Suppose $F: C \rightarrow D$ is a covariant functor and $G: C \rightarrow D$ is contravariant. Does it make sense to define natural transformations $\eta_1 : F \Rightarrow G$ or $\eta_2 : G \Rightarrow F$ or are there any obstructions?

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    $\begingroup$ Well, no. Some of the arrows point the wrong way in the obvious "commutative diagram" you'd want to write down. $\endgroup$ Mar 15, 2012 at 2:58
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    $\begingroup$ Natural transformations are only defined between functors of the same type. Since a contravariant functor from $\mathcal{C}$ to $\mathcal{D}$ is really just a functor $\mathcal{C}^\textrm{op} \to \mathcal{D}$, you can't really talk about natural transformations from a functor $\mathcal{C} \to \mathcal{D}$ to $\mathcal{C}^\textrm{op} \to \mathcal{D}$. That said, there is a notion of "extraordinary natural transformation", which in a degenerate case lets you transform a functor $\mathcal{C} \to \mathcal{D}$ to a functor $\mathcal{C}^\textrm{op} \to \mathcal{D}$. $\endgroup$
    – Zhen Lin
    Mar 15, 2012 at 7:25

1 Answer 1


It does make sense. If $F$ is covariant and $G$ is contravariant, a transformation $F \to G$ is a family of maps $F(x) \to G(x)$ such that for every morphism $x \to y$ the diagram

$$\begin{array}{ccc} F(x) & \rightarrow & G(x) \\ \downarrow && \uparrow \\ F(y) & \rightarrow & G(y) \end{array}$$

commutes. In fact, this is a special case of the more general notion of dinatural transformations.

  • $\begingroup$ Sure, that's what I said in my comments. It's not really a natural transformation, however. $\endgroup$
    – Zhen Lin
    Mar 25, 2013 at 22:53

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