Why is every category not isomorphic to its opposite? This is a beginner category theory question:
I'm trying to wrap my head around the fact that we do not have $\mathbf{Sets} \cong \mathbf{Sets}^\mathsf{op}$, i.e., the category of sets is not isomorphic to its opposite.
As far as I have understood, an isomorphism between categories is a functor $F : \mathcal{C} \rightarrow \mathcal{D}$ for which there is an inverse $G : \mathcal{D} \rightarrow \mathcal{C}$ such that $FG = 1_\mathcal{D}$ and $GF = 1_\mathcal{C}$, i.e., an isomorphism in the category of categories.
Now, consider the functor $-^\mathsf{op}$, i.e., the functor sending its category to its opposite. Why is this functor not an isomorphism (for any category!), with itself as the inverse? I know that I have missed something, since this would imply that $\mathbf{Sets}$ is isomorphic to its opposite.
 A: Here's a good hands-on example.
Posets can be thought of as categories where any two objects have at most one arrow between them. If you start with a poset $(P,\leqslant)$ then you can form a category $\mathscr{C}_P$ whose object set is $P$ itself and $u,v\in P$ have an arrow $u\to v$ if and only if $u\leqslant v$.
Conversely, if $\mathscr{C}$ is a small category with at most one arrow between objects, you can form a poset $(P_\mathscr{C},\leqslant_\mathscr{C})$ by saying $u\leqslant_\mathscr{C} v$ if and only if there is an arrow $u\to v$.
Moreover, isomorphisms between posets and their corresponding categories are the same thing. I think that should be easy to see.
So, start with your favorite poset $(P,\leqslant)$. Interpret this then as a category $\mathscr{C}_P$. When you take the opposite category $\mathscr{C}_P^\text{op}$ you clearly get another poset (still at most one arrow!). Which poset? The opposite poset! Namely, $(P,\leqslant_\text{op})$ where $x\leqslant_\text{op}y$ if and only if $y\leqslant x$.
In particular, it's easy to see that $\mathscr{C}_P^\text{op}\cong\mathscr{C}_P$ if and only if $(P,\leqslant_\text{op})\cong(P,\leqslant)$
Now, why is this relevant for us? Take any poset that has a minimal element, but no maximal element (say $\mathbb{N}$ with the normal ordering). Then, it's not hard to see that the $(P,\leqslant_\text{op})$ has a maximal element, but no minimal element! But, these are properties that are preserved under isomorphisms of posets. So, $(P,\leqslant_\text{op})\not\cong (P,\leqslant)$ and so $\mathscr{C}_P^\text{op}\not\cong\mathscr{C}_P$.
Of course, the problem is, as others have pointed out to you, the "isomorphism" flips everything (e.g. the minimal element became the maximal element).
Thinking about posets is a good way to test understanding. Keep it in your back pocket.
A: 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.
A: The arguments above can be rephrased saying that ${\bf Sets}$ has a single initial objet (the empty set), and infinitely many final objects (all singleton sets), while for ${\bf Sets^{op}}$ this is all reversed.
An arrow of ${\bf Set}$ can be viewed as a family of sets indexed by the codomain, while in $\bf Set^{op}$, an arrow is a family of sets indexed by the domain. Still, the two categories are closely related, and would it not be legitimate to say that they are 'anti-isomorphic'?
In a debate between Bénabou, Cartier and others available on the web, it is pointed out that this makes the category ${\bf Set^{op}}$ "non-set-like", and more "topos-like". I didn't really understand, but this might be related to differences in the behavior of subobjects (which are quotient objects) and "truth values" in ${\bf Set^{op}}$. 
