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.
Best Answer
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.