What is the meaning of a monomorphism in $S/X$ being "fiberwise"?

Question

Notation: $S$:the category of abstract sets; $S/X$: the slice category of $S$ over a set $X$; $A_x$: the fiber of a set $A$ over an element $x$ of the codomain of a function $A\rightarrow X$.

My question is the following: If $\alpha:f\rightarrow g$ is a monomorphism in $S/X$, what does it mean for it to be "fiberwise"? The question appears as exercise 2.43 in "Sets for Mathematics" by F.W. Lawvere and R. Rosebrugh and I have included it below for convenience's sake:

An important case of slice categories (see Exercise 1.30(e)) is the category of X-indexed families of abstract sets $S/X$. Recall that in $S/X$ objects are mappings with codomain $X$ and arrows are commutative triangles. The name “family” arises as follows: For any object $f:A\rightarrow X$ of $S/X$ and any element $x:1\rightarrow X$ the inverse image of $x$ along $f$ is a part of A denoted $A_x$ and is called the “fiber of $A$ over x”, A is the “sum” of the family of all its fibers. This is a very simple example of a variable set.

1. Show that the category $S/X$ has binary sums that are computed “fiberwise”.
2. Show that monomorphisms in $S/X$ are also “fiberwise” and have characteristic morphisms taking values in the object $\Omega$ of $S/X$, which has each fiber equal to $2$.

I understand that a monomorphism in $S/X$ precisely corresponds to an injective function that respects the fiber-structure induced on the domain by morphisms into $X$. Further more, if $f:A'\rightarrow X$ and $g:A''\rightarrow X$ then given any element $x\in X$,

$$\alpha A'_x \subseteq A''_x$$

This implies that in order to specify a monomorphism in $S/X$, it involves for each fiber $A'_x$, choosing a 'part' of the fiber $A''_x$. Is that a correct interpretation of the adjective "fiberwise"?

Answer 1

Given $A\stackrel f \to X$ and $B\stackrel g \to X$, for every morphism $\alpha\colon A\to B$ in $S/X$ and every element $x\colon 1\to X$, there is a commuting diagram as below.

$\hskip2in$

The leftmost map $\alpha_x$ exists thanks to the universal property of the lower square, combined with the commutativity of the upper square and the rightmost triangle (notice that we need not use the universality of the upper square).

The usual terminology is that a property is fibrewise when it holds on $\alpha$ if and only if it holds on $\alpha_x$ for all $x$.

Now, consider the property of being a monomorphism. Since the top square is a pull-back, $\alpha$ being a monomorphism implies that $\alpha_x$ is also a monomorphism, for all $x$.

Conversely, if $\alpha$ is not a monomorphism, in particular it is not an injective map of sets over $X$, and there is an $x$ for which the corresponding $\alpha_x$ is not injective.

In conclusion, we can check the property of being a monomorphism on the fibres - this is the meaning of being fibrewise.