Hi Martin, Oh right, I see! That's a very interesting question. I think I have a source of examples of algebras, though I'm not sure if they are free or not (I would guess not in general). Write L for the partial map classifier monad. Write P for the power-object monad. The unit map X >--> LX is monic for any object, and each PX is injective. So for each X, we can extend the map P(eta^L_X): PX --> PLX along the monic eta^L_{PX}: PX >---> LPX to get a map LPX --> PLX. If we do this extension in a random way, there would be no particular reason for it to be well-behaved. But there is an "obvious" way to take extensions into P, and I *think* if we take this way, it gives a distributive law of the endofunctor P over the monad L. The unit axiom is obvious (it's how we defined things). For the axiom expressing compatibility with the multiplication of L, my argument involves figuring out what postcomposition with LPX ---> PLX does. A map A ---> LPX is a partial map A <--< A' ---> PX, which means giving a subobject A' <= A together with a relation R <= A' x X. On the other hand, a map A ---> PLX is a relation S <= A x LX. If I have this right, postcomposition with LPX --> PLX acts by sending (A', R) to the relation R >---> A' x X >---> A x X >--1 x eta--> A x LX and now we can use this to show that both composites LLPX ---> PLX are the same map. Assuming this is right, P lifts to an endofunctor on the category of L-algebras (it doesn't lift to a monad, because the distributive law axiom expressing compatibility with the unit of P doesn't hold -- which doesn't seem that surprising, since the power-set monad on Set doesn't distribute over itself for the same reason). Now my guess (based off feels) is that if you take the terminal L-algebra 1 and hit it with P, you get Omega with the meet L-algebra structure. The analogy is with the distributive law of P over the commutative monoid monad in Set: if you see P as adding all meets, then for a monoid X the monoid structure on PX is given by \bigwedge U . \bigwedge V = \bigwedge {uv : u \in U, v in V}. Similarly in this case I would guess that the L-algebra "join" operation on P1 (indexed by a subobject phi <= 1) is given by "Join"_{x in phi} psi_x = "Join"_{x in phi} \bigwedge_{y in psi_x} {*} = \bigwedge_{f: \bigwedge_{x in phi} psi_x} "Join"_{x in phi} {*} = \bigwedge_{f: \bigwedge_{x in phi} psi_x} {*} = \bigwedge_{x in phi} psi_x (here line 2 to line 3 is the application of the distributive law, and 3 to 4 is the trivial L-algebra structure of 1). Note that the terminal L-algebra 1 = L0 is free. So potentially applying P to other free L-algebras gives more examples of non-free L-algebras! Richard Martin Escardo <escardo.martin@gmail.com> writes:
Hi Richard,
Thanks! This is very interesting and nice.
Although I didn't say this explicitly in my original message, I am looking for algebras that can be defined in any (elementary) topos (by diagrams or in the internal language) that are not free in all toposes.
Do you also have examples of those? (Other than the one I gave in the opening message of this thread.)
I am sure there should be a plentiful supply of algebras that one can define in all toposes, but which are not free in all toposes.
I guess that such an exploration would lead to a better understanding of the algebras of this monad in an arbitrary topos. In particular, we may ask whether Anders Kock's characterization of the algebras as posets that are complete in a certain way may help to shed light on this question.
Best, Martin
On 07/01/2026 01:07, Richard Garner wrote:
Hi Martin,
Maybe one can look at the Sierpinski topos. In there, if my calculations are correct, the partial map classifier takes A ---> B to 1+A+B ---> 1+B, and an algebra for this monad is a map p: A-->B equipped with a section s together with points in A and B which are preserved by p and s.
In this case, Omega seems to have exactly two non-isomorphic algebra structures, which can be realised by union and intersection.
But for a general p: A-->B, I think there are either:
- no algebra structures (if B is empty or p is non-surjective), or else - exactly one non-isomorphic algebra structure for each distinct cardinality possessed by one of the fibres of p
Indeed, in the second case, pick an element of B whose fibre has the given cardinality. This is your B-point *. Then use AC to pick a section s of p, and take the point in A to be s(*). If you chose a different section, you get an isomorphic algebra. If you chose a different basepoint in B of the same cardinality, you will again get an isomorphic algebra.
In particular, Omega(Omega(1)) has one fibre of cardinality 1, one of cardinality 2 and one of cardinality 3, and so should have three distinct algebra structures.
All the best,
Richard
Martin Escardo <escardo.martin@gmail.com> writes:
Dear 1-topos theorists,
Anders Kock has a nice paper from the last millennium (1990), about
Algebras for the partial map classifier monad https://link.springer.com/chapter/10.1007/BFb0084225⚠️⚠️ https://tildeweb.au.dk/au76680/jonna5.pdf⚠️⚠️
As he remarks and is well known, and also trivial, in a boolean topos, all algebras are free. Then he goes on to say many interesting things that hold in all 1-toposes.
Jon Sterling recently conjectured that, in an arbitrary topos, not all algebras are free.
I came up with an example. My question is whether this example is well known, and, moreover, whether more examples are known.
First of all, the subobject classifier Ω is a free algebra on one generator.
If you think of Ω as the powerset of the terminal object 𝟙, then the structure map of Ω as a free algebra is *union*.
(†) But, you can check, also *intersection* exhibits Ω = 𝓟 𝟙 as an algebra.
I have proved that this algebra is free if and only if the principle of excluded middle holds, that is, the topos is boolean.
Is this known?
Then I wanted to find more counter-examples to "every algebra is a free algebra".
I tried, first, exponential powers of Ω. But they are free in all toposes.
Then I tried, more generally, arbitrary products of free algebras. But, again, they are free in all toposes.
Does anybody know a source of more counter-examples? At the moment, the only counter-example I know is (†).
It is embarrasing to know only one counter-example.
Best wishes, Martin PS. I have written my proofs (on paper and) in a proof assistant, namely Agda (and this is publicly available and advertised in various forums). So this gives some confidence regarding the above claims. I still have to write a human-readable version for public consumption, but here I am more interested in knowning what people already know about this.