This question is not really about elementary topoi, it is much more about a category $(\mathcal{E}, \Omega)$ admitting a subobject classifier, or about a category with power objects, you can choose the context that inspires you the most. Of course, elementary topoi are the go-to example.

Q: Can we prove, or provide a counter-example, that the opposite of such a category cannot have a subobject classifier, or have power objects, or be an elementary topos?

Rem. It is well known that the opposite of a Grothendieck topos cannot be a Grothendieck topos, but the proof relies on the fact that the opposite of a locally presentable category cannot be locally presentable. This completely obscures the importance of $\Omega$, which is what I would like to know about.

Rem. Even in the most trivial case, why do we know that $\mathsf{Fin}^\circ$, or $(\mathsf{Fin}^C)^\circ$ does not have a subobject classifier?

Clarification. The title mentions elementary topoi because "category with a subobject classifier" would not sound as good.

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    $\begingroup$ Just a comment because I don't know what to say about the general case: For the opposite of Fin, we would get a "quotient coclassifier" in Fin, i.e. an epimorphism such that every other epimorphism is uniquely a pushout along it. But that's absurd, for example pushouts along a fixed epimorphism of finite sets can't reduce the cardinality by more than a fixed amount. $\endgroup$ – Achim Krause Apr 12 at 21:11
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    $\begingroup$ In a cartesian closed category $C$, the functor $-\times\emptyset$ preserves the initial object. Thus the set $X=C(c,\emptyset)$ satisfies that the projection $X\times C(c,d)\to X$ is a bijection for any $d$, which implies that $X=\emptyset$ unless $c$ is initial. In particular, the subobject classifier would have to be initial, which is probably enough to get a contradiction (and deals with the examples you gave). $\endgroup$ – Bertram Arnold Apr 12 at 21:39
  • $\begingroup$ @BertramArnold Is this a reformulation of Tom's answer? Anyway, I like it much more! If you spell out it and make it work for the case that $\mathcal{E}$ has a subobject classifier and is monoidal closed, I will accept it as an answer. $\endgroup$ – Ivan Di Liberti Apr 12 at 21:52

The answer to the question in the title is no, assuming you want to exclude the trivial case of the terminal category.

Let $E$ be an (elementary) topos whose opposite is also a topos. The initial object of a topos is strictly initial (any map into it is an iso), so the terminal object $1$ of $E$ is strictly terminal. Since there is a map from $1$ to the subobject classifier $\Omega$ of $E$, this implies that $\Omega$ is terminal. Hence every object of $E$ has precisely one subobject. In particular, $1$ has only one subobject. But the initial object $0$ of $E$ is a subobject of $1$. So $0 \cong 1$ in $E$, so $E$ is trivial.

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    $\begingroup$ Thanks Tom, this is a very good starting point. I upvoted, but for the moment I will not accept, as you rely on an exactness property of the elementary topos which might be lost in more general but similar contexts. $\endgroup$ – Ivan Di Liberti Apr 12 at 21:36
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    $\begingroup$ People may be disappointed to find out that an elementary topos is not a profound truth. $\endgroup$ – Will Sawin Apr 12 at 21:37

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