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

Question

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

Answer 1

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.

Answer 2

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 !

Answer 3

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.