Exercise with subobject classifier Suppose that $\bf C$ is a category with a subobject classifier $t:T\to \Omega$. If $A$ is a subobject of $B$, and $B$ is a subobject of $X$, $A$ is also a subobject of $X$. Denote the characteristic functions of $A$ as a subobject of $B$ and $X$, respectively, $\chi:B \to \Omega$ and $\bar \chi:X \to \Omega$. I must describe the relation between $\chi$ and $\bar \chi$.
The proof that $A\hookrightarrow B\xrightarrow \chi \Omega$ is the same arrow as $A\hookrightarrow B\hookrightarrow X\xrightarrow{\bar\chi} \Omega$ is straightforward; however it is likely that $\chi$ factorizes in $B\hookrightarrow X\xrightarrow{\bar\chi} \Omega$. Can you give me an hint to prove it?
I tried to use the fact that if a rectangle formed by two squares is a pullback and the right square is a pullback, the left square is a pullback too, but it is of no use; in fact I would need that if the left square is a pullback the right square is a pullback too (or at least is commutative) but these facts aren't true in general. I tried to involve the characteristic function of $B\hookrightarrow X$ but it didn't seem useful neither.
Thanks for any suggestion
 A: The relationship is that $\chi$ is $\overline{\chi}$ restricted to the domain of $B$. In other words, if $i_B : B \to X$ is the subobject map, then $\chi = \overline{\chi} \circ i_B$.
To prove this, it suffices to show that the pullbacks of $t$ along $\chi$ and along $\overline{\chi} \circ i_B$ are the same.
Of course, the pullback of $t$ along $\chi$ is $A \hookrightarrow B$.
Now pulling back is functorial. The pullback of $t$ along $\overline{\chi}$ is $A \hookrightarrow B \hookrightarrow X$. So we must compute the pullback of this map along $i_B$.
To do this, note that the pullback of $i_B$ along itself is the identity map $1_B : B \to B$ since $i_B$ is monic. And the pullback of Therefore, the pullback of $A \hookrightarrow B \hookrightarrow X$ along $i_B$ will be $A \hookrightarrow B$, as required.
In terms of diagrams, we simply show that this diagram is a pullback:
$$\require{AMScd}
\begin{CD}
A @= A @>>> 1 \\
@VVV   @VVV   @VVV \\
B @=   B    \\
@|  @VVV \\
B @>>> X @>>> \Omega
\end{CD}$$
By showing all 3 of the smaller rectangles contained therein are pullbacks and using the pasting properties of pullback squares.
(just pretend the arrow stretches all the way from $1$ to $\Omega$ - not sure how to format this).
A: Since this question is still open, I'll add a slightly more pedestrian take on Mark Saving's argument. Of course, if you unwind everything, we're really doing the same thing.
First remember that the characteristic function associated to a subobject is unique. So to show that
$\chi = \overline{\chi} \circ \iota$ (where $\iota : B \rightarrowtail X$), it suffices to show that $\overline{\chi} \circ \iota$ is a characteristic function for $A \rightarrowtail B$. That is, it suffices to show

is a pullback.
Of course, we can show a bit more information in this diagram, and that will make it easy to check this directly:

We know that the right hand square is a pullback (since $\overline{\chi}$ is the characteristic function for $A \rightarrowtail X$) and the left hand square is a pullback (a simple exercise), but this makes the outer square a pullback, as desired.

I hope this helps ^_^
