I'm working on learning some category theory and the book claims that $\mathbf{Rel} \stackrel{\sim}{=} \mathbf{Rel}^\mathsf{op}$ where $\mathbf{Rel}$ is the category of binary relations. The functor that is supposed to show this is $F : \mathcal{Rel} \rightarrow \mathcal{Rel}^\mathsf{op}$ where it takes objects to themselves and relations to its opposite relation. That is, for an arrow $f$ in $\mathbf{Rel}$, $f \subseteq A \times B$ is mapped to $f^\mathsf{op} := \left\{ <b, a> \in B \times A | <a, b> \in f \right\}$.

My problem is I don't see how this makes it past one of the axioms of being a functor, given in the book as

$F(f: A \rightarrow B) = F(f):F(A) \rightarrow F(B)$

First, I really just don't think I understand how I'm supposed to understand this part of the definition. Am I supposed to be taking the left side of the equality as mapping the arrow, and the right side as asserting that the arrow it is mapped to has as domain and codomain the functor applied to the domain and codomain of the original arrow?

It seems like on this current example, it doesn't even give us the arrow $f^\mathsf{op}$ because it doesn't exchange the domain and codomain. And there seems to be no way of doing this exchange by mapping the objects to different objects.

In short, I'm probably hopelessly confused. How is it that this is a functor? And how am I supposed to understand this part of the definition of a functor? It seems wishy washy.

Thank you for any help you may provide.

  • 3
    $\begingroup$ In the first place: persevere. I also experienced this as a confusing subject. Note that $f^{op}$ as defined in your question satisfies $f^{op}\in\mathbf{Rel}\left(B,A\right)$ (denoting the homset). Then automatically $f^{op}\in\mathbf{Rel}^{op}\left(A,B\right)$ where $A$ and $B$ are switched. $\endgroup$ – drhab Feb 28 '14 at 9:36
  • $\begingroup$ The functor axiom you give is for covariant functors. You know you want to reverse the arrow, so you should expect a contravariant functor, giving $F(B) \rightarrow F(A)$ as the final clause. $\endgroup$ – Eric Towers Feb 28 '14 at 9:46
  • $\begingroup$ @drhab: You could make that into an answer. $\endgroup$ – Hurkyl Feb 28 '14 at 9:51
  • $\begingroup$ @Hurkyl So I did. Feel free to check. $\endgroup$ – drhab Feb 28 '14 at 13:18
  • $\begingroup$ What book is that? Awodey perhaps? $\endgroup$ – magma Feb 28 '14 at 23:38

Also see my comment.

Based on category $\mathcal{C}$ we have category $\mathcal{C}^{op}$ determined by:

1) $\mathcal{C}^{op}$ has the same set of objects as the original category $\mathcal{C}$

2) For every pair of objects $a,b$ we have $\mathcal{C}^{op}\left(a,b\right)=\mathcal{C}\left(b,a\right)$

3) If we denote the composition on $\mathcal{C}$ by $\circ$ and the composition on $\mathcal{C}^{op}$ by $\circ^{op}$ then $f\circ^{op}g$ is defined iff $g\circ f$ is defined and this by $f\circ^{op}g:=g\circ f$

If $g\in\mathcal{C}^{op}\left(a,b\right)$ and $f\in\mathcal{C}^{op}\left(b,c\right)$ then $f\circ^{op}g$ should be defined, wich is indeed the case since $g\in\mathcal{C}\left(b,a\right)\wedge f\in\mathcal{C}\left(c,b\right)$ tells us that $g\circ f$ is defined.

Here $f\circ^{op}g=g\circ f\in\mathcal{C}\left(c,a\right)=\mathcal{C}^{op}\left(a,c\right)$ as we would expect.

You could say that $\mathcal{C}^{op}$ has exactly the same objects and arrows as $\mathcal{C}$ and the only thing different is the composition. Consequently domain and codomain of an arrow interchange. Observing an arrow $f$ then we simply must ask ourselves the question in what context we are observing: is it an arrow of $\mathcal{C}$ here, or an arrow of $\mathcal{C}^{op}$?

Functor $F:\mathbf{Rel}\rightarrow\mathbf{Rel}^{op}$ as described in your question sends arrow $f\in\mathbf{Rel}\left(A,B\right)$ to an arrow $f^{op}$ that belongs to $\mathbf{Rel}\left(B,A\right)=\mathbf{Rel}^{op}\left(A,B\right)=\mathbf{Rel}^{op}\left(F\left(A\right),F\left(B\right)\right)$ as it should be.

Quite often arrow $f$ in $\mathcal{C}$ is denoted as $f^{op}$ if it is looked at as an arrow in $\mathcal{C}^{op}$. This is not really necessary, can clear up things (it tells us what context we are working in) but can also cause confusion. For instance the arrow $F\left(f\right)$ in your question is also denoted as $f^{op}$ but it is not this $f$ in the other context. If it should be then we would have $f^{op}\in\mathbf{Rel}\left(A,B\right)=\mathbf{Rel}^{op}\left(B,A\right)$ which is not the case.

  • $\begingroup$ It seems to make more sense to look at the functor as acting on the homsets. It seems I have another confusion in this area. If we are in the dual category of sets, $\mathbf{Set}^\mathsf{op}$, then an arrow (which are functions) $f: A \rightarrow B$ from $\mathbf{Sets}$ ends up in $\mathcal{C}^{op}\left(b,a\right)=\mathcal{C}\left(a,b\right)$. Does this mean that the ordered pairs in the function also get flipped around, and thus that it may not necessarily be a function anymore? $\endgroup$ – e.istre91 Feb 28 '14 at 18:53
  • 1
    $\begingroup$ No. In the following make a good distinction between function and arrow. A function $f:A\rightarrow B$ (i.e. a triple $\left(A,G_{f},B\right)$ where $G_{f}\subset A\times B$ satisfies a certain condition) belongs to homset $\mathbf{Set}\left(A,B\right)$ and in that sense it is an arrow from $A$ to $B$ in $\mathbf{Set}$. It also belongs to homset $\mathbf{Set}^{op}\left(B,A\right)$ and in that sense it is an arrow from $B$ to $A$ in $\mathbf{Set}^{op}$. In both cases $f$ is exactly the same thing: the triple $\left(A,G_{f},B\right)$. $\endgroup$ – drhab Feb 28 '14 at 19:23
  • 2
    $\begingroup$ Confusing is here that - when it comes to a function - we also say that it has a domain and a codomain. If it comes to its membership of $\mathbf{Set}\left(A,B\right)$ then no ambiguity arises, since arrow and function have the same (co)domain. However, looking at it as a member of $\mathbf{Set}^{op}\left(B,A\right)$ you are (still) dealing with a function having $A$ as domain and $B$ as codomain, wich is an arrow having $B$ as domain and $A$ as codomain. $\endgroup$ – drhab Feb 28 '14 at 19:23

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