Let $A$ be a homotopy ring spectrum. Then the homology theory $A\_\ast : Spectra \to GrAb$ lifts to a homology theory valued in $GrMod(\pi\_\ast A)$. If $A$ is homotopy commutative, then this functor $A\_\ast : Spectra \to GrMod(\pi\_\ast A)$ is lax monoidal. But it makes sense to ask for a lax monoidal structure on this functor even if $A$ is not homotopy commutative, so long as $\pi\_\ast A$ is graded-commutative. And the answer can be *yes* --e.g. at the prime 2, Morava $K$-theory $K(n)$ is not homotopy commutative, but $K(n)\_\ast$ is strong monoidal.


**Question:** Let $A$ be a homotopy ring spectrum such that $\pi\_\ast A$ is graded-commutative. Then is the functor $A\_\ast : Spectra \to GrMod\_{\pi\_\ast A}$ lax monoidal? Even lax symmetric monoidal?



If $\pi_\ast A$ is graded-commutative, then is $A_\ast$ a lax monoidal functor?
As pointed out in the comments, the functor $A\_\*$ cannot in general be lax symmetric monoidal without making some alterations.


Here is an incomplete discussion of when $A\_\*$ can be lax monoidal.


The first observation is that, for any homotopy associative ring spectrum $A$, the functor $A\_\*$ naturally takes values in $\pi\_\* A$-bimodules. The left and right actions of $A$ on itself produce natural left and right actions of $A$ on $A \otimes X$ for any $X$, and lax monoidality of $\pi\_\*$ show that $A\_\* X$ is then a $\pi\_\* A$-bimodule.


The second observation is that this makes the functor $A\_\*$ a lax monoidal functor to the category of $\pi\_\* A$-bimodules. The homotopy associativity of $A$ ensures that the two composites $A \otimes A \otimes A \to A \otimes A \to A$ are homotopic, and so for any $X$ and $Y$ the two composites
$$
(A \otimes X) \otimes A \otimes (A \otimes Y) \to (A \otimes X) \otimes (A \otimes Y) \to A \otimes (X \otimes Y)
$$
are homotopic. The lax monoidality of $\pi\_\*$ then tells us that the two composites
$$
A\_\* X \otimes \pi\_\* A \otimes A\_\* Y \to A\_\*(X \otimes Y)
$$
are equal, establishing $\pi\_\* A$-bilinearity. However, tracking which maps actually appear on the spectrum level, this specifically uses the right $\pi\_\* A$-module structure on $A\_\* X$ and the left $\pi\_\* A$-module structure on $A\_\* Y$.


We now assume that the coefficient ring $\pi\_\* A$ is graded-commutative. We would like to show that, for any $X$, the left and right module structures "coincide": $a \cdot x = \pm x \cdot a$ for any $a \in \pi\_\* A$ and $x \in A\_\* X$ according to the Koszul sign rule. If we can do this, then bimodule bilinearity collapses to module bilinearity.


Suppose $x \in A\_d X$ comes from a map $S^d \to A \otimes X$. We can express $A$ as a filtered hocolim of finite spectra $A\_i$, and get a lift to a map $S^d \to A\_i \otimes X$, with an adjoint map $DA\_i \otimes S^d \to X$ using Spanier-Whitehead duality. This gives us a lift of $x$ to a factorization
$$
S^d \to \xrightarrow{\eta} A \otimes (DA\_i \otimes S^d) \to A \otimes X.
$$
Therefore, it suffices to check this in the "universal" cases where $x \in A\_\* (DA\_i)$ comes from the canonical map $\eta: S^0 \to A \otimes DA\_i$.


This lifts to the limit over $i$, however: the canonical unit map $S^0 \to map(A,A)$ to the function spectrum. In these terms, we are asking if the canonical element $1 \in [A,A]$ is sent to the same element under post-multiplication by $a$ on either the left or the right; or equivalently, if "multiply by $a$ on the left" and "multiply by $a$ on the right" are always homotopic maps $A \to A$, rather than merely giving equal maps $\pi\_\* A \to \pi\_\* A$.




---


An example of this can be constructed from the following differential graded algebra. Start with the discrete ring $\Bbb Z[x]$. We first form a bimodule with a single generator $y$, annihilated by $x$ on the left and right; we form a second bimodule with a single generator $z$, annihilated by $x^2$ on the left and $x$ on the right (so it has a basis $\{z,xz\}$). Form the mapping cone $M$ of the map $z \mapsto 2y$: it has $H\_0 = \Bbb Z/2$, generated by $y$, and $H\_1 = \Bbb Z$, generated by $xz$. Let $A$ be the square-zero extension $\Bbb Z[x] \times M$ by this differential graded bimodule $M$. The coefficient ring is graded-commutative because almost all products are zero. However, in mod-$2$ homology we see the element $z$ which no longer commutes with the generator $x$ from $A\_\*$.


