3

The List Monad is defined as a triple $< L , \mu, \eta >$.

$L: Set \rightarrow Set$

$L$ takes a set to the set of all lists on that set.

$\mu : L \cdot L \rightarrow L$

$\mu$ takes a list of lists to a list by just concatenating all the internal lists.

$\eta : I \rightarrow L$

$\eta$ takes every set element and produces the list with just that element.

What, precisely, is the Eilenberg-Moore Category for the List monad? How do we know this? Is it computable from just the data I have given about the List monad?

Ben Sprott
  • 1,359
  • 4
    It’s the category of monoids. The list monad is the free monoid monad. – Qiaochu Yuan Aug 30 '20 at 04:01
  • 1
    As Qiaochu Yuan says modules over the list monad are monoids. Try writing out the definition of an algebra over this monad and verify that the map $\alpha_X : LX\to X$ gives an associative product on $X$, with its action on a list being to multiply everything in the list together. Note that the product of the empty list will be the identity of the monoid. What does the unitality requirement of the algebra correspond to? – jgon Aug 30 '20 at 04:02
  • @QiaochuYuan I don't understand. Doesn't the category of monoids contain the commutative monoids? I would think that the EM category would be the category of free monoids. – Ben Sprott Aug 30 '20 at 04:12
  • @jgon Why is everyone using Monoid and Free monoid interchangeably? – Ben Sprott Aug 30 '20 at 04:14
  • 3
    @BenSprott Monoid and free monoid are not interchangeable terms. The monad here is often called the free monoid monad because it comes from the (free monoid) left adjoint to (forgetful) adjunction between $\mathbf{Set}$ and $\mathbf{Monoid}$. However, the algebras for this monad are all monoids, not just free monoids. – jgon Aug 30 '20 at 04:17
  • @jgon ok thank you. – Ben Sprott Aug 30 '20 at 04:18
  • 2
    @Ben: it's the Kleisli category that is the category of free monoids. – Qiaochu Yuan Aug 30 '20 at 05:28

0 Answers0