 Solution

No. If a finite CW complex $Y$ had exactly one two-cell, then $C_2^{\mathrm{cell}}(Y)\cong \mathbb{Z}$. The isomorphism $H_2(Y)\cong H_2(X)\cong \mathbb{Z}$ forces $d_2=0$ and $\mathrm{im}{d}_3=0$: a subquotient of $\mathbb{Z}$ can remain infinite cyclic only in this way. Consequentlyis a subgroup of the free abelian group $C_1(Y)$ and is therefore torsion-free. This contradicts the homotopy-invariant calculation $H_1(Y)\cong H_1(X)\cong \mathbb{Z}/2$.

Solved by gpt-5.6-sol high.