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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.

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1 Answer 1

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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.

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    $\begingroup$ How can I know that the domain of t is the terminal object 1, or do I need to? $\endgroup$ Commented Jul 23, 2021 at 12:58
  • $\begingroup$ @AprilGrimoire Good question. In fact, you can deduce this from the definition of subobject classifier! So if you replace $1$ in the definition tot quoted by some object $X$ (and stipulate that $t$ must be a monomorphism) then you can deduce that $X$ is the terminal object. $\endgroup$ Commented Jul 23, 2021 at 13:47
  • $\begingroup$ Please see my edit. $\endgroup$
    – asdq
    Commented Jul 23, 2021 at 14:20
  • $\begingroup$ @asdq In your edit: how do you know there is a morphism $E \to X$. $\endgroup$ Commented Jul 23, 2021 at 14:25
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    $\begingroup$ I mean the map into the subobject (which is defined as the pullback of $t$ along $tf$) that is induced from the pair $(f, \operatorname{id}_E)$ via the universal property of pullback. $\endgroup$
    – asdq
    Commented Jul 24, 2021 at 9:07

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