# Associativity of a homology ring of an H-space.

Let $$X$$ be a based topological space with $$\mu:X\times X\rightarrow X$$ which is homotopy associative and unital wrt the basepoint. We call such $$(X,\mu)$$ an associative H-space. The multiplication on $$H_*(X)$$ is defined as $$H_*(X)\otimes H_*(X)\rightarrow H_*(X\times X)\rightarrow H_*(X)$$ with the second map being induced by $$\mu$$ and the first one by the Kunneth theorem. Is there a way to prove the associativity of this multiplication without referring to the definition of a Eilenberg-Zilber map which is used in the Kunneth theorem? I'm looking for a proof which uses naturality or some other properties of a map $$H_*(X)\otimes H_*(X)\rightarrow H_*(X\times X)$$ if it is possible.

This question comes from section 3.C of Hatcher where he proves that homology and cohomology of an $$H$$-space are Hopf algebras. He proves just the counit axim for the cohomology and I was trying to fill all details for other axioms without going to the actual definition of Eilenberg-Zilber map which is rather technical.

Any references also would be helpful.

• Well the homology cross product is associative $(x\times y)\times z=x\times (y\times z)$, but I think this is proved using the EZ theorem. You also have that two maps which are homotopic induce the same map on homology, and here $\mu\circ(1\times\mu)\simeq\mu\circ(\mu\times 1)$ by assumption. Commented Apr 21, 2019 at 14:37
• Alternatively it will suffice by duality to prove that the coproduct on the cohomology of $X$ is coassociative, that is $(1\times\mu)^*\mu^*=(\mu\times 1)^*\mu^*$. Here the external cross product is just cup product $x\times y=(x\times 1)\cup (1\times y)$, and I know Hatcher proves that the cup product is associative at some point in chapter 3. Commented Apr 21, 2019 at 14:40
• @Tyrone I'm still stuck. The diagram for the coassociativity of a coproduct on a cohomology tells me the following: Given $\mu^*a=\sum (p_1^*(a_i) \otimes p_2^*(b_i)$ I have to show that there exist $\sum(c_j)\otimes (d_j) \otimes (e_j)$ s.t. $\sum \mu^*(a_i) \otimes (b_i)=\sum p_1^*(c_j)\otimes p_2^* (d_j) \otimes (e_j)$ and $\sum (a_i) \otimes \mu^*(b_i)= \sum (c_j)\otimes p_1^* (d_j) \otimes p_2^*(e_j)$. Any hints? Commented Apr 21, 2019 at 15:35
• You also have that two maps which are homotopic induce the same map on homology. This is one of the ES axioms. Hence $(1\times\mu)^*\mu^*=(\mu(1\times\mu))^*=(\mu(\mu\times 1))^*=(\mu\times1)^*\mu^*$. This then gives you the constants $c_j,d_j$. Commented Apr 21, 2019 at 15:40
• Do you mean that I have to take the image of $a$ under $(1\times\mu)^*\mu^*$ split it twise using inverse of Eilenberg-Zilber map, call the resulting object $\sum (c_j)\otimes (d_j) \otimes (e_j)$. Then we say that doing the same by using$(\mu\times1)^*(\mu)$ will not matter because of the associativity of $\mu$. This constitutes to the commutativity of the diagram. Meanwhile, we used several times the naturality of the Eilenberg-Zilber. Commented Apr 21, 2019 at 16:14

Let $$X$$ be a homotopy associative $$H$$-space with multiplication $$\mu$$. All homology and cohomology is taken with coefficients in a fixed field. I was unsure in the comments if it was the associativity of the homology cross product that you were unsure of, or something else. I suggested approaching the dual problem of coassociativity of the coproduct in cohomology since I know that Hatcher proves that cup products are associtive (and these are the same as the cohomology cross product). Anyway, I'll use homology in this post, and if I've misunderstood the content of your question please let me know.

I'll write $$m$$ for the Pontryagin product, which is the composition $$m:H_*X\otimes H_*X\xrightarrow\times H_*(X\times X)\xrightarrow{\mu_*}H_*X$$ where $$\times$$ is the homology cross product and $$\mu:X\times X$$ is the homotopy associative H-space multiplication. Then directly from the definitions we have a commutative diagram of abelian groups and homomorphisms $$\require{AMScd}$$ $$\begin{CD} H_*X\otimes H_*X\otimes H_*X@>1\otimes m>> H_*X\otimes H_*X@>m>> H_*X\\ @V\cong V 1\otimes\times V @V= V V&@V= V V\\ H_*X\otimes H_*(X\times X) @>1\otimes\mu_*>> H_*X\otimes H_*X@>m>>H_*X\\ @V\cong V\times V @V\cong V\times V&@V= V V\\ H_*(X\times X\times X) @>(1\times\mu)_*>> H_*(X\times X)@>\mu_*>>H_*X. \end{CD}$$ The only place that commuativity is not obvious is the bottom-left square, and here it follows from the naturality of the homology cross product, a statement of which can be found as Lemma 3B.2 (pg. 270).

Now we know that homotopic maps induce the same maps on homology, and as we have assumed the existence of an associating homotopy

$$\mu\circ(1\times\mu)\simeq\mu\circ(\mu\times 1)$$

we can use functorality of the induced homomorphisms to get

$$\mu_*(1\times\mu)_*=(\mu\circ(1\times\mu))_*=(\mu\circ(\mu\times1))_*=\mu_*(\mu\times1)_*$$

for the maps on the bottom row of our diagram.

Therefore if we return to our first diagram and change the bracketing to get a second commuative diagram $$\require{AMScd}$$ $$\begin{CD} H_*X\otimes H_*X\otimes H_*X@>m\times 1>> H_*X\otimes H_*X@>m>> H_*X\\ @V\cong V \times\otimes 1 V @V= V V&@V= V V\\ H_*(X\times X)\otimes H_*X @>\mu_*\otimes 1>> H_*X\otimes H_*X@>m>>H_*X\\ @V\cong V\times V @V\cong V\times V&@V= V V\\ H_*(X\times X\times X) @>(\mu\times 1)_*>> H_*(X\times X)@>\mu_*>>H_*X \end{CD}$$ then we find that we can paste this new diagram and our first diagram together along their bottom rows using the commutative square $$\require{AMScd}$$ $$\begin{CD} H_*(X\times X\times X)@>(1\times\mu)_*>> H_*(X\times X)\\ @VV(\mu\times_1)_* V @VV \mu_* V\\ H_*(X\times X) @>\mu_*>> H_*(X). \end{CD}$$

At this stage we're flying so far above my AMScd skills that I'm not going to attempt any further diagrams, but hopefully I've done enough to convince you that the result is true.

• Thank you, everything is perfectly clear now! There might be a typo at the second objects in the bottom rows of the first two diagrams. $H_*(X\times X)$ instead of $H_*(X)\otimes H_*(X)$ Commented Apr 22, 2019 at 20:32
• Well spotted, @Mihail! I've fixed them now. Commented Apr 23, 2019 at 10:06