Let $B$ be a strict monoidal category, and $\left \langle c,\mu ',\eta ' \right \rangle$ a monoid in $B$. Now suppose we consider the simplex category $\left \langle \triangle ,+,0 \right \rangle$, the monoid $\left \langle 1,\mu ,\eta \right \rangle$and define $F:\Delta \rightarrow B$ in the obvious way:
On objects take
$Fn=c^{n}$ (so that in particular $F0=\eta $ and $F1=c$.
On arrows, note that if $f:m\rightarrow n=$ then $f=\sum_{i=0}^{n-1}\mu ^{m_{i}}$ with $\sum_{i=0}^{n-1}m_{i}=m$ and $f(j)=i$ if $j\in m_{i}$, for all $0\leq j\leq m-1$, so it makes sense to set
$Ff=\sum_{i=0}^{n-1}(\mu' )^{m_{i}}$
Now let $g:n\rightarrow k$ so that $g=\sum_{i=0}^{k-1}\mu ^{n_{i}}$, with $\sum_{i=0}^{k-1}n_{i}=n$ and $g(j)=i$ if $j\in n_{i}$, for all $0\leq j\leq k-1$.
My problem is proving that $F$ is a functor. That is, that $F(g)F(f)=F(gf)$. This amounts to showing that
$(\sum_{i=0}^{n-1}\mu ^{m_{i}})(\sum_{i=0}^{k-1}\mu ^{n_{i}})$ composes in the same way as $(\sum_{i=0}^{n-1}(\mu ') ^{m_{i}})(\sum_{i=0}^{k-1}(\mu ')^{n_{i}})$.
I thought this would be an easy application of the associative law, which holds in any strict monoidal category, namely:
Let $k_{1},k_{2},\cdots, k_{n}$ be any set of $n$ integers. Then,
$\mu ^{n}(\left ( \mu ^{k_{1}} \otimes \mu ^{k_{2}}\right )\otimes \cdots \mu ^{k_{n}})=\mu ^{k_{1}+k_{2}+\cdots +k_{n}}$.
I have tried using the associative law on this, but it is clear that, by itself, it won't do because each side is an arrow with codomain $c$.
I have tried considering the product $Fgf$ written $Fgf:c^{m}\rightarrow c^{k}$, $Fgf=\sum_{i=0}^{k-1}(\mu ')^{m'_{i}}$, with $\sum_{i=0}^{k-1}m'_{i}=m$ and $Fgf(j)=i$ if $j\in m'_{i}$, for all $0\leq j\leq k-1$. Then I need to show that
$\sum_{i=0}^{k-1}(\mu ')^{m'_{i}}=(\sum_{i=0}^{n-1}(\mu ') ^{m_{i}})(\sum_{i=0}^{k-1}(\mu ')^{n_{i}})$. But I have not been able to prove this, although there are interesting relations here using the definitions of the $m _{i},m' _{i}$ and $n _{i}$.
I do have $\mu ^{k}(\mu '^{m'_{0}}\otimes \cdots \otimes \mu '^{m'_{k-1}})=\mu ^{k}(\mu '^{m_{0}}\otimes \cdots \otimes \mu '^{m_{n-1}})(\mu '^{n_{0}}\otimes \cdots \otimes \mu '^{n_{k-1}})$ and so I tried induction on $k$, but to no avail.
Can anyone point me in the right direction?
Note: The associative law follows from setting $\mu _{e}=\eta, \mu _{c}= 1_{c}, \mu _{c\otimes c}=\mu $ and thereafter $\mu _{u_{c}\otimes v_{c}}=\mu(\mu _{u_{c}}\otimes \mu _{v_{c}})$ Then, the multiplication is, using the notation for powers of $c$, $\mu ^{0}= \eta $, $\mu ^{1}=1_{c}$, $\mu ^{2}=\mu $, and thereafter, $\mu ^{n+1}=\mu \left ( \mu ^{n} \otimes 1\right )$. From this you get the result in a strict monoidal category.
Mac Lane showed on page 171 of "categories for working mathematicians" that any combination of multiplications lead to the same function, if they have the same domain, and their codomain is the monoid, via coherence. Since both $F(g \circ h)$ and $F(g) \circ F(h)$ are combinations of multiplications, they are equal via that theorem.
His proof goes along the lines of that, since the basic associative laws are in terms of the distributor and two unitors, any two iterated products would be various combinations of them, making the result a canonical arrow.