In Grayson's 'Higher Algebraic K-theory II', leading up to the categorical generalisation of the plus construction, he considers $\pi_0(S) = \pi_0(BS)$, where $S$ is a (small, symmetric) monoidal category and $BS$ is its classifying space. It is then tacitly assumed that $\pi_0(S)$ is itself an abelian monoid... but I can't see how this is true.

How is $\pi_0(S)$ a monoid, explicitly?

The only possible candidate for a monoidal structure that I can think of is the following. Let $C_1 = [ [(A_0 \to A_m, s)]]$ and $C_2 = [ [(B_0 \to B_n, t)]]$ be path components of $BS$ with, say, $n \leq m$. If $n < m$, extend the sequence of the $A_i$ to a sequence $A_0 \to \cdots \to A_n$ by adding $0$'s - $0$ being the identity element in $S$ - and extend $s$ to an element $s' \in \Delta_n$ by adding components of zero. Then define $C_1 + C_2$ by $[[(A_0 + B_0 \to \cdots \to A_n + B_n, \frac{s' + t}{2})]]$. But I doubt this is even associative (let alone well-defined)!

In Grayson's 'Higher Algebraic K-theory II', leading up to the categorical generalisation of the plus construction, he considers $\pi_0(S) = \pi_0(BS)$, where $S$ is a (small, symmetric) monoidal category and $BS$ is its classifying space. It is then tacitly assumed that $\pi_0(S)$ is itself an abelian monoid... but I can't see how this is true.

How is $\pi_0(S)$ a monoid, explicitly?

In Grayson's 'Higher Algebraic K-theory II', leading up to the categorical generalisation of the plus construction, he considers $\pi_0(S) = \pi_0(BS)$, where $S$ is a (small, symmetric) monoidal category and $BS$ is its classifying space. It is then tacitly assumed that $\pi_0(S)$ is itself an abelian monoid... but I can't see how this is true.

How is $\pi_0(S)$ a monoid?

The only possible candidate for a monoidal structure that I can think of is the following. Let $C_1 = [ [(A_0 \to A_m, s)]]$ and $C_2 = [ [(B_0 \to B_n, t)]]$ be path components of $BS$ with, say, $n \leq m$. If $n < m$, extend the sequence of the $A_i$ to a sequence $A_0 \to \cdots \to A_n$ by adding $0$'s - $0$ being the identity element in $S$ - and extend $s$ to an element $s' \in \Delta_n$ by adding components of zero. Then define $C_1 + C_2$ by $[[(A_0 + B_0 \to \cdots \to A_n + B_n, \frac{s' + t}{2})]]$. But I doubt this is even associative (let alone well-defined)!

In Grayson's 'Higher Algebraic K-theory II', leading up to the categorical generalisation of the plus construction, he considers $\pi_0(S) = \pi_0(BS)$, where $S$ is a (small, symmetric) monoidal category and $BS$ is its classifying space. It is then tacitly assumed that $\pi_0(S)$ is itself an abelian monoid... but I can't see how this is true.

How is $\pi_0(S)$ a monoid, explicitly?

The only possible candidate for a monoidal structure that I can think of is the following. Let $C_1 = [ [(A_0 \to A_m, s)]]$ and $C_2 = [ [(B_0 \to B_n, t)]]$ be path components of $BS$ with, say, $n \leq m$. If $n < m$, extend the sequence of the $A_i$ to a sequence $A_0 \to \cdots \to A_n$ by adding $0$'s - $0$ being the identity element in $S$ - and extend $s$ to an element $s' \in \Delta_n$ by adding components of zero. Then define $C_1 + C_2$ by $[[(A_0 + B_0 \to \cdots \to A_n + B_n, \frac{s' + t}{2})]]$. But I doubt this is even associative (let alone well-defined)!