Spaces homotopy dominated by $S^2 \times S^2\times S^2$ 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:

*

*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?


*Is there a space $A$ homotopy dominated by $X$ with $H_2 (A)\cong H_4 (A)\cong \mathbb{Z}$?
 A: 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.
A: You're actually right. The cohomology algebra of $X=S^2\times S^2\times S^2$ with coefficients in $\mathbb{Z}$ is $H^*(X)=\mathbb{Z}[x,y,z]/(x^2,y^2,z^2)$ with $|x|=|y|=|z|=2$. The cohomology algebra of the homotopy retract $H^*(A)$ must be a retract of $H^*(X)$, i.e. the image of an idempotent endomorphism $\Phi$ of $\mathbb{Z}[x,y,z]/(x^2,y^2,z^2)$, namely $\Phi=(fg)^*$.
It is easy to check that the only square-zero elements in $\mathbb{Z}[x,y,z]/(x^2,y^2,z^2)$ are the multiples of $x,y,z$. Since $\Phi$ is idempotent, $\Phi^2=\Phi$, we deduce that $\Phi$ maps $\{x,y,z\}$ to $\{x,y,z,0\}$. Therefore, up to isomorphism, the only retracts of $\mathbb{Z}[x,y,z]/(x^2,y^2,z^2)$ are:

*

*$\mathbb{Z}[x,y,z]/(x^2,y^2,z^2)$ itself.

*$\mathbb{Z}[x,y]/(x^2,y^2)$.

*$\mathbb{Z}[x]/(x^2)$,

*and $\mathbb{Z}$.

The space $X$ is simply connected, hence so is $A$. Moreover, both have free homology groups. Therefore, in case 1, $X\simeq A$ by the homological Whitehead theorem. In the remaining cases, $H^*(A)$ and $H_*(A)$ (they are duals) are concentrated in dimensions $\leq 4$. This proves that $A$ is homotopy equivalent to a 1-connected CW-complex of dimension $\leq 4$ with cells concentrated in degrees 2 and 4 (apart from the base point). The homotopy type of such a CW-complex is determined by the cohomology algebra (the cup-product from dimension 2 to 4 determines the attaching map). Finally, 2 is $H^*(S^2\times S^2)$, 3 is $H^*(S^2)$, and 4 is $H^*(*)$. Hence, we are done.
