If you want that an equivalence relation is the kernel pair of some arrow, take the arrow to be the classifying map $X \to PX$ of the relation $R$ as subobject of $X \times X$. (You can also take the arrow to be the coequalizer, but it's more work.) A detailed proof is in Johnstone's Sketches of an Elephant (p. 97); try googling that +djvu and you should find what you need. It's not very hard.

Phil: yes. This is a famous exactness property of toposes: that every equivalence relation is the kernel pair of the coequalizer (of the two projection maps restricted to the equivalence relation). One says that in a topos, "equivalence relations are effective". I wasn't able to find this fact for general toposes in Mac Lane-Moerdijk, but see for example Categories, Allegories by Freyd and Scedrov, 1.951 (page 173).

(I should have paused; this looks like homework.)

This type of expression is very well known. See Concrete Mathematics by Graham, Knuth, and Patashnik, particularly table 174 (p. 174).

That could be interesting, Romeo. In the meantime, it sounds as if James Griffin is a good go-to guy for information.

+1 on your first comment, James. And your second comment sounds highly plausible.

For example, free groups have interesting automorphism groups; they contain braid groups for instance (cf. Artin representation). 3. I am not absolutely sure, but I think the automorphism group of the operad whose algebras are monoids might be Z mod 2. The nontrivial automorphism would send an operation of arity n, namely a total ordering of the elements 1, 2, ..., n, to the reverse ordering.

1. The groupoid is equivalent to a sum over isomorphism classes of trees of the automorphism groups of class representatives, and each such automorphism group is an iterated wreath products of symmetric groups. A useful picture might be to think of a tree as a hereditarily finite multiset. Then an automorphism of a multiset consists of a permutation of multiple copies of an element together with an automorphism of each element (as a multiset). 2. E.g., a monoid or group can be viewed as an operad where each operation has arity one. Pick a monoid with an interesting automorphism group. (Cont.)

Do your semirings have a multiplicative identity? So in other words, do you mean the underlying additive structure is a commutative monoid, the underlying multiplicative structure is a monoid, and multiplication distributes over addition?

I'm still having trouble understanding the question. I don't suppose you want the same conditions to apply to all three spaces X, Y, Z? If not, why do you expect a unique answer (you say the minimal conditions)? For example, by slightly strengthening a condition on X one could slightly weaken a condition on Y, so that the two sets of conditions are incomparable. (Finally, may I ask what is the motivation for demanding the compact-open topology, if some other conceptually similar but slightly different topology works?)

I'd need more clarification then. I'll ask under your question.

The homomorphism R --> R/Z x R/Z which takes t to (t mod Z, st mod Z) where s is irrational. This is a continuous injective homomorphism, with dense image.

Since the category of locally compact Hausdorff abelian groups is self-dual, we may as well ask whether every monomorphism is a kernel. And indeed that's false: kernels are closed subgroups, so the irrational line on a torus would be a counterexample.

@Timothy: thanks. I appreciate that you were at MIT and may have spoken Rotaese well, and I'm glad to hear you defend him. But Rota expresses himself a lot more aggressively than you are making out! "Those mathematicians who knew some lattice theory watched with amazement as the algebraic geometers of the Grothendieck school clumsily reinvented the rudiments of lattice theory in their own language." Nice! @Yemon: that's a good observation. My appetite for gnomic, cryptic utterances is less than it used to be. Sometimes they're profound, but often times not. I'm still undecided in this case.