3
$\begingroup$

I'm reading Steve Awodey's lecture notes on category theory, and am confused about the notation for covarian/contravariant representable functors:

For a locally small category $\mathcal{C}$ with objects $A, B$, the covariant representable functor is defined as $\mathcal{C}(A, -): \mathcal{C} \to \texttt{Set}$, where

$$ \begin{align} &\mathcal{C}(A, B) = \{ f \in \mathcal{C}_{1} \mid f: A \to B \} \\ &\mathcal{C}(A, g) : f \mapsto g \circ f \end{align} $$

The contravariant representable functor is defined as $\mathcal{C}(-, B): \mathcal{C}^{op} \to \texttt{Set}$. Here, I'm not sure if the following is correct:

$$ \begin{align} &\mathcal{C}(A, B) = \{ f \in \mathcal{C}^{op}_{1} \mid f: B \to A \} \end{align} $$

If so, how can we differentiate between $\mathcal{C}(A, B)$ for the covariant representable functor and that of the contravariant representable functor?

$\endgroup$
3
  • 1
    $\begingroup$ Possible duplicate of "Good" notation for $\hom$-functors $\endgroup$ Commented Nov 8, 2016 at 4:30
  • 1
    $\begingroup$ $\mathcal{C}(-,B)=\mathcal{C}^{\text{op}}(B,-)$ $\endgroup$
    – Nex
    Commented Nov 8, 2016 at 5:35
  • $\begingroup$ Is there a difference between $\mathcal{C}(A,g)$ and $\mathcal{C}(B, g)$? Or are they the same (both being composition by $g$)? $\endgroup$ Commented Nov 8, 2016 at 5:52

1 Answer 1

5
$\begingroup$

No, $\mathcal{C}(A,B)$ is always the set of maps $A\to B$ in $\mathcal{C}$. The difference is simply whether we consider $A$ or $B$ as the variable; if $A$, then we have $\mathcal{C}(A,B)=[\mathcal{C}(-,B)](A)$, contravariant in $A$; if $B$, then we have $\mathcal{C}(A,B)=[\mathcal{C}(A,-)](B)$, covariant in $B$. It's perhaps least ridiculous to simply recognize that both of these are shadows of the full object, $\mathcal{C}(-,-)$, a functor of two variables $\mathcal{C}^{\mathrm{op}}\times \mathcal{C}\to\mathtt{Set}$. Regardless, $\mathcal{C}(A,B)$ is always the same thing: it's a value of at least three different functors, and which one, if any, the author has in mind can't be determined just from the notation, since the notation only indicates a set.

$\endgroup$
2
  • $\begingroup$ Yes, I think I get it. What I was confused by is the $\mathcal{C}(-, B): \mathcal{C}^{op} \to \texttt{Set}$ notation. I thought it meant that the morphisms are reversed regardless of the variable, but it seems that the morphisms are reversed only when the variable is a morphism (if that makes any sense...). $\endgroup$ Commented Nov 8, 2016 at 5:48
  • 1
    $\begingroup$ The part about thinking of $\mathcal{C}(A,-)$ and $\mathcal{C}(-, B)$ as shadows of $\mathcal{C}(-, -)$ is very helpful! $\endgroup$ Commented Nov 8, 2016 at 5:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .