TITLE: The category Set seems more prominent/important than the category Rel. Why is this?
QUESTION [11 upvotes]: There's a lot of talk about Set, but less about Rel. As an outsider to category theory, this surprises me, because Rel seems "more closed." In particular,

The converse of a function needn't be a function; whereas the converse of a relation is always a relation, and
The complement of a function is (almost) never a function; whereas the complement of a relation is always a relation.

Thus, from my naive point of view, Rel seems particularly nice. So, why is Set the more prominent?
I'm going to pre-empt one possible answer, namely that, "The structure-preserving maps between algebraic structures are functions; thus, Set and its subcategories are particularly interesting." My counter-question is this: why do structure-preserving maps need to be functions?

REPLY [4 votes]: 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 $X^{\uparrow}$