The Eilenberg–Zilber theorem in singular homology, relating the monoidal structure of the category of chain complexes with the chain complex of the cartesian product of the underlying spaces, is used in proving the Künneth theorem.

When I read about it online, I often see it's referred to in much more abstract contexts, where it's usually stated along the lines of:

The singular chain functor is lax-monoidal.

What is the structural significance of this? I understand extra structure is always nice, but what are some consequences that are easy to see in this light?

  • 2
    $\begingroup$ Lax monoidality doesn't quite capture everything about the Eilenberg–Zilber theorem. I think it is much more important to know that the comparison is a quasi-isomorphism. $\endgroup$
    – Zhen Lin
    Feb 21, 2016 at 12:05
  • $\begingroup$ I'll point out that one inverse map to the E-Z map, the Alexander-Whitney map is the starting point of what is called an $E_\infty$ algebra structure on cochains, which leads to things like Steenrod-operations. $\endgroup$ Feb 22, 2016 at 12:49

1 Answer 1


As Zhen Lin says, a really important part of the Eilenberg–Zilber theorem is that the lax-monoidal structure map is a quasi-isomorphism. Nevertheless, reformulating the first part of the theorem as "the functor $C_*(-)$ is lax-monoidal" is helpful for applying general results to it. For example:

Proposition. Let $\mathsf{C}$ and $\mathsf{D}$ be symmetric monoidal categories, and $F : \mathsf{C} \to \mathsf{D}$ a lax monoidal functor. Then $F$ induces a functor from monoid objects in $\mathsf{C}$ to monoid objects in $\mathsf{D}$.

Proof. Let $M$ be a monoid object in $\mathsf{C}$. The product morphism on $F(M)$ is given by $$F(M) \otimes F(M) \xrightarrow{\text{lax monoidal}} F(M \otimes M) \xrightarrow{F(\mu)} F(M)$$ and the unit by $$1_\mathsf{D} \xrightarrow{\text{lax monoidal}} F(\mathsf{1}_C) \xrightarrow{F(e)} F(M).$$ Then it's easy to check that this gives $F(M)$ the structure of a group object, and that $F$ then defines a functor from group objects to group objects.

Corollary. If $M$ is a topological monoid, then $C_*(M)$ is a monoid in the category of chain complexes, i.e. a dg-algebra.

The proof is immediate from the Eilenberg–Zilber theorem, and it's helpful to split the proof in two parts in order not to get bogged down in technical details. More generally, if $\mathtt{P}$ is a topological operad, then $C_*(\mathtt{P})$ is a dg-operad, and if $A$ is a $\mathtt{P}$-algebra, then $C_*(A)$ is a $C_*(\mathtt{P})$-algebra.

Basically, any sort of "product" structure can be transported from topological spaces to chain complexes; in general, this is "the point" of determining if a functor is lax monoidal or not. You would need an oplax monoidal functor to transform topological "coproduct" structures into coproduct dg-structures. The EZ functor isn't oplax monoidal, so in general you cannot expect it to turn "coalgebras" into "coalgebras".

  • $\begingroup$ In your proposition, you should replace "group object" by "monoid object": which is in fact what you prove, as you don't talk (and couldn't in full generality) about the inverse map. In fact, if I don't mistake, a lax-monoidal functor between cartesian monoidal categories does not necessarily preserve group object (although a strong one would). $\endgroup$
    – Pece
    Feb 21, 2016 at 13:10
  • $\begingroup$ @Pece It seems you're right (and groups aren't even algebra over an operad, not sure what I was thinking). $\endgroup$ Feb 21, 2016 at 13:50

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