Verifying a Construction Satisfies the $\Omega$-axiom. I'm stuck on Exercise 4.5.1 of Goldblatt's, "Topoi: A Categorial Analysis of Logic".
It's in the topos $\mathbf{Bn}(I)$ of bundles over a set $I$. Goldblatt asks the reader to verify that
$\tag{1}$
satisfies the $\Omega$-axiom.${}^\dagger$ The construction is defined in the first link above.
For convenience: here $(A, f)\stackrel{k}{\rightarrowtail}(B, g)$ is an arbitrary, monic $\mathbf{Bn}(I)$-arrow, taken as an inclusion; $(I, \operatorname{id}_I)$ is the terminal object in $\mathbf{Bn}(I)$; $p_I$ is the projection $p_I(\langle x, y\rangle)=y$; $\top$ is defined by $\top(i)=\langle 1, i\rangle$; and $\chi_k$ is the product map $\langle\chi_A, g\rangle$, i.e., $$\chi_k(x)=\begin{cases}\langle 1, g(x)\rangle &: x\in A \\ \langle 0, g(x)\rangle &: x\notin A.\end{cases}$$
Thoughts: What I've done so far is replace $\chi_k$ with an arbitrary $\mathbf{Bn}(I)$-arrow $h: \langle B, g\rangle\to \langle 2\times I, p_I\rangle$ in $(1)$, supposing what I get is a pullback. Then I've run it though the definition of a pullback quite easily. I've had numerous stupid ideas about what to do next (with all manner of confusing diagrams) but to no avail.

I'd like a detailed solution, please.

It should be easier than I think it is. Maybe my problem is with bundles themselves. This is my second attempt at reading Goldblatt's book: last time I thought I had'm but got up to "11.4: Models in a Topos" - right where I wanted to be - before other commitments made me lose track entirely; now I'm about to read "4.8: $\Omega$ and comprehension".
Please help :)

$\dagger$: The $\Omega$-axiom is given on page 81, ibid., via the definition of a subobject classifier:

Definition: If $\mathbb{C}$ is a category with a terminal object $1$, then a subobject classifier for $\mathbb{C}$ is a $\mathbb{C}$-object $\Omega$ with a $\mathbb{C}$-arrow $\text{true}: 1\to\Omega$ that satisfies the following axiom.
$\Omega$-axiom: For each monic $f:a\rightarrowtail d$ there is one and only one $\mathbb{C}$-arrow $\chi_{f}:d\to\Omega$ such that $\chi_f\circ f=\text{true}\circ !$ is a pullback square.

 A: $\require{AMScd}$
First, recall that $\mathbf{Bn}(I)$ is just a notation for the slice category $\mathsf{Set}_{/I}$.

Lemma. For any category $\mathscr C$ and any object $c$ of $\mathscr C$, the forgetful functor $\mathscr C_{/c} \to \mathscr C$ commutes with fibered products.

So if you have a pullback as in your question, the square 
$$ \begin{CD}
A @>k>> B \\
@VfVV @VV\chi_k V \\
I @>>\top> 2 \times I
\end{CD} $$ 
is a pullback in $\mathsf{Set}$.
Then, remark that the square
$$ \begin{CD}
I @>\top>> 2\times I \\
@VVV @VVV \\
1 @>>\mathrm{true}> 2
\end{CD} $$ 
is also a pullback ($\mathrm{true}$ being the map selecting $1 \in 2$).
So, concatenating the two squares makes the outer square of
$$ \begin{CD}
A @>k>> B \\
@VfVV @VV\chi_k V \\
I @>>\top> 2 \times I \\
@VVV @VVV \\
1 @>>\mathrm{true}> 2
\end{CD} $$ 
a pullback again. But then $2$, equipped with the map $\mathrm{true} \colon 1 \to 2$, is a subobject classifier for $\mathsf{Set}$. From here, you can easily derive the uniqueness of $\chi_k$ (remember that $p_I \circ \chi_k$ is fixed to be $g$ by hypothesis). 
A: Just to fill in the gaps of the great answer above.  Understand how PBL (pullback lemma) works in Goldblatt.  And understand that for products there exists a unique arrow $\langle p, q \rangle$ for any third object $a \xleftarrow{p} c \xrightarrow{q} b$ onto the argument objects (of the product).

Note that $! \circ f = !$ and that in set $\chi_A$ is already known to exist (regardless of uniqueness) since $\textbf{Set}$ has a subobject classifier $\Omega$.  Thus after pasting the two pullbacks, you get a pullback diagram on the right (the square).  But more importantly, we know that the morphism $\chi_A$ is the unique such morphism that creates that pullback square by the $\Omega$ axiom in $\text{Set}$.  
Thus $\chi_k = \langle m, g \rangle$ where $m = \chi_A$.  Now apply the product rule to the object $B$ with its two projectors onto the arguments of $2 \times I$, namely $\chi_A : B \to 2$ and $g:B\to I$.  Then by definition of $\langle \chi_A, g\rangle$ it is the unique such arrow (we call it $\chi_k$) such that $p_2 \circ \chi_k = \chi_A$ and $p_I \circ \chi_k = g$.  That's using the universal property of product.  
But that is the same as saying it is the unique arrow such that the square on the top left (together with arrows into $I$) above is a pullback in $\text{Bn}(I)$ because one of the conditions that the whole thing commutes or in particular that $p_I \circ \chi_k = g$. 
As is often the case in mathematics, we don't always make full use of a condition, i.e. we only said "such that $p_I \circ \chi_k = g$" and didn't mention all the other conditions going on in the pullback.  One condition was enough in this case.
