We say that a topological space $A$ is homotopy dominated by a topological space $X$ if there exist continuous maps $f:A\to X$ and $g:X\to A$ such that $g\circ f\simeq 1\_A$.


Let $X$ be $S^2 \times S^2 \times S^2$. I'm trying to show by the Whitehead Theorem that if $A$ is homotopy dominated by $X$, then $A$ is homotopy equivalent by one of $\*$, $S^2$, $S^2 \times S^2$ or $X$. We know that $H\_2 (X)\cong X\_4 (X)\cong \mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}$ and $H\_0 (X)\cong H\_8 (X)\cong \mathbb{Z}$. Since $A$ is homotopy dominated by $X$, then $H\_p (A)$ is a direct summand of $H\_p (X)$ for all $p\geq 0$. Obviously, $\*$, $S^2$, $S^2 \times S^2$ and $X$ are retracts of $X$ so they are homotopy domintaed by $X$. Now I have two questions:


1. Let $H\_2 (A)\cong \mathbb{Z}\oplus \mathbb{Z}$ and $H\_0 (A)\cong H\_4 (A)\cong \mathbb{Z}$. Consider the map $\pi \circ f:A\to X\to S^2 \times S^2$, where $\pi :S^2 \times S^2 \times S^2$ is the projection map. I wanted to show that $H\_2 (\pi \circ f)$ is epimorphism, but I couldn't. Can $H\_2 (\pi \circ f)$ be an epimorphism generally?
2. Is there a space $A$ homotopy dominated by $X$ with $H\_2 (A)\cong H\_4 (A)\cong \mathbb{Z}$?



Spaces homotopy dominated by $S^2 \times S^2\times S^2$
Put
$$ R(n)=H^\*((S^2)^{\times n}) = \mathbb{Z}[x\_1,\dotsc,x\_n]/(x\_1^2,\dotsc,x\_n^2) $$
A key point is that if $u\in R(n)$ with $|u|=2$ then $u^2=0$ iff $u=0$ or $u=m\,x\_i$ for some $m\in\mathbb{Z}\setminus\{0\}$ and some $i$; this is easy to check.


Let $\phi\colon R(n)\to R(n)$ be a map of graded rings with $\phi^2=\phi$. Then $\phi(x\_i)^2=\phi(x\_i^2)=0$ so $\phi(x\_i)=0$ or $\phi(x\_i)=m\,x\_j$ for some $m\neq 0$ and $j$. Using $\phi^2=\phi$ we then get $m(x\_j-\phi(x\_j))=0$, so $\phi(x\_j)=x\_j$. Using this, we see that there is a subset $J\subseteq\{1,\dotsc,n\}$ such that $\phi(x\_j)=x\_j$ for all $j\in J$, and for $i\not\in J$ we have $\phi(x\_i)=0$ or $\phi(x\_i)=m\,x\_j$ for some $j\in J$ and $m\in\mathbb{Z}\setminus\{0\}$. It follows that $\text{img}(\phi)$ is the subring generated by $\{x\_j:j\in J\}$, and this is isomorphic to $R(m)$, where $m=|J|$. More precisely, there is an evident inclusion $i\colon(S^2)^{\times m}\to (S^2)^{\times n}$ corresponding to $J$, and the resulting map $R(n)\to R(m)$ restricts to give an isomorphism $\text{img}(\phi)\to R(m)$.


Now suppose we have maps $A\xrightarrow{f}(S^2)^{\times n}\xrightarrow{g}A$ with $g\circ f\approx 1$. Then the map $\phi=(f\circ g)^\*$ is an idempotent ring endomorphism of $R(n)$, and $g^\*$ induces an isomorphism from $H^\*(A)$ to $\text{img}(\phi)$. It follows that there is an inclusion $i\colon(S^2)^{\times m}\to(S^2)^{\times n}$ such that the composite $g\circ i$ gives an isomorphism $H^\*(A)\to H^\*((S^2)^{\times m})$. As $A$ is a homotopy retract of $(S^2)^{\times n}$ it is also simply connected and of finite type, so we conclude that $g\circ i\colon (S^2)^{\times m}\to A$ is an equivalence.


In particular, we see from this that we cannot have $H^2(A)\simeq H^4(A)\simeq\mathbb{Z}$. Here is a more direct argument for this particular point. In $R(n)$, the product map $R(n)^2\otimes R(n)^2\to R(n)^4$ is surjective, but the squaring map $R(n)^2\to R(n)^4$ is zero mod $2$. If $A$ is a homotopy retract of $(S^2)^{\times n}$, it follows that the product map $H^2(A)\otimes H^2(A)\to H^4(A)$ is surjective, but the squaring map $H^2(A)\to H^4(A)$ is zero mod $2$. This is inconsistent with $H^2(A)\simeq H^4(A)\simeq\mathbb{Z}$.


To push the analysis a bit further, I claim that every ring map $\alpha\colon R(n)\to R(m)$ arises from a continuous map $f\colon (S^2)^{\times m}\to (S^2)^{\times n}$. Indeed, for each $i\leq n$ we must have $\alpha(x\_i)=0$ or $\alpha(x\_i)=\mu(i)x\_{\sigma(i)}$ for some $\mu(i)\neq 0$ and $j\leq m$. If $\alpha(x\_i)=0$ then we define $f\_i\colon(S^2)^{\times m}\to S^2$ to be the constant map to the basepoint. If $\alpha(x\_i)=\mu(i)x\_{\sigma(i)}$ then we define $f\_i$ to be the composite of the projection $\pi\_{\sigma(i)}\colon(S^2)^{\times m}\to S^2$ and a map $S^2\to S^2$ of degree $\mu(i)$. The maps $f\_i$ can be combined to give a map $f\colon(S^2)^{\times m}\to(S^2)^{\times n}$, and this has $f^\*=\alpha$ as required.


Next, recall that there is a fibration sequence
$$ S^1 =K(\mathbb{Z},1) \xrightarrow{} S^3 \xrightarrow{\eta} S^2 \xrightarrow{} BS^1=\mathbb{C}P^\infty = K(\mathbb{Z},2) \to BS^3=\mathbb{H}P^\infty $$
This gives exact sequences
$$ H^1(X) \to [X,S^3] \to [X,S^2] \to H^2(X) \to [X,\mathbb{H}P^\infty]. $$
Here the first two terms are groups but the last three are only pointed sets. The group $[X,S^3]$ acts on $[X,S^2]$ with stabilisers given by the image of $H^1(X)$ in $[X,S^3]$, and the orbit set maps injectively to $H^2(X)$. Taking $X=(S^2)^{\times m}$ and using $[X,(S^2)^{\times n}]=[X,S^2]^{\times n}$ and noting that $H^1(X)=0$ we arrive at the following conclusion: the group $G(m,n)=[(S^2)^{\times m},(S^3)^{\times n}]$ acts freely on the set $P(m,n)=[(S^2)^{\times m},(S^2)^{\times n}]$, and two maps from $(S^2)^{\times m}$ to $(S^2)^{\times n}$ have the same effect in cohomology iff they lie in the same orbit of this action.


Finally, we note that $S^3=\Omega\mathbb{H}P^\infty$ so $[X,S^3]\simeq[\Sigma X,\mathbb{H}P^\infty]$. Using the standard splitting $\Sigma(X\times Y)\simeq\Sigma X\vee\Sigma Y\vee\Sigma(X\wedge Y)$, we get
$$ [X\times Y,S^3] \simeq [X,S^3] \times [Y,S^3]\times [X\wedge Y,S^3]. $$
Using this repeatedly, we obtain a bijection
$$ G(m,n) \simeq \prod\_{i=1}^n [(S^2)^{\times m},S^3] \simeq
 \prod\_{J\subseteq\{1,\dotsc,m\}}\prod\_{i=1}^n[S^{2|J|},S^3].
$$
However, this is not obviously an isomorphism of groups.


