# Understanding contravariant functors and their notation

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".

• 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? Commented Feb 12, 2023 at 16:43