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I want to define the List-Permuation (co)monad as follows:

$List-Perm = (L, \mu, \eta, \nu, \zeta)$.

$$L:Set \rightarrow Set$$

such that, L returns the set of Lists of a given set.

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

by returning a list from a list of lists by concatenation.

$$\eta : 1_{Set} \rightarrow P$$

by sending a set to the set of lists with just those elements (eg $\eta:(\{a,b,c \}) \rightarrow \{ [a],[b],[c] \}$)

$$\nu : L \rightarrow L \cdot L$$

This works by sending each list, $w$ to a list of permutations of $w$. This can be done by defining for every set $S$

  1. a function $f_S$ such that, if $n$ is the cardinality of $S$, $f_{S}^{m} =I$ iff mModn = 0 and $I$ is the identity function on $S$.
  2. a map from $N$ to the set of functions $f^{i}$ which we use to define a list of functions $[f^{0}, f^1, f^2 \cdots]$.
  3. then define a list of permutations as, $A \in Set$, $w \in L(A)$, $L \cdot L (A) = [[f^0(w)], [f^1(w)], [f^2(w)], \cdots ]$

[Edit] I have defined the functions, $f$, to act on the set $S$. This is probably not going to work. We want the coproduct to take a list to a list of all of its permutations. Instead, given a list, $w$,of length $l$, we want to define a function on the set $W$ of natural numbers less than $l $. Then do as we did before so that $f_{w}$, rather than permuting set elements, it permutes the list positions.

  1. a function $f_w$ such that, if $l$ is the length of $w$, $f_{w}^{m} =I$ iff mMod$l $ = 0 and $I$ is the identity function on $W$.

    1. a map from the set of functions $f_{w}^{i}$ to $N $ which we use to define a list of functions $[f_{w}^{0}, f_{w}^1, f_{w}^2 \cdots]$.

[Edit] You might want to define a set of functions, one for every integer which we take as list length. Each function with the properties above.

[Edit] I realize now that this doesn't work, simply by cardinality: the list generated by $f_{l}$ has $l $ elements, and there are $n! $ permutations. We need to somehow have the permutation group act on the list in a natural way.

The counit:

$$\zeta : L \rightarrow I$$

This may not admit a counit. My only guess is that you take a List, $w = [a,b,c]$ and map it to $a \in S$, ie you map the head of the list to its itself in $S$.

Does this work?

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    $\begingroup$ I don't understand this 'comultiplication' $\eta$. Shouldn't it go $L\to L\cdot L$? How do the permutations of lists enter the picture here? $\endgroup$ – Berci Mar 7 '18 at 0:38
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    $\begingroup$ Sorry, it's about the $\nu$. $\endgroup$ – Berci Mar 7 '18 at 0:44
  • $\begingroup$ Yes that was a mistake thank you $\endgroup$ – Ben Sprott Mar 7 '18 at 1:22
  • $\begingroup$ Hi @Berci, I have been convinced that this will not work for the following reason: the species of linear orderings is not naturally isomorphic to the species of permutations. $\endgroup$ – Ben Sprott Mar 7 '18 at 21:40
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Yes, the 'head' function $\zeta:=[a,b,c,\dots]\mapsto a\ $ works as counit, provided that the permutation $f$ for a list of length $n$ is the specific cycle $(1\,2\,\dots\,n)$.
The empty set and the empty list should not be considered here.

But, the two counit identities are satisfied: $$[a_1,a_2,\dots,a_n]\ \overset\nu\longmapsto\ \big[\overbrace{[a_1,a_2,\dots]},\ [a_2,a_3,\dots],\ \dots \big] \ \overset{\zeta L}\longmapsto\ [a_1,a_2,\dots] \\ [a_1,a_2,\dots,a_n]\ \overset\nu\longmapsto\ \big[[\underline{a_1},a_2,\dots],\ [\underline{a_2},a_3,\dots],\ \dots \big] \ \overset{L\zeta}\longmapsto\ [a_1,a_2,\dots]$$

Also, the comultiplication $\nu$ so defined is coassociative: the common ternary comultiplication will be $$[a_1,a_2,\dots,a_n]=:w\ \longmapsto\ \Big[ [w,fw,..];\ [fw,f^2w,..];\ [f^2w,f^3w,..];\ \dots \Big] $$

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