# What is the mathematical name for Haskell's Alternative Functor

Haskell's Applicative Functor is called Lax Monoidal Functor in mathematics.

What is Haskell's Alternative Functor called in mathematics?

Recall that Haskell's Alternative Functor is defined as follows:

class Applicative f => Alternative f where
-- | The identity of '<|>'
empty :: f a
-- | An associative binary operation
(<|>) :: f a -> f a -> f a


Edit: In other words, an Alternative Functor f is an Applicative Functor such that for every type a, f a is a monoid, where <|> is the binary operation of type f a -> f a -> f a and empty is the identity of type f a. Uncurrying the type of <|>, we get $$f\ a \times f\ a \rightarrow f\ a$$ and empty can be equivalently defined to have type $$1 \rightarrow f\ a$$ In the Haskell definition, the type variable a is universally quantified. As far as I know, this translates to the requirement that they are natural in a.

• I guess for an average user here, to understand your question is much more difficult than to give an answer. Could you please provide some explanations for those who are not familiar with the background? What is meant by functor in Haskell? What is the Applicative Functor? What is standard notation? Which part is code and which part is comment? What does <|> mean? Jan 9 '17 at 6:00
• The concepts are described in Haskell documentation: stackoverflow.com/documentation/haskell/3800/functor, stackoverflow.com/documentation/haskell/8162/…. The lines starting with -- are comments. Other lines are code. <|> is a user-defined infix operator in Haskell.
– xuh
Jan 9 '17 at 7:54
• There might be users who know the name you are looking for but are not familiar with Haskell. The way you phrase the question might be suboptimal for getting an answer. Jan 9 '17 at 9:44
• See also this answer for explanation of 'Applicative' in more mathematical terms: cstheory.stackexchange.com/q/12412/3431 To summarise: an applicative functor is in more usual terminology a lax monoidal endofunctor equipped with a strength (for the latter, see ncatlab.org/nlab/show/tensorial+strength). The monoidal product is taken to be cartesian. See also: semantic-domain.blogspot.com.au/2012/08/… Jan 9 '17 at 12:19
– xuh
Jan 9 '17 at 12:35

This is not a complete answer, just a translation of Haskell to math so that the resident category theorists can tell what the question is.

As is usual in such cases, the terminology and the concepts in Haskell are slightly modified concepts from mathematics. It is best in the first iteration to look just at the broad similarities and not worry so much about the details.

Recall from Section 7 of the paper by Connor McBride and Ross Paterson that Haskell's Applicative is a lax monoidal functor (kind of as the Haskell definition uses internal Hom-sets where one would expect the external ones).

The definition of Alternative translates into: a functor $F : \mathcal{C} \to \mathcal{C}$, where $C$ is at least cartesian, together with natural transformations $e : 1 \to F$ and $m : F({-}) \times F({-}) \to F$, or concretely, for every object $A \in \mathcal{C}$ we have arrows $$e_A : 1 \to F(A)$$ and $$m_A : F(A) \times F(A) \to F(A)$$ such that $(F(A), e_A, m_A)$ is a monoid for all $A$, naturally in $A$. What would you call such a thing?

The next question to ask is whether there is anything extra about having a lax monoidal functor which also has the above structure of monoids? I don't know off the top of my head. If there is another view of the same situation, it will surely be useful for various Haskell hacking tricks. Haskell people are very good at using category-theoretic algebra for all sorts of cool purposes.

Let me also explain on the difference between parametricity and naturality. When we define a "functor" $F$ in Haskell, that is not really a functor. It is acertain mapping from types to types which acts on internal hom-sets as a functor would. In addition, this mapping has a strong property known as parametricity which says that $F$ behaves "in the same way" on all types. Parametricity is quite similar to naturality, and there is a way to view it semantically as such, but it is not precisely naturality. This is not a detail into which we should worry about here, though.

• Haven't you just described a factorization of the functor through the category of monoids in C? So an alternative functor is just a functor F: C---> Mon(C)? I guess we have to encode the lax-monoidal part... Jan 9 '17 at 13:45
• Thanks for the clarification. I'm not a mathematician. So it is much appreciated. I have one question: shouldn't the type of $e_A$ be $1 \rightarrow F(A)$?
– xuh
Jan 9 '17 at 13:46
• Cartesian monoid object? Can't think of a better term for it at the moment. (Unfortunately you can't just call it a monoid object since there is also endofunctor composition as monoidal product. Remark that endofunctor composition $- \circ F$ preserves cartesian products, but obviously not so for $F \circ -$. There is some sort of weak/lax distribution for $F \circ -$ though if we add in Applicative.) Jan 9 '17 at 14:04
• But do values of $F$ on morphisms come out monoid homomorphisms? I don't see it from what is given. Jan 9 '17 at 21:37
• @მამუკაჯიბლაძე: that's a good question. Staring at the official definition there does not seem to be that requirement. On the other hand they also have the some and many things which do have some connection to the Applicative class. Those might actually be more interesting than the monoid structure. Jan 10 '17 at 14:06

A slightly different set of primitive operations that we can look at is the following:

class Functor f => Alternative' f
where
unit :: () -> f Void
zip :: (f a, f b) -> f (Either a b)


This represents a lax monoidal functor from the monoidal category of -> under the tensor Either to the monoidal category of -> under the tensor (,). The laws are the usual laws for the coherence maps of lax monoidal functors, including those that demand conformance with the functor (i.e. you can crib them with minor adjustments from the Applicative typeclass).

By contrast, the operations and laws of the Alternative typeclass of today seem to amount to type Alternative f = Monoid (forall a. f a). We can witness a pair of maps that convert between the primitive operations of Alternative' and Alternative:

instance Alternative f => Alternative' f
where
unit _ = empty
zip (fa, fb) = (Left <$$> fa) <|> (Right <$$> fb)

instance Alternative' f => Alternative f
where
empty = absurd <$$> unit () fa1 <|> fa2 = either id id <$$> zip (fa1, fa2)


However, it seems likely that this is not truly an isomorphism, due to the difference in constraints and laws between the two typeclasses.

On one hand, Alternative is more permissive than Alternative', since it does not demand any form of compatibility between the monoid and the functor, whereas the commutative diagrams of a lax monoidal functor do.

On, the other hand, Alternative demands Applicative, a constraint that Alternative' does not demand. When viewed through the lens of lax monoidal functors, Applicative f simply means that f coheres the (,) monoidal structure to (,), and Alternative f means that it coheres the Either monoidal structure to (,). These are orthogonal claims, so there is no need for Alternative' f to demand a Applicative f constraint.

That said, I conjecture that most types that support Alternative instances also support a lawful instance of Alternative', and so you can imagine them to be "applicatives for different tensor".

This might not be the right approach, but here's my take on the question:

If $$(\mathcal{C}, \otimes, I, \dots)$$ is a (strict, for the sake of readability) monoidal category, we can endow $$[\mathcal{C}, \mathcal{C}]$$ with two different monoidal structures:

1. The "usual" monoidal structure, where $$\circ$$ is the product and $$\mathbf{1}_\mathcal{C}$$ is the unit. This is the monoidal category whose monoids are called monads. Notice how this doesn't actually use the fact that $$\mathcal{C}$$ is endowed with a monoidal structure.

2. The "other" (I sometimes refer to it as "pointwise") monoidal structure has "pointwise $$\otimes$$" (let's call it $$\boxtimes$$) as product, and $$\Delta_I$$ (the constant functor on the unit object of $$\mathcal{C}$$) as monoidal units. In this category, monoids are related to what Haskell programmers call "alternative" functors, but without the part about being "applicative" and with one additional property. This can be further generalized to the case of functors $$\mathcal{B} \to \mathcal{C}$$ where $$\mathcal{B}$$ is not required to be monoidal, and everything that follows should work just fine (but I won't venture there).

Let me elaborate on the latter, since the former is (probably) already well-known to most haskell programmers here (other than being significantly less relevant):

First we'll need to properly define $$\boxtimes$$: if we fix functors $$\mathcal{F}, \mathcal{G} : \mathcal{C} \to \mathcal{C}$$ we can define $$(\mathcal{F} \boxtimes \mathcal{G}) (a) := \mathcal{F}(a) \otimes \mathcal{G}(a)$$ (the definition for morphisms is analogous). This, together with $$\Delta_I$$ as a unit, defines a monoidal category: all axioms (and further properties, such as symmetry and strictness) follow trivially from the monoidal structure on $$\mathcal{C}$$.

Now, monoids in this monoidal category are what haskell programmers call "alternative" functors, but that's not all: they are so in a functorial way, meaning that if $$f : a \to b : \mathcal{C}$$ and $$\mathcal{M}$$ is a monoid in $$([\mathcal{C}, \mathcal{C}], \boxtimes, \Delta_I)$$ then $$\mathcal{M}(f)$$ is a morphism of monoids. To see this, note how this diagram needs to commute in a pointwise fashion, and the same applies to this other diagram too. We're in the strict case, but nothing significant changes in the general one (except that you need to keep track of the coherence laws).

A more abstract (but equivalent) formulation is that each object $$\mathcal{M} : \mathbf{Mon}([\mathcal{C}, \mathcal{C}])$$ naturally defines a functor $$\hat{\mathcal{M}} : \mathcal{C} \to \mathbf{Mon}(\mathcal{C})$$, and this point of view leads to the observation that this actually defines a functor $$(\hat{-}) : \mathbf{Mon}([\mathcal{C}, \mathcal{C}]) \to [\mathcal{C}, \mathbf{Mon}(\mathcal{C})]$$.

What this amounts to is just that a "monoidal" natural transformation $$\nu : \mathcal{M} \to \mathcal{N} : \mathbf{Mon}([\mathcal{C}, \mathcal{C}])$$ between monoid object in this "pointwise" monoidal category of (endo)functors also defines a natural transformation $$\hat{\nu} : \hat{\mathcal{M}} \to \hat{\mathcal{N}} : [\mathcal{C}, \mathbf{Mon}(\mathcal{C})]$$: the argument is similar but you need to look at the diagrams for morphisms of monoids.

Things are not over yet: it seems that $$(\hat{-})$$ is actually an equivalence of categories! You can convince yourself of this if you squint hard enough, and the argument is roughly "brute force":

Let's construct a strict inverse to $$(\hat{-})$$, call it $$(\tilde{-})$$: take $$\mathcal{F} : \mathcal{C} \to \mathbf{Mon}(\mathcal{C})$$, compose it with the forgetful functor $$\mathbf{U} : \mathbf{Mon}(\mathcal{C}) \to \mathcal{C}$$. We'll now use the "forgotten structure" on $$\mathcal{F}$$ to endow $$\mathbf{U} \circ \mathcal{F}$$ with the structure of a monoid in $$[\mathcal{C}, \mathcal{C}]$$: to do this, notice how the monoid structure on $$\mathcal{F}(a)$$ (for $$a : \mathcal{C}$$) needs to be such that each morphism $$\mathcal{F}(f) : \mathcal{F}(a) \to \mathcal{F}(b)$$ is a morphism of monoids: this amounts to saying that the correspondences $$a \mapsto \mathbf{m}^{\mathcal{F}(a)}$$ and $$a \mapsto \mathbf{e}^{\mathcal{F}(a)}$$ are natural in $$a$$ and hence natural transformations (its now trivial to check that these endow $$\mathbf{U} \circ \mathcal{F}$$ with the structure of a monoid in $$[\mathcal{C}, \mathcal{C}]$$).

We now need to check that $$(\hat{-}) \circ (\tilde{-}) = \mathbf{1}_{[\mathcal{C}, \mathbf{Mon}(\mathcal{C})]}$$ and $$(\tilde{-} \circ \hat{-}) = \mathbf{1}_{\mathbf{Mon}([\mathcal{C}, \mathcal{C}])}$$: they should both be long and trivial computations, and I'll (boldly) omit them.

Edit: this can probably be "merged" to the other answer by Andrej Bauer to give a more complete answer

• Isn't $\mathcal{M}(f)$ only a morphism of monoids if $f$ is? Feb 5 '20 at 14:30
• 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 '20 at 19:56
• 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 23 '20 at 17:45
• 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 26 '20 at 9:11
• 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 '20 at 9:25