Timeline for What is the mathematical name for Haskell's Alternative Functor
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2020 at 22:08 | comment | added | LorenzoPerticone | Let us continue this discussion in chat. | |
| Feb 26, 2020 at 22:07 | comment | added | Asad Saeeduddin | I'm probably just not understanding your notation, but to make sure we're on the same page, the family of functors I'm describing is $(\mathbf{Set}, +) \to (\mathbf{Set}, \times)$, i.e. the monoidal structure of interest in the domain is that of coproducts, but in the codomain that of products. | |
| Feb 26, 2020 at 22:00 | comment | added | LorenzoPerticone | Sorry, choosing different $f, g$ won't actually satisfy associativity, but if $f = g \neq 1_X$ everything should work smoothly. | |
| Feb 26, 2020 at 21:55 | comment | added | LorenzoPerticone | Now, I'm not sure these "twisted" monoids are preserved functorially by an alternative functor, and I'm too tired to check now: I'll try as soon as possible (probably tomorrow), but I suspect it won't work. | |
| Feb 26, 2020 at 21:52 | comment | added | LorenzoPerticone | I now see why you kept mentioning the cocartisian monoidal structure, but you should be careful: there is a functor $(\mathbf{Set}, +) \to \mathbf{Mon}(\mathbf{Set}, +)$ and it's surjective on objects, but it isn't an equivalence of categories. This just means there are "sets" with multiple (cocartesian) monoid structures, and actually every (non-empty, non-singleton) set does admit such a structure: fix two (different) functions $f,g : X \to X$, and use it to "twist" your multiplication, setting $\mu : X + X \to X$ to be $\mu(x_l) = f(x)$ and $\mu(x_r) = g(x)$. | |
| Feb 26, 2020 at 17:45 | comment | added | Asad Saeeduddin | Of course what I was missing was whether you demand it or not makes no difference, monoids in the domain category are trivial, and hence so are monoids in the image of the monoidal functor. Another way to view this (through the perspective of the Day convolution structure you mentioned) is to say that the Day convolution of two functors with respect to the coproduct tensor on $\mathbf{Set}$ is isomorphic to their cartesian product (under the "pointwise" cartesian structure on the functor category being described). | |
| Feb 26, 2020 at 17:43 | comment | added | Asad Saeeduddin | Yes, sorry about the confusion. You didn't mention the cocartesian monoidal structure, but my understanding of Alternative functors is of them being lax monoidal functors from the category of sets under the cocartesian monoidal structure to the category of sets under the cartesian monoidal structure (similarly to how applicative functors are lax monoidal functors from the cartesian monoidal structure to itself). I was trying to understand your representation through this lens, and started looking for somewhere that you demand a monoid on the domain category of $\mathcal{M}$. | |
| Feb 26, 2020 at 9:25 | comment | added | LorenzoPerticone | I think the misunderstanding is about the monoidal structure we are endowing $[\mathcal{C}, \mathcal{C}]$ with: if we use day convolution our monoids are monoidal functors; of we use composition, our monoids are monads. Here I'm considering a different structure, that I called "pointwise": monoids there (as sketched in my answer above) are precisely functors with "values in the category of monoids" $\mathcal{C} \to \mathbf{Mon}(\mathcal{C})$. This is even true if such functors are not monoidal! | |
| Feb 26, 2020 at 9:11 | comment | added | LorenzoPerticone | I didn't mention neither $\mathbf{Set}$ nor the cocartesian monoidal structure, so I'm a bit confused about your last comment (of course, the relevant case is indeed $\mathcal{C}=\mathbf{Set}$, but in my answer I only assumed a monoidal category). Moreover, as you mentioned, the cocartesian monoidal structure isn't very interesting: every set is a monoid under it, in a canonicale way. This means that even when considering sets and functions we usually work with monoids in the cartesian monoidal structure, not in the cocartesian one. | |
| Feb 23, 2020 at 17:45 | comment | added | Asad Saeeduddin | Oh right, I was confused. When $\mathcal{M}$ is interpreted as a monoidal functor, it is indeed the case that $\mathcal{M}(f)$ is only a morphism of monoids if $f$ is, but the relevant monoidal structure on the codomain is the cocartesian structure on $\mathbf{Set}$, under which every object is trivially a monoid and every morphism trivially a monoid morphism. | |
| Feb 5, 2020 at 19:56 | comment | added | LorenzoPerticone | If you combine the functoriality of $\mathcal{M}$ with the fact (mentioned in the text above) that the two diagrams have to commute in a pointwise fashion, you should get "monoidality" that for any such $\mathcal{M}(f)$. | |
| Feb 5, 2020 at 14:30 | comment | added | Asad Saeeduddin | Isn't $\mathcal{M}(f)$ only a morphism of monoids if $f$ is? | |
| Jan 24, 2020 at 0:24 | history | edited | LorenzoPerticone | CC BY-SA 4.0 |
minor correction and remark
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| Jan 23, 2020 at 23:39 | history | answered | LorenzoPerticone | CC BY-SA 4.0 |