According to this definition a category consists of a class of objects, a class of morphisms, a class function $\text{dom}:\text{hom}(C) \to \text{Ob}(C)$ a codomain function $\text{cod}:\text{hom}(C) \to \text{Ob}(C)$ and a composition operation that satisfies the usual axioms. Using this, the opposite category $C^{op}$ consists of the same objects and morphisms, but the domain class function $\text{dom}_{op}$ is defined to be $\text{cod}$ and the codomain function $\text{cod}_{op}$ is defined to be $\text{dom}$. I adopt the notation $f:X \to Y$ if a morphism $f$ has domain $X$ and codomain $Y$ in its category. Thus for morphisms $f:X \to Y, g:Y \to Z$ in $C$, we get $f:Y \to X$, $g:Z \to Y$ in $C^{op}$ and the composition $f \cdot_{op} g$ in $C^{op}$ is defined to be the morphism $g \cdot f$ in $C$ but with $\text{dom}_{op}(g \cdot f)=Z, \text{cod}_{op}(g \cdot f)=X.$

Now, one can define a contravariant functor from $C$ to $D$ to be a covariant functor $F:C^{op} \to D$, that is, there is an object $F(c)$ in $D$ for every $c$ in $C$ and for every morphism $f:X \to Y$ in $C$ we get a morphism $F(f):F(Y) \to F(X)$ in $D$ such that for any $f:X \to Y, g:Y \to Z$ in $D$ we have $F(g \cdot f)=F(f \cdot_{op} g)=F(f) \cdot F(g)$ and $F(1_c)=1_{F(c)}.$

$(1)$ Is the object $F(c)$ unique when $c$ is in $C$? If so, what in the definition ensures this? Many common examples give that $F(c)$ is unique, but I can't deduce this from the definition of a functor.

Consider the following example. Let $C$ be locally small and $c$ in $\text{Ob}(C)$, then there is a contravariant functor that maps $f:x \to y$ in $C$ to $f^*:C(y,c) \to C(x,c), g \mapsto f \cdot g.$ Using the convention that morphisms from $C^{op}$ are depicted as morphisms in $C$, this situation can be depicted as follows:

\begin{matrix} C^{op} & \longrightarrow & \mathrm{Set}\\ \ \ \ x & \mapsto &C(x,c) \\ f \downarrow & \mapsto & \uparrow \ \ \ f^*\\ \ \ \ y & \mapsto & C(y,c) \end{matrix}

$(2)$ Why isn't it more natural to use the diagram

\begin{matrix} C^{op} & \longrightarrow & \mathrm{Set}\\ \ \ \ y & \mapsto &C(y,c) \\ f \downarrow & \mapsto & \downarrow \ \ \ f^* \\ \ \ \ x & \mapsto & C(x,c) \end{matrix}

where $f:y \to x$ is a morphism of $C^{op}$? In Emily Riehl's Category Theory in Context she writes "To avoid unnatural arrow-theoretic representations, a morphism in the domain of a functor $F:C^{op} \to D$ will always be depicted as an arrow $f:x \to y$ in $C$, pointing from its domain in $C$ to its codomain in $C$". However, I neither see why this is necessary nor why it is natural, or rather, why the alternative diagram would be "unnatural and lead to unnatural arrow-theoretic representations".


1 Answer 1


(1) Yes, functors act on objects via a function, so each object has a uniquely determined value. Some definitions may not literally write the word “function” here since they use terminology widely understood to define a function throughout mathematics.

(2) It’s an aesthetic preference, but I very much agree with Riehl that your version is more likely to be confusing.

  • $\begingroup$ Thanks. $(1)$ Is a functor basically a class function on the objects and morphisms that satisfies the usual axioms? $(2)$ Hm okay. I suspect that the confusion might occur when one considers categories whose morphisms are actual functions which will give that a function $f:X \to Y$ has (categorical) domain defined to be $Y$ in the opposite category whereas its actual domain for input values is still $X$, even in the opposite category. Is this what is meant? $\endgroup$
    – user3118
    Commented Feb 12, 2023 at 16:43
  • 1
    $\begingroup$ (1) yep (2) That’s a particularly natural way to get confused. $\endgroup$ Commented Feb 12, 2023 at 18:32
  • $\begingroup$ Thanks a lot for the insights. $\endgroup$
    – user3118
    Commented Feb 12, 2023 at 20:13
  • $\begingroup$ Sure, happy if that helped. $\endgroup$ Commented Feb 13, 2023 at 2:02

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