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Stream: learning: questions

Topic: coalgebras of constant functors


view this post on Zulip Bruno Gavranović (May 20 2021 at 23:17):

Are coalgebras of constant endofunctors a sensible thing to study? I'm trying to get some intuition about what kind of systems they describe, and what kind of behavior the final coalgebra do in this setting.

For instance, for a constant functor FF at X:CX : \mathcal{C} (and idXid_X on morphisms) it seems there's always a final coalgebra (X,α:XX)(X, \alpha:X \to X) (Lambek's theorem), where from any other coalgebra (Y,f:YX)(Y, f:Y \to X) the unique coalgebra homomorphism is given by ff itself.

It seems like this is trivial in some way, but I don't know how to think about this

view this post on Zulip Nathanael Arkor (May 21 2021 at 00:00):

Coalgebras for FF are just going to be slices over XX. The final coalgebra is therefore just the identity on XX.

view this post on Zulip Bruno Gavranović (May 21 2021 at 10:02):

Right. I suppose I'm trying to interpret what this means in terms of behaviour. If I have a "stream" functor FX=A×XFX = A \times X, then its coalgebra is a map XA×XX \to A \times X, taking in a state and producing an output and a new state. And I can take the final coalgebra, which will give me the stream.

In the case of the the constant functor I suggested (mapping any object to just AA), there is no new state. That is, if we try to iterate this thing, we quickly notice that the only time we can do that is when the carrier of this algebra is AA - which is exactly the final coalgebra.

view this post on Zulip Bruno Gavranović (May 21 2021 at 10:11):

While this seems trivial now, I suppose I should mention that I'm thinking about this in the base category of simple lenses, which adds this bidirectional flow and it looks like something nontrivial should happen. I'm not entirely sure.

That is, I'm thinking of a coalgebra of the constant functor at (A,A)(A, A)

F:Lens(C)Lens(C)(X,X)(A,A)\begin{align*} F : Lens(\mathcal{C}) &\to Lens(\mathcal{C})\\ (X, X) &\mapsto (A, A) \end{align*}

A particular coalgebra here with the carrier (X,X)(X, X) is a lens f:(X,X)(A,A)f : (X, X) \to (A, A). The final coalgebra story is the same as before, I suppose, but somehow the fact that things are now bidirectional, and that I can in principle iterate them is making me think something nontrivial should happen with behaviour.

But I think my intuition is wrong here (since I unpacked what this means in my previous message)

view this post on Zulip Bruno Gavranović (May 21 2021 at 10:12):

This might have something to do with the fact that lenses are themselves coalgebras of the costate comonad