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I have to give a talk about some categorical things in a student seminar soon. As this is an introductory talk I cannot assume much knowledge and need very basic arguments.

For example I want to present that the category of sets is not self-dual, i.e. there is no equivalence of categories $F : Sets \rightarrow Sets^{opp}$. I will define an equivalence of categories as a fully faithful functor which is also essentially surjective as this is easier to explain in a short amount of time than natural transformations. Could you give me a simple argument? One argument I found online (actually here on stacksexchange) is that $Sets$ is a distributive category and $Sets^{opp}$ is not but this is too difiicult to explain.

Btw: Does one know a more concrete category that is equivalent to $Sets^{opp}$?

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I guess this one should be easy enough. Suppose $F:\textbf{Set} \rightarrow \textbf{Set}^{\text{op}}$ is an equivalence of categories. Recall, that $\text{Hom}_{\textbf{Set}}(M,N) = \emptyset$ iff $M$ is non-empty and $N$ is empty. Now let $M$ be a non-empty set. We get $$\emptyset = \text{Hom}_{\textbf{Set}}(M,\emptyset) \cong_{\textbf{Set}} \text{Hom}_{\textbf{Set}^{\text{op}}}(F(M),F(\emptyset)) = \text{Hom}_{\textbf{Set}}(F(\emptyset),F(M)),$$ such that $F(M) = \emptyset$ and $F(\emptyset) \neq \emptyset$. Thus $F$ is not essentially surjective in contradiction to our assumption.

Yes, the category $\textbf{Caba}$ of complete atomic boolean algebras. This is sometimes referred to as the Lindenbaum-Tarski duality and the equivalence is given by the power set functor.

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    $\begingroup$ $F$ is not essentially surjective because no set of cardinality different than $|F(\emptyset)|$ and 0 is of the form $F(X)$? $\endgroup$ – Josh Jun 18 '19 at 18:34
  • $\begingroup$ Yes, exactly. You got it. $\endgroup$ – TMO Jun 18 '19 at 18:38
  • $\begingroup$ Thanks a lot for the example, but it probabaly also takes too long to explain the details of your cabas. $\endgroup$ – Josh Jun 18 '19 at 18:43
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$\textbf{Set}$ has an initial object (empty set) and a terminal object (any one-element set). There is a morphism from any initial object to any terminal object but no morphism from any terminal object to any initial object.

Dually, $\textbf{Set}^{\textrm{op}}$ has an initial object and a terminal object. There is a morphism from any terminal object to any initial object but no morphism from any initial object to any terminal object.

So $\textbf{Set}$ and $\textbf{Set}^{\textrm{op}}$ cannot be equivalent categories.

Now when you talk about "a more concrete category" do you mean a concrete category? That Wikipedia pages shows how $\textbf{Set}^{\textrm{op}}$ is a concrete category.

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  • $\begingroup$ No, I just wanted an example of a category that is equivalent to $Sets^{opp}$ to show how we can understand $Sets^{opp}$ in a more concrete way without just reversing arrows. $\endgroup$ – Josh Jun 18 '19 at 18:36
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    $\begingroup$ Your argument is essentially the same as the one of @ThorWittich right? Just for singletons? $\endgroup$ – Josh Jun 18 '19 at 18:44

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