3
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • 5
    $\begingroup$ It would be best to better learn the details before asking questions, so your details are best ignored. Conceptually, the functors B and \pi_0 commute with products: now follow your nose. $\endgroup$
    – Peter May
    Commented Feb 17, 2013 at 23:22

1 Answer 1

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • 4
    $\begingroup$ Perhaps it's worth being explicit and saying that it is the monoidal product $\otimes$ in the monoidal category $S$ that descends to the binary operation of the monoid $\pi_0 S$. $\endgroup$
    – Zhen Lin
    Commented Feb 17, 2013 at 23:30
  • 2
    $\begingroup$ The concept of taking the elements of $\pi_0(S)$ to actually be whole path-components of $S$ is not helpful in this instance, since $\pi_0(S)$ is only defined by a bunch of universal properties. In particular, nothing should depend on the representation of elements of $\pi_0(S)$ chosen. So pick the simplest one you can think of... $\endgroup$
    – David Roberts ♦
    Commented Feb 18, 2013 at 0:37
  • $\begingroup$ @David: Thanks for the answer (and then the further edit). $\endgroup$
    – Joshua Seaton
    Commented Feb 18, 2013 at 1:24
  • $\begingroup$ @David: silly question: why would a path-component of $BS$ necessarily contain a 0-cell? $\endgroup$
    – Joshua Seaton
    Commented Mar 17, 2013 at 15:59
  • 1
    $\begingroup$ Each object gives a 0-cell in the geometric realisation. Every point belongs to some n-simplex, which corresponds to a finite list of arrows. The vertices of the n-simplex are identified with the points given by objects contained in this list, so every point in the geometric realisation is connected by a path to a 0-cell. $\endgroup$
    – David Roberts ♦
    Commented Mar 17, 2013 at 20:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .