Definition 8.16. Let $\mathcal{E}$ be a category with all finite limits. A subobject classifier in $\mathcal{E}$ consists of an object $\Omega$ together with an arrow $t : 1 \to \Omega$ that is a "universal subobject," in the following sence:
Given any object $E$ and any subobject $U \to E$ (injective, but mse doesn't support tikzcd), there is a unique arrow $u: E \to \Omega$ making the following diagram a pullback:
$\cdots$
It is easy to show that a subobject classifier is unique up to isomorphism: the pullback condition is clearly equivalent to requiring the contravariant subobject functor,
$$\mathrm{Sub}_{\mathcal{E}}(-): \mathcal{E}^{op} \to \mathbb{Sets}$$
(which acts by pullback) to be representable,
$$\mathrm{Sub}_{\mathcal{E}}(-) \cong \mathrm{Hom}_{\mathcal{E}}(-, \Omega)$$
It is easy to go left to right, but I'm having trouble going the other way, from the isomorphism of the two functors to the desired UMP.
Thanks.
Best Answer
This is the usual argument how to recover a universal property from a representable presheaf. Plugging in $\Omega$ into the isomorphism of functors yields a universal subobject $t$ that corresponds to the identity on $\Omega$, and then the desired universality follows immediately since every map into $f\colon E\to\Omega$ is the image of the identity on $\Omega$ along $f^\ast$ and since $f^\ast$ acts on subobjects by means of pullback along $f$.
Edit: To see that the domain $X$ of $t$ is the terminal object, note that any morphism $f\colon E\to X$ determines a map $tf$ and therefore a subobject of $E$. By construction, $E$ must be a retract of this subobject, which is only possible if the subobject is the maximal one (i.e. the identity on $E$). As the identity on $E$ defines a subobject of $E$ and therefore gives rise to such a map $f$, this shows that there is a unique such map $f$, hence $X$ is terminal.