Suppose that I have 2 CW-complexes $A\subset B $ such chat thé inclusions is a retract.


Let $H\_{\ast}$ be the ordinary homology with integral coefficients.
Let
$$\mu: B\times B \rightarrow B $$
be continiuous map such that:


* the restriction map $\mu:A\times A\rightarrow A$ is well defined.
* There exists an element $b\in B$ such that $\mu(b,b')=\mu(b',b)=b'$ for any $b'\in B$
* there exists an element $a\in A$ such that $[a]=[b] $ in $H\_{0}(B)$.


**Question:** Clearly the map $\mu$ induces a map in homology $\bullet:H\_{\ast}(A)\otimes H\_{\ast}(A)\rightarrow H\_{\ast}(A)$. Does it follow that $$[a]\bullet[a\_{n}]=[a\_{n}]\bullet[a]= [a\_{n}]$$
for any $[a\_{n}]\in H\_{n}(A)$



Retractions, homology and multiplication
The role of $C$ in the problem is irrelevant.


Let $r : C \to B$ be a retraction, and set
$$m := r\circ \mu : B \times B \to B .
$$ Set $\ast := b$ and think of it as the basepoint of $B$. Then the restriction of $m$ to the wedge $B\vee\_\ast B$ is the fold map $B \vee\_\ast B\to B$ (so $B$ is an $H$-space).


Since $[a] = [b]$ in $H\_0(B)$, the restriction of $m$ to
$$
B\vee\_a B := a \times B \cup B\times a
$$
is homotopic to the the fold map on $B \vee\_a B$. Choose
a retraction $s: B\to A$. Then the composition
$$
A \vee\_a A \to B \vee\_a B \overset{m}\to B \overset{s}\to A
$$
is homotopic to the fold map. But this last composite coincides with the restriction of the map $\mu: A\times A \to A$ to $A\vee\_a A$.


The result you seek follows immediately from this by taking homology.


