I am having some trouble proving exercise 4.23 in the book "Sets for Mathematics" by Lawvere and Rosebrugh.
The problem is to prove that from a partition $p:A \rightarrow I$, which is defined there as a morphism $p$ which is a surjection/epimorphism with $A$ as domain. Taking a pullback, labelled $R_p$, of $p$ along itself should then give a monomorphism $(p_0,p_1):R_p \rightarrow A \times A$, which is the definition of being a relation on $A$, where $p_0:R_p \rightarrow A$ and $p_1:R_p \rightarrow A$ are the respective morphisms coming from taking the pullback of $p$ along itself such that $p p_0 = p p_1$:
My work so far is:
Let $t_0,t_1 : T \rightarrow R_p$ be morphisms such that $(p_0,p_1)t_0=(p_0,p_1)t_1$ and prove that $t_0=t_1$.
I then use the UMP of the product $A \times A$ in order to prove that there is a $\phi : T \rightarrow A \times A$ such that $p_0 t_0 = \pi_0 \phi$ and $p_1 t_1 = \pi_1 \phi$.
From here I don't know how to continue since I cannot see where/if I must use the UMP of the pullback at all?
Many thanks in advance!
You want to prove that $t_0=t_1$. But $t_0,t_1 : T \to R_p$ have a pullback as codomain. Your best bet is to use uniqueness in the UMP of $R_p$.
Meaning: try to find arrows $f,g : T \to A$ such that $pf=pg$ and such that $$ p_0t_i = f, \quad p_1t_i = g \qquad \forall i \in \{0,1\}.$$
(If you draw a diagram with all your data, you'll see you don't really have a choice for $f$ and $g$.)