$\mathbf {The \ Problem \ is}:$ Let $X$ be a simply connected CW-space with base point as a single $0$-cell . Let $C$ be the reduced cellular chain complex of $X$ (quotient of cellular chain complex of $X$ by the cellular chain complex of the base point). Let $D$ be another chain complex of free abelian groups where $D_k= 0$ for $k < 2.$
Show if $D$ is chain homotopy equivalent to $C$, then there exists a based CW-space $Y$ such that $X$ is homotopy equivalent to $Y$ and the reduced cellular chain complex of $Y$ is isomorphic to $D.$
$\mathbf {My \ approach}:$ This type of problem is quite new for me. $X$ is simply connected hence by cellular approximation, $X$ has no $1$-cell $\implies H_1(X)=0.$
Here, $C_k(X)=H_k(X^k,X^{k-1})$ thus $C_0(X)=\mathbb Z$ and $C_1(X)=0.$
Let $f,g : C\to D$ be chain homotopy equivalent maps. So there are morphisms $\Psi : D_n\to D_{n+1}, \Phi : C_n\to C_{n+1}$ such that :
$f_ng_n-id_{D_n}=\partial^D_{n+1}\Psi_n+\Psi_{n-1}\partial^D_n$ and analogously
$g_nf_n-id_{C_n}=\partial^C_{n+1}\Phi_n+\Phi_{n-1}\partial^C_n$ for all $n.$
And $f_ig_i=id_{D_i} ; g_if_i=id_{C_i}$ for $i=0,1.$
But, then how to construct such a space $Y$ ? I can't use any of there known theorems for homotopy theory . Thanks in advance for a modest hint.