Why isn't every category isomorphic to its opposite?

James Leslie - Why isn't every category isomorphic to its opposite?

Whenever one comes across a mathematical object, such as a group, topological space etc, it is important to look at how it interacts with other similar objects. The same goes for categories with functors being how they interact.

Given two categories $\mathscr{A}, \mathscr{B}$ , a (covariant) functor $F$ from $\mathscr{A}$ to $\mathscr{B}$ assigns to each object $A \in \mathscr{A}$ an object $F(A) \in \mathscr{B}$ and to each $\mathscr{A}$ -morphism $f:A \rightarrow A'$ , a $\mathscr{B}$ -morphism $F(f):F(A) \rightarrow F(A')$ . The assignements must respect composition and identities, so

$F(g \circ h) = F(g) \circ F(h)$ , for $\mathscr{A}$ -maps $f,g$ ,
$F(1_A) = 1_{F(A)}$ , for all $A \in \mathscr{A}$ .

We write functors as $F: \mathscr{A} \rightarrow \mathscr{B}$ . This then leads to the category $\mathbf{CAT}$ whose objects are categories and whose morphisms are functors. We say that two categories are isomorphic if they are isomorphic in $\mathbf{CAT}$ . Given a category $\mathscr{A}$ , we can form its opposite category $\mathscr{A}^{op}$ , whose objects are the same as those in $\mathscr{A}$ , with $\mathscr{A}^{op}$ -morphisms $f:A' \rightarrow A$ in one to one correspondence with $\mathscr{A}$ -morphisms $f:A \rightarrow A'$ . This means we can think of $\mathscr{A}^{op}$ as being $\mathscr{A}$ with it’s morphisms turned around. It is clear from the definition of a functor that the obvious correspondence between a category and its opposite is not a functor. However, we can define a new type of functor that makes this possible.

A contravariant functor $F$ from $\mathscr{A}$ to $\mathscr{B}$ is a (covariant) functor $F:\mathscr{A}^{op} \rightarrow \mathscr{B}$ .

The key property of a contravariant functor is that it reverses composition. If $F$ from $\mathscr{A}$ to $\mathscr{B}$ is a contravariant functor, then for $\mathscr{A}$ -morphisms $f, g$ , we have $F(f \circ g) = F(g) \circ F(f)$ .

A question one may ask is whether or not a category is isomorphic to its opposite. In general this is false; however, to newcomers this isn’t always obvious why. A typical false proof goes along the following lines:

For any category $\mathscr{A}$ , there are canonical contravariant functors $F:\mathscr{A}^{op} \rightarrow \mathscr{A}$ and $G:\mathscr{A} \rightarrow \mathscr{A}^{op}$ .
Clearly $F\circ G= 1_{\mathscr{A}}$ and $G\circ F= 1_{\mathscr{A}^{op}}$ .
$\mathscr{A} \cong \mathscr{A}^{op}$ .

The reason this doesn’t hold is because the functors defined are contravariant, not covariant. To show such an isomorphism exists, one would need to construct a covariant functor $F:\mathscr{A}^{op} \rightarrow \mathscr{A}$ , so in particular the composition rule in Definition 1 must hold. Clearly for an arbitrary category, the canonical (contravariant) functor breaks this rule. A counter-example is to have a category with a terminal object and but no initial object. Its opposite has an initial object but no terminal, hence the two aren’t isomorphic.

Take the following two categories $\mathcal A$ and $\mathcal B$ , where $\mathcal B = \mathcal A ^{\text{op}}$ .

Suppose we have a covariant functor from $\mathcal A \rightarrow \mathcal B$ that fixes the objects. The issue is that there are no arrows in $\mathcal B$ that we can map $f$ or $g$ to. Since such a functor doesn’t exist, $\mathcal A$ and $\mathcal B$ cannot be isomorphic. For such a functor to exist, it must be contravariant.