# The Pontrjagin product for homology at the chain level

The singular cochain complex of a space $X$ with coefficients in a ring $R$ can be endowed with a product, turning it into a differential graded $R$-algebra, such that in cohomology it gives the familiar cup product. However, at the chain level one has to be careful, since different choices of products can have different properties: e.g., an arbitrary one will not necessarily be associative, whereas the one induced by the Alexander–Whitney map will be. This indeterminacy is parametrized by the Eilenberg–Zilber theorem.

Now, consider an H-space $X$. Then in homology (with whichever coefficients) we have a product, the Pontrjagin product, which is associative or commutative if the product of $X$ is so too, up to homotopy. But what about at the chain level? The product at the chain level is given by $$C_*(X)\otimes C_*(X)\to C_*(X\times X)\to C_*(X),$$ where the first map is a fixed chain homotopy equivalence, and the second one is induced by the multiplication.

My question is:

If we take the first map to be the shuffle map and $X$ to be homotopy associative, is the resulting differential graded algebra $C_*(X)$ associative?

I'm guessing the analogous question for commutativity has a negative answer, since it doesn't work for cohomology; any comments on this are also welcome.

If $X$ is truly associative (not just up to homotopy), then since the AW map is associative too, the composite will be associative.
Otherwise, choose $x,y,z \in X$ such that $(xy)z \neq x(yz)$. If $u \in X$ is some point, let $\sigma(u) : \Delta^0 \to X$ be the associated zero-simplex (constant map equal to $u$). Then the image of $(\sigma(x), \sigma(y))$ under the composite of the Alexander–Whitney map and the product of $X$ is simply $\sigma(xy)$ (the AW map is especially simple in degree zero...). It follows that $$\bigl(\sigma(x) \sigma(y)\bigr) \sigma(z) = \sigma((xy)z) \neq \sigma(x(yz)) = \sigma(x) \bigl(\sigma(y) \sigma(z)\bigr)$$ and so the map you described isn't associative.