### Dold–Kan Question

#### Posted by John Baez

Lately I’ve been writing a whole bunch of issues of This Week’s Finds in parallel, trying to tell a long story and make sure it’s told in detail.

Right now I just want to make sure I understand one thing. In their book *Rational Homotopy Theory*, Félix, Halperin and Thomas claim:

**Proposition 26.4**: If $G$ is a topological monoid, then the normalized singular chain complex $C_*(G,\mathbb{Q})$ is a differential graded Hopf algebra.

Now, there’s something puzzling about this right away. Following your instincts,
when $G$ is a topological *group* you’d naturally expect $C_*(G,\mathbb{Q})$ to be a differential graded *Hopf* algebra. But when $G$ is just a topological *monoid*, you’d only expect $C_*(G,\mathbb{Q})$ to be a differential graded *bialgebra* — since where would the antipode come from, if not the inverse operation in $G$?

This puzzle is easily resolved: Félix *et al* define a differential graded Hopf algebra in a way that omits any mention of the antipode! So, they’re really getting a differential graded bialgebra.

But my real question concerns the Dold–Kan theorem, which is underlying the proof of this result.

We’ve got a functor

$X \mapsto C_*(X,\mathbb{Q})$

from spaces to rational chain complexes, sending any space to its normalized singular chain complex with rational coefficients. *If this functor were strongly monoidal*, it would send monoids to monoids, comonoids to comonoids, and bimonoids to bimonoids. In other words: it would send topological monoids to differential graded algebras, topological comonoids to differential graded coalgebras, and topological bimonoids to differential graded bialgebras.

Then we’d be in luck, since *every topological space is a topological comonoid in a unique way*, with comultiplication being the diagonal

$\Delta : X \to X \times X$

And as a spinoff, *every topological monoid is a topological bimonoid in a unique way*.

So, *if our functor were strongly monoidal*, it would send topological monoids to differential graded bialgebras.

However, in the $n$Lab, a wonderful entry on the monoidal Dold–Kan correspondence is gradually taking shape. This concerns the extent to which the ‘normalized Moore complex’ functor from simplicial abelian groups to chain complexes is monoidal. I believe that’s the meat of the question here! And this entry seems to suggest that this functor is merely *lax* monoidal. Perhaps I’m reading too much into it, or the wrong things… in which case I’d like to be corrected!

If our functor

$X \mapsto C_*(X,\mathbb{Q})$

were merely lax monoidal, it would send monoids to monoids, but not necessarily comonoids to comonoids. So then — even though any space $X$ is a topological comonoid — we’d have no right to expect that $C_*(X,\mathbb{Q})$ is a differential graded coalgebra… unless some miracle intervenes, which I’d then like to understand.

In Section 4b of *Rational Homotopy Theory*, Félix *et al* describe the Alexander–Whitney map

$AW : C_*(X \times Y, \mathbb{Q}) \to C_*(X,\mathbb{Q}) \otimes C_*(Y,\mathbb{Q})$

and they say this makes $C_*(X,\mathbb{Q})$ into a differential graded coalgebra.

So what’s going on? Is our functor

$X \mapsto C_*(X,\mathbb{Q})$

strongly monoidal, or just lax monoidal? And if the latter, why is $C_*(X,\mathbb{Q})$ a differential graded coalgebra?

## Re: Dold–Kan Question

I should emphasize that I’m following Félix

et al— but not the $n$Lab — in using $C_*$ to denote thenormalizedsingular chain complex, where we mod out by degenerate simplices.As they point out, authors often neglect to mod out by degenerate simplices when defining the singular chain complex. For the purposes of defining homology this is harmless. But if we neglected to do this, the chain complex for a point, $C_*(point, \mathbb{Q})$, would not be just $\mathbb{Q}$ concentrated in grade zero. So then our functor

$X \mapsto C_*(X, \mathbb{Q})$

would

definitely fail to be strongly monoidal.