Path components of a monoidal category form a monoid? 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?

 A: The result will follow if you can show that $\pi_0\colon Cat \to Set$ preserves finite products, because monoidal categories can be defined diagrammatically in the 2-category $Cat$, and these diagrams are sent, under the assumption of product preservation, to the diagrams defining a monoid in $Set$.
But $\pi_0$ can be defined via the following sequence of functors: $N\colon Cat \to sSet$, followed by $|-|\colon sSet \to CGHaus$, followed by $\pi_0\colon CGHaus \to Set$. Here $CGHaus$ has the k-space product, namely $X\times_k Y := k(X\times Y)$. The functors $N$ and $|-|$ preserve finite products, so we only need to know that $\pi_0$ sends $\times_k$ to the product of sets. Since $I$ is compact Hausdorff, a function $I \to X\times_k Y$ is continuous iff the function $I \to X\times Y$ is continuous, hence two points in $X\times_k Y$ are in the same path component iff they are in the same path component in $X\times Y$. Then we use the fact that $\pi_0$ preserves the ordinary product of topological spaces.
EDIT: In light of the subtle edit, here is some more detail. Don't try to write down the product of a pair of path components. A element of $\pi_0(S)$ is represented by an object of $S$. The product of $[a]$ and $[b]$ is then $[a\otimes b]$. That's it. The above two paragraphs serve to show that this is well-defined on equivalence classes, associative and unital.
