Proof that $ \textbf{Set}\ $ is not isomorphic to its dual I am trying to understand in an intuitive and straightforward way why the category $ \textbf{Set}\ $ is not isomorphic to its dual. An answer that I find appealing is [here] (Why is every category not isomorphic to its opposite?) where one respondent states:
"Perhaps the easiest way to see that the category of sets isn't isomorphic to its dual is to observe that there is an object, namely the empty set, such that all morphisms into it are isomorphisms, but there is no object such that all morphisms out of it are isomorphisms."
However, I fail to see the reasoning here.  To prove or disprove isomorphism between categories we need to look at functors between the two categories.  And any functor is of course just a pair of functions that links objects in one category to objects in the other, and morphisms in one category to morphisms in the other.
How, then, do morphisms and isomorphism within a category figure into the problem. Obviously, I am missing a link somewhere.
 A: This is a way of thinking that you have to get used too: if two categories are equivalent and one has a categorical property then so has the other. All the goal is to understand what is a categorical property. (Of course we could say that they are the properties perserved by equivalence, but hum...)

An example of such a categorical property: there is an terminal object.
A counterexample: there is a unique object.

The property you describe is categorical. A precise proof follows.

Suppose $F : \mathbf{Set} \to \mathbf{Set}^{\rm op}$ is an equivalence with pseudo inverse $G:\mathbf{Set}^{\rm op} \to \mathbf{Set}$. 
Remark that $\emptyset \in \mathbf{Set}$ has the curious property that any map $f: X \to \emptyset$ forces $X$ to be isomorphic to $\emptyset$ and $f$ to be such an isomorphism$^{\text{1}}$. Let us then prove that this property should also be satisfied by $F(\emptyset)$ in $\mathbf{Set}^{\rm op}$: 

If $g : Y \to F(\emptyset)$ in $\mathbf{Set}^{\rm op}$, then $G(g)$ is a map $G(Y) \to GF(\emptyset)$ in $\mathbf{Set}$. But there is an natural isomorphism $\alpha: GF \Rightarrow \mathrm{Id}_\mathbf{Set}$ (by definition of a pseudo inverse), so that $\alpha_\emptyset \circ G(g)$ is a map $GF(X) \to \emptyset$. The curious property says that this is an isomorphism. But as $\alpha_\emptyset$ already is an iso, it means that $G(g)$ is an isomorphism. Now you should know that an equivalence (in fact a fully faithful functor is enough) reflects isomorphisms (if you do not, prove it as an exercise!): that means that $G(g)$ is iso if and only if $g$ is. So we get that $g$ is an isomorphism.

Now you just have to prove that such an object cannot exist in $\mathbf{Set}^{\rm op}$. But it the same as proving that in $\mathbf{Set}$ we can not have an object $A$ such that any morphism $A \to X$ is a isomorphism. Which is obvious: one always have an injection $A \to A + 1$ which is never a surjection ($A+1$ is the set $A$ in which we add a new element distinct from any other and the map $A \to A+1$ is just the inclusion of $A$ in that new set). 



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*depending on your fondations, every set $X$ isomorphic to $\emptyset$ could be equal to $\emptyset$, but this is irrelevant here and reasoning up to isomorphism is usually better in category theory.

