# 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 $\mathbf{Sets} \stackrel{\sim}{=} \mathbf{Sets}^\mathsf{op}$, 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.

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That functor is contravariant rather than covariant. –  Tobias Kildetoft Sep 19 '13 at 9:45
Ah, I see, because it reverses the direction of morphisms! But why must an isomorphism be a covariant functor then? –  Ulrik Rasmussen Sep 19 '13 at 9:47
@UlrikRasmussen It's the distinction between "preserve composition" and "reverse composition". Some call a "composition-reversing isomorphism" a "dual isomorphism". Others insert $\sf op$ as needed. –  Lord_Farin Sep 19 '13 at 9:51
In fact, the functor $(-)^\mathrm{op} : \mathbf{Cat} \to \mathbf{Cat}$ is an involution. But that's a whole different story. –  Zhen Lin Sep 19 '13 at 10:01
An isomorphism must be a covariant functor because this is how it has been defined. An invertible contravariant functor is usually called either an anti-isomorphism or an anti-equivalence. –  Tobias Kildetoft Sep 19 '13 at 10:35

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.

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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.

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Right, needless to say that the only morphism going into the empty set is its own identity, which is indeed an iso. –  magma Sep 20 '13 at 9:12

As a new stackexchange user (and beginning student of category theory) the site will not let me comment, but this is aimed at the answer by @andreas-blass:

Is it in fact the case that the empty set is an object such that all morphisms into it are isomorphisms? It seems there would be a morphism $f:A\to\emptyset$ from any set A into the empty set, each one the empty function, but each with a different dom(f)=A and as such distinct as arrows of Set? It then seems that of these, only $id_\emptyset:\emptyset\to\emptyset$ is an isomorphism, seeing as how $\#A=\#\emptyset=0$ for $A=\emptyset$ only?

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There is no function from $A\to\varnothing$ if $A$ is nonempty. –  Nishant Nov 22 at 1:09
This is a question, not an answer. Questions should be asked separately. –  Qiaochu Yuan Nov 22 at 9:18