Doubt in proof showing in topoi, equivalence relations are effective

Question

This is proposition 1.4 (pg. 30) in van Oosten's lecture notes on Topos Theory.
Prop: In a topos, every equivalence relation is effective, i.e. a kernel pair.
Following is the first half of the proof.
Let $\phi : X \times X \to \Omega$ classify the subobject $\langle r_0, r_1 \rangle : R \to X \times X$, and let $\overline{\phi} : X \to Ω^X$ be its exponential transpose. We claim that the square,

is a pullback, so that $R$ is the kernel pair of $\overline{\phi}$. To see that it commutes, we look at the transposes of the compositions $\overline{φ}r_i$, which are arrows,
$$R \times X \xrightarrow{r_i \times id} X \times X \xrightarrow{\phi} \Omega$$
Both these maps classify the object $R′$ of $R$-related triples, seen as subobject of $R \times X$, so they are equal.
I was able to understand this proof until the last line. How are we able to consider $R'$ as a subobject of $R \times X$? And how is the arrow $\phi (r_i \times id)$ the classifying arrow of $R'$.
$R'$ is defined as part of pullback square:

According to this answer, there is typo. Instead of $X \times X$, it should be $X$.
I have been stuck for hours trying to understand this. I would appreciate some help.

Answer 1

Consider first the diagram $$ \require{AMScd} \begin{CD} R'@>{t}>> R @>>> *\\ @V{\langle s, r_1t\rangle}VV @VV{\langle r_0,r_1\rangle}V @VV{\top}V\\ R\times X @>{r_1\times\mathrm{id}_X}>> X\times X@>{\varphi}>>\Omega\\ @V{\mathrm{pr}_0}VV @VV{\mathrm{pr}_0}V\\ R @>{r_1}>>X \end{CD} $$ where $*$ is the terminal object of your topos and the top right square is a pullback square. The bottom left square is also a pullback, as is the large left vertical rectangle, both by direct inspection. By pullback pasting, this means that the upper left square is a pullback. In particular, the map $R'\to R\times X$ is the pullback of a mono, and hence a mono as well.

A similar argument shows that the map $X\times R\xrightarrow{\mathrm{id}_X\times r_0} X\times X\xrightarrow{\varphi}\Omega$ classifies $R'$, but note that we have swapped the terms in the categorical products. To show that the map $R\times X\xrightarrow{r_0\times\mathrm{id}_X}X\times X\xrightarrow{\varphi}\Omega$ classifies $R'$, it suffices to show that $X\times X\xrightarrow{\langle\mathrm{pr}_1,\mathrm{pr}_0 \rangle}X\times X\xrightarrow{\varphi}\Omega$ equals $X\times X\xrightarrow{\varphi}\Omega$, where the map $\langle\mathrm{pr}_1,\mathrm{pr}_0 \rangle\colon X\times X\to X\times X$ is the transposition map.

To show this, we can use that maps into $\Omega$ classify subobjects. $\varphi$ classifies $\langle r_0,r_1\rangle\colon R\to X\times X$, and by pullback pasting we find that $\varphi\circ\langle\mathrm{pr}_1,\mathrm{pr}_0\rangle$ classifies $\langle r_1,r_0\rangle\colon R\to X\times X$. However, we know by symmetry of $R$ that the transposition $\langle\mathrm{pr}_1,\mathrm{pr}_0 \rangle\colon X\times X\to X\times X$ restricts to a self-inverse isomorphism $T\colon R\to R$. This isomorphism satisfies that $T\circ\langle r_0,r_1\rangle = \langle r_1,r_0\rangle$. This shows that both subobjects $\langle r_0,r_1\rangle,\langle r_1,r_0\rangle\colon R\to X\times X$ are equal in $\mathrm{Sub}(X\times X)$. Therefore, the maps $X\times X\xrightarrow{\langle\mathrm{pr}_1,\mathrm{pr}_0 \rangle}X\times X\xrightarrow{\varphi}\Omega$ and $X\times X\xrightarrow{\varphi}\Omega$ are equal, with which we are done.