# Morphism in both the category and dual category is isomorphism?

If I consider the category of sets with injective functions as morphisms, denote $$Set_{\text{inj}}$$, and $$Set_{\text{sur}}$$ the category of sets with surjective functions as morphisms, I think it is true that $$\Big( Set_{\text{inj}} \Big)^{op}=Set_{\text{sur}}$$, and that if $$h$$ is a morphism in both categories then it is an isomorphism.

Is it true that if a morphism is both in the category and the dual category, then it is an isomorphism? Is it perhaps true when we also assume that the category $$\mathcal{C}$$ satisfies the Cantor-Schroeder-Bernstein property?

• It’s not true that the category of sets/injections is (or even is equivalent to) the opposite of the category of sets/injections. For example there is a set (the empty sets) with no surjections from any non-isomorphic set, but there is no set with the “dual” property of having no injections to any non-isomorphic set. Oct 8 '20 at 15:13

So let's make your claim more precise in this way: you claim that there is an isomorphism of categories $$F: (\mathbf{Set}_{inj})^{op} \to \mathbf{Set}_{sur}$$ that is the identity on objects. This cannot be true. In $$\mathbf{Set}_{sur}$$ the empty set is disconnected from the rest of the category: there are no arrows with domain or codomain $$\emptyset$$, other than the identity arrow. On the other hand, in $$\mathbf{Set}_{inj}$$ we have an arrow $$\emptyset \to X$$ for every set $$X$$. In fact, this means that there cannot even be a functor $$F: (\mathbf{Set}_{inj})^{op} \to \mathbf{Set}_{sur}$$ such that $$F(\emptyset) = \emptyset$$.
Unfortunately, the rest of your question is not precise enough to answer. You seem to be interested in the fact that in $$\mathbf{Set}$$ we have that an injective and surjective function is a bijection. So in category-theoretic terms: an arrow that is a monomorphism and epimorphism is an isomorphism. This is not generally true in categories, but there are reasonable assumptions where it does work (e.g. one of the morphisms is regular). See this question for more details.