Retractions, homology and multiplication

Question

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 $[a_{n}]\in H_{n}(A)$

Answer 1

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.