# Has the opposite category exactly the same morphisms as the original?

This is actually a question about categories; not only about the category that I mention here specifically. I only use category $\mathsf{Rel}$ as an example.

How to describe a morphism that belongs to $\mathsf{Rel}^{op}\left(B,A\right)=hom_{\mathsf{Rel}^{op}}\left(B,A\right)$?

As a triple $\left(A,R,B\right)$ or as a triple $\left(B,R,A\right)$ where in both cases $R\subseteq A\times B$?

ACC tells me: "...Thus $\mathcal{A}$ and $\mathcal{A}^{op}$ have the same objects and, except for their direction, the same morphisms..."(3.5.).

It comes to $\mathcal{A}^{op}\left(B,A\right)=\mathcal{A}\left(A,B\right)$. Denoting the compositions in $\mathcal{A}$ and $\mathcal{A}^{op}$ by $\circ$ and $\circ^{op}$ respectively we have: $f\circ^{op}g$ is defined iff $g\circ f$ defined and this as $f\circ^{op}g=g\circ f$.

CWM tells me: "... the arrows of $\mathcal{A}^{op}$ are arrows $f^{op}$ in a one-one correspondence $f\mapsto f^{op}$ with the arrows $f$ of $\mathcal{A}$..."(page 33).

Here we arrive at $f^{op}\circ^{op}g^{op}=(g\circ f)^{op}$

Category $\mathsf{Rel}$ has (small) sets as objects and relations as morphisms. If $R\subseteq A\times B$ then it can be recognized as a morphism $A\rightarrow B$ in $\mathsf{Rel}$. The demand that homsets are disjoint 'tells' us that formally we must do it with triples $\left(A,R,B\right)$.

ACC makes me tend to: $\left(A,R,B\right)\in\mathsf{Rel}\left(A,B\right)=\mathsf{Rel}^{op}\left(B,A\right)$.

CWI makes me tend to: $\left(B,R,A\right)=\left(A,R,B\right)^{op}\in\mathsf{Rel}^{op}\left(B,A\right)$.

• Below CWM's definition, it should be $(g \circ f)^{op}$ intead of $g \circ f$, right ? – Pece Aug 19 '14 at 15:14
• @Pece Yes, indeed. Thanks very much. I repaired. – drhab Aug 19 '14 at 15:18

## 3 Answers

To clarify a point that may be confusing, $\mathbf{Rel}$ is one of the relatively rare categories equivalent to its own opposite, essentially by the isomorphism $A\times B\cong B\times A$. That's why you find a natural representative for a morphism in $\mathbf{Rel}^{op}$: you're actually demonstrating this equivalence.

In a general category, there wouldn't be any concrete expression for the opposite of a morphism $f:A\to B$. For instance in $\mathbf{Set}$, there's no natural function $B\to A$ corresponding to a function $A\to B$. (We can get around this by passing to powersets, and this is another case in which the opposite category had a concrete description-but it's not equivalent to $\mathbf{Set}$!) In such categories $f^{op}$, instead of a meaningful "reverse" of $f$, is just a formal symbol.

In the case of $\mathbf{Rel}$, ACC's definition would have $\mathbf{Rel}^{op}(B,A)=\{(A,R,B):R\subset A\times B\}$, while CWM's would have $\mathbf{Rel}^{op}(B,A)=\{(A,R,B)^{op}:R\subset A\times B\}$. But here $(A,R,B)^{op}$ has no interpretation as any particular relation from $B$ to $A$. Noting that there is one, namely $(B,R,A)$, is a strictly further step unique to this case.

• I said: CWM makes me tend to $\left(B,R,A\right)=\left(A,R,B\right)^{op}$. If I understand you well, then I am going a step to far there. I should just do it with $\left(A,R,B\right)^{op}$ as some symbol. Correct? – drhab Aug 19 '14 at 15:48
• Yes, that's correct. – Kevin Carlson Aug 19 '14 at 16:02

Strict equality is something that category theory really don't (shoudn't?) care about.

The two opposite categories you construct are equivalent, and it is all that matters !

• When I first met 'opposite category' I was not yet on a level high enough to oversee that this indeed 'makes no essential difference' when it comes to categories. It was a source of confusion for me and I think I am not only one in this. That is mainly the reason to ask this question. I think that this topic deserves some attention in spite of the fact that it is not really relevant. This especially to avoid confusion. Something like 'it does not matter' is not enough. – drhab Aug 19 '14 at 15:33
• @drhab I know it is confusing for the beginner, but struggling with that kind of confusion is probably better than trying to avoid it. It ultimately make you feel intuitive with the objects you first have trouble to understand. If you concentrate on trying to find a set-theoretic definition of the opposite category, at the end, you just miss the point. What would you answer to an algebra beginner asking the following : is the elements of the group $\mathbb Z/2$ defined as $\{\mathbb Z,1+\mathbb Z\}$, $\{0,1\}$ or $\{1,-1\}$ ? – Pece Aug 19 '14 at 16:14
• I agree: struggling with confusion is indeed a good thing (luctor et emergo) and develops mathematical maturity. Avoiding confusion is a good thing too, though. What's best?... The answer differs off course by subject and by person. Thank you sincerely for your comments and your answer. Cheers. – drhab Aug 19 '14 at 17:08

I don't have CWM here right now but ACC gives you an exact definition for the opposite category, and that answers your question at least partially. Here is the definition in ACC:

For any category $\mathbf{A}=(\mathcal{O},\hom_\mathbf{A},id,\circ)$ the dual category of $\mathbf{A}$ is the category $\mathbf{A}^\mathrm{op}=(\mathcal{O},\hom_{\mathbf{A}^\mathrm{op}},id,\circ^\mathrm{op})$, where $\hom_{\mathbf{A}^\mathrm{op}}(A,B)=\hom_\mathbf{A}(B,A)$ and $f\circ^\mathrm{op} g=g\circ f$.

So, if you define $\mathbf{Rel}$ such that $$\hom_\mathbf{Rel}(A,B)=\{(A,R,B)\mid R\subseteq A\times B\},$$ then that definition immediately gives you that $$\hom_{\mathbf{Rel}^\mathrm{op}}(B,A)=\hom_\mathbf{Rel}(A,B)=\{(A,R,B)\mid R\subseteq A\times B\}.$$

Note that in this case a morphism $A\to B$ in $\mathbf{Rel}$ is not a subset of $A\times B$ which you just somehow formally think of as a triple $(A,R,B)$. Instead, a morphism $A\to B$ is such a triple and you informally think of it just as a subset of $A\times B$.

You cannot take a morphism $A\to B$ to be a subset of $A\times B$ because, as you mentioned, you want the hom-sets to be disjoint. And you want the hom-sets to be disjoint because then every morphism has a unique domain and codomain. Indeed, now given any morphism $f$ it belongs to exactly one hom-set $\hom(A,B)$, and this gives you the domain $A$ and codomain $B$ of $f$. If you would just take "pure" relations as morphisms then there would be no way to assign each morphism a unique domain and codomain.