9.5.2 Contravariant Functors

Let $\mathcal{C}$ and $\mathcal{D}$ be categories, and let $\mathcal{C}^{\mathsf{op}}$ denote the opposite category of $\mathcal{C}$ of .

A contravariant functor from $\mathcal{C}$ to $\mathcal{D}$ is a functor from $\mathcal{C}^{\mathsf{op}}$ to $\mathcal{D}$.

In detail, a contravariant functor from $\mathcal{C}$ to $\mathcal{D}$ consists of:

  1. Action on Objects. A map of sets
    \[ F \colon \text{Obj}\webleft (\mathcal{C}\webright ) \to \text{Obj}\webleft (\mathcal{D}\webright ), \]

    called the action on objects of $F$.

  2. Action on Morphisms. For each $A,B\in \text{Obj}\webleft (\mathcal{C}\webright )$, a map
    \[ F_{A,B} \colon \textup{Hom}_{\mathcal{C}}\webleft (A,B\webright ) \to \textup{Hom}_{\mathcal{D}}\webleft (F\webleft (B\webright ),F\webleft (A\webright )\webright ), \]

    called the action on morphisms of $F$ at $\webleft (A,B\webright )$.

satisfying the following conditions:

  1. Preservation of Identities. For each $A\in \text{Obj}\webleft (\mathcal{C}\webright )$, the diagram

    commutes, i.e. we have

    \[ F\webleft (\text{id}_{A}\webright ) = \text{id}_{F\webleft (A\webright )}. \]
  2. Preservation of Composition. For each $A,B,C\in \text{Obj}\webleft (\mathcal{C}\webright )$, the diagram
    commutes, i.e. for each composable pair $\webleft (g,f\webright )$ of morphisms of $\mathcal{C}$, we have
    \[ F\webleft (g\circ f\webright ) = F\webleft (f\webright )\circ F\webleft (g\webright ). \]

Throughout this work we will not use the term “contravariant” functor, speaking instead simply of functors $F\colon \mathcal{C}^{\mathsf{op}}\to \mathcal{D}$. We will usually, however, write

\[ F_{A,B}\colon \textup{Hom}_{\mathcal{C}}\webleft (A,B\webright )\to \textup{Hom}_{\mathcal{D}}\webleft (F\webleft (B\webright ),F\webleft (A\webright )\webright ) \]

for the action on morphisms

\[ F_{A,B}\colon \textup{Hom}_{\mathcal{C}^{\mathsf{op}}}\webleft (A,B\webright )\to \textup{Hom}_{\mathcal{D}}\webleft (F\webleft (A\webright ),F\webleft (B\webright )\webright ) \]

of $F$, as well as write $F\webleft (g\circ f\webright )=F\webleft (f\webright )\circ F\webleft (g\webright )$.


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