Classifying map of a graph of a morphism Let $\epsilon$ be a topos. Let $f:Y\longrightarrow X$ be a morphism in $\epsilon$. By the graph of f we mean the mono $<id_Y,f>:Y\longrightarrow Y\times X$. Let $\Delta$ be the classifying map of the diagonal subobject $\delta:X\longrightarrow X\times X$ ($\delta=<id_X,id_X>$). Show that $\Delta o (f\times id_X)$ classifies the graph of f.
I tried to use the pullback lemma but I couldn’t prove that the other square with arrows $<id_Y,f>, f\times id_X,f, \delta$ is a pullback.
 A: By our favorite pullback lemma, if the top square in
$$\require{AMScd}\begin{CD} \bullet @>>> Y\times X \\
@VVV @VV{f\times id_X}V \\
X @>\delta_X>> X\times X \\
@VVV @VV\Delta_XV \\
1 @>>\top>\Omega \end{CD}$$
is a pullback then the outside rectangle is a pullback and the top horizontal morphism is the subobject classified by the right side. It's easy to see that $X\overset{f}{\longleftarrow}Y\overset{\langle id_Y,f\rangle}{\longrightarrow}Y\times X$ makes the top square commute, so we only need to show that it's a pullback to prove the theorem.
Let $X\overset{j}{\longleftarrow}J\overset{\langle k_1,k_2\rangle}{\longrightarrow}Y\times X$ be such that it makes the top square commute. Then $\langle j,j\rangle=\langle fk_1,k_2\rangle=\langle fk_1,fk_1\rangle$. But this means $j=fk_1$ and $\langle k_1,k_2\rangle =\langle k_1,fk_1\rangle=\langle id_Y,f\rangle\circ k_1$. So we have that there exists a map, namely $k_1$, $J\to Y$ with the requisite property.
So we show that this $k_1$ is unique. Let $i:J\to Y$ be any morphism such that  $fi=j$ and $\langle id_Y,f\rangle\circ i=\langle k_1,k_2\rangle$. But $\langle i,fi\rangle=\langle k_1,k_2\rangle$ immediately implies $i=k_1$, so our $k_1$ is unique.
Hence $X\overset{f}{\longleftarrow}Y\overset{\langle id_Y,f\rangle}{\longrightarrow}Y\times X$ makes the top square a pullback, and $\langle id_Y,f\rangle$ is the map classified by $\Delta_X\circ(f\times id_X)$.
