EnjoysMath

One way to understand these maps is to think of them as tuples of set maps $f: A \to C, g : B \to D$.  So if you're having trouble despite, the following categorical explanation, just think of them that way - the way from which the categorical construction is derived. Suppose that the products $A\times B$ and $C\times D$ exist in our category.  This means that by definition: Or in English: $A \xleftarrow{p_1} A \times B \xrightarrow{p_2} B$ is a product diagram if and only if for every glued in diagram $A \xleftarrow{x_1} X \xrightarrow{x_2} B$, there exists a unique map (i.e. UMP property here) $u: X \to A\times B$ such that everything commutes, i.e. $p_i u = x_i$ for each $i=1,2$. Now, we simply glue in the arrows $f$ and $g$: as well as the product $C \times D$ and its  projection maps: Can you finish the derivation from here?  If not, then please continue. Now, instead of the UMP for $A\times B$, use the UMP of $C\times D$ to get that there exis...

  Proposition  Let $\mathcal{E}$ be a topos with subobject classifier $\Omega$.  If $f: X \to \Omega$, then $f = \chi_m$ for some mono  $m: Y \rightarrowtail X$. Proof A topos is closed under taking pullbacks.  We already have the diagram, Taking its pullback, we get a diagram, But a pullback of $\text{true}$ along any morphism $f$ is always monic, By the definition of subobject classifier, namely the part about uniqueness of $\chi_m$, we must have that $\chi_m = f$. $\blacksquare$