Why isn't every category isomorphic to its opposite?

Posted on July 7, 2018 by James Leslie

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:

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

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.