The category Set seems more prominent/important than the category Rel. Why is this?

Solution 1:

In mathematics we study that which is relevant, not (necessarily) that which is more symmetric, pretty, cool, or nice. It is (perhaps not) a coincidence that most of those things that are relevant are also symmetric, pretty, beautiful, cool, and nice, but that is besides the point.

We study sets and functions because these are relevant for modeling many many things. Physical processes are very adequately described using functions rather than relations. Same goes for modeling in many other situations. The less symmetric notion of a function simply seems to be more common then general relations in the sense that while the complement of a relation is a relation and that of a function is not, I don't really care since who wants to take the complement of a function anyway?

As for why sets are more commonly used to talk about structure preserving maps is of the same flavour. The concepts we care about from applications are things like groups, rings, fields etc. and the corresponding notions of structure preserving maps are the function based ones. E.g., to a pointed topological space corresponds a fundamental group. To a continuous pointed mappings corresponds a group homomorphism.

Moreover, mathematics is highly non-symmetric in the sense that it is far from self dual. Thus, in some sense, finding a self dual category should bring about a reaction of "huh?!?! weird", rather than "Wow, let's adopt it".

Solution 2:

If your question is literally why is there more talk in category theory about Set than about Rel, I think the answer is simply that categories are by definition the same as enriched categories that are enriched over Set. This is why Set plays a prominent role in category theory, for example, it is why the Yoneda embedding of C lands in functors from C^op to Set.

Now, maybe you really meant to ask why do we study the special case of enriched category theory where the categories are enriched in Set so much, why don't we use instead categories enriched in Rel (or maybe not "instead" but "also") or some other structure that is to categories as Rel is to Set (I think allegories are probably the only reasonable other candidate for this)? That I'm not so sure about. Certainly there are a huge number of very important and interesting examples of categories enriched in Set, this alone justifies studying them. We could ask (1) are there also lots of naturally occuring examples of categories enriched in Rel or of allegories (that are not just categories), (2) if there aren't that many is it for a good reason or just lack of interest in Rel-enriched categories or in allegories? I don't know the answers to either question. I suspect the answer to (1) is no, but that might just be ignorance.

Solution 3:

Something that deserves to be said here: $\mathsf{Rel}$, unlike $\mathsf{Set}$, is far from complete or cocomplete! Below I'll give an example of an equalizer which fails to exist in $\mathsf{Rel}$. Since $\mathsf{Rel}$ is self-dual, it also doesn't have coequalizers. The picture does not improve much if we consider $\mathsf{Rel}$ as a 2-category: it has products and (trivially) equifiers, but not inserters (so it doesn't have the PIE limits which form the appropriate 2-categorical notion of completeness).

This is not a complete answer, but it's at least a practical one: the basic, primordial categories tend to have good completeness properties, and that's an important part of the way we work with them. So we can't work with $\mathsf{Rel}$ in the same way.

Here are two relations $2 \to 1$ which don't have an equalizer in $\mathsf{Rel}$ (where $2 = \{0,1\}$ and $1 = \{0\}$): let $\alpha$ be the total relation (so $\alpha(0,0)$ and $\alpha(1,0)$ both hold), and let $\beta$ be the adjoint of the function $1 \to 2$ which picks out $0$ (so that $\beta(0,0)$ holds but $\beta(1,0)$ does not). Consider the set $X = \mathrm{Eq}(\mathsf{Rel}(1,\alpha), \mathsf{Rel}(1,\beta))$ of relations $1 \to 2$ which equalize $\alpha$ and $\beta$. It has 3 elements: the empty relation, the total relation, and the function which picks out 0 (but not the function which picks out 1). Suppose were an equalizer $E=\mathrm{Eq}(\alpha,\beta)$. Then we would have $\mathsf{Rel}(1,E) \cong X$. But for any set $E$, $\mathsf{Rel}(1,E)$ has cardinality $2^{|E|}$, so it can't be 3, a contradiction.

As for inserters, suffice it to say that there's a natural guess for what the inserter should be, but it doesn't work, and as soon as you understand why you can construct two relations $2 \to 2$ which can be shown to have no inserter in much the same way as the equalizer argument goes. It's even easier to see that $\mathsf{Rel}$ doesn't have powers with the arrow category -- $\mathsf{Cat}(\uparrow, \mathsf{Rel}(1,X))$ is the set of pairs of included subsets $A \subseteq B \subseteq X$, which is not in bijection with a powerset $\mathsf{Rel}(1,X^{\uparrow})$ for any $X^{\uparrow}$