Cup Multiplication in H' spaces is trivial In the book Homotopic Topology by Fomenko, Fuchs and Gutenmacher, an H' space is defind to be a topological space $Y$, equipped by two functions: 
$\mu:Y\to Y\lor Y$ which is called the comultiplication, and $\nu:Y\to Y$ which is called coinverse, such that:
 -  $(Id_Y\lor\mu) \circ \mu : Y\to Y\lor Y\lor Y$ is homotopic to $(\mu \lor Id_Y)\circ \mu :Y\to Y\lor Y\lor Y$ (coassociativity)
 - $pr_1\circ \mu$, and $pr_2\circ \mu$ are homotopic to $Id_Y:Y\to Y$ where $pr_1,pr_2:Y\lor Y\to Y$ are the projections.
 - $(Id_Y\lor \nu) \circ \mu :Y\to Y$ is homotopic to the constant mapping.  
This is an H' space. Now, the question is to prove that cup product on cohomology groups of H' space is trivial: I.e. the cup product of every elements in the graded cohomology of an H space is $0$.
I don't know how to use the comultiplication and coinverse wisely in that context, can you please help?
Thank you very much.
 A: Hint:
1) Check that the diagonal map $\Delta:Y\rightarrow Y\times Y$ is homotopic to a map that factors through through the inclusion $i:Y\vee Y\rightarrow Y\times Y$. 
Note that $i(y)=(pr_1(y),pr_2(y)$. Hence $i\mu(y)=(pr_1(\mu(y)),pr_2(\mu(y))$. But the assumptions then give that $i\mu$ is homotopic to $\Delta$ where $\Delta(y)=(y,y)$. 
2) What does the inclusion $i$ do on cohomology?
Let $\pi_1,\pi_2:Y\times Y\rightarrow Y$ be the projections. Then $\pi_1i=pr_1$ and $\pi_2i=pr_2$. 
By the K\"unneth formula all cohomology classes of $Y\times Y$ are given as combinations of $\pi_1^*(\alpha)\cup \pi_2^*(\beta)$ for classes $\alpha,\beta$ on $Y$. But then $i^*(\pi^*_1(\alpha)\cup \pi_2^*(\beta))=pr_1^*(\alpha)\cup pr_2^*(\beta)$.
It is a standard calculation that $H*(Y\vee Y)\cong H^*(Y)\oplus H^*(Y)$ where the cup product vanishes between cohomology classes of both terms, i.e.~$pr_1(\alpha)\cup pr_2(\beta)=0$ if $\alpha$ and $\beta$ both have non-zero degree. 
3) Recall how $\Delta$ is used in the definition of the cup product. 
I think you can manage this. 
