. So if the proposition was true, it would imply that every surjective function from a set to a topological space is extremally generating. But as mentionned in Example 8.17(3), this is only true for discrete topological spaces. Indeed, if $(A,\tau)$ is a topological space with a non-discrete topology, then any surjective function $X\to A$ factors through the underlying function of the non-invertible continuous monomorphism  $id_A\colon (A,\mathcal{P}(A))\to (A,\tau)$ (note that this is an example of the situation in the first paragraph).
I suspect the proposition was meant to apply to concretely generating arrows instead. Indeed, if you ask that $m$ be initial in the sense of Definition 8.6(1) then you can easily prove that $U(m)$ being an iso does imply that $m$ is an iso, because the inverse of $m$ must also be an $\mathbf{A}$-morphism. And note that this is compatible with Examples 8.17(3) and (4), which says that surjective functions are concretely generating.
{"set_name": "stack_exchange", "score": 1, "question_id": 3624872}