Let $X$ be CW-complex and $\pi_1(X) = 0$. I want to prove that there is $Y \simeq X$ with only one $0$-cell and without $1$-cells.
Geometrically it's obvious, but I have no idea for rigorous proof. I know result about evaluating $\pi_1$ for CW-complexes: $\pi_1(X) \simeq \langle c_1,\dots,c_k \mid \phi_1\dots\phi_r\rangle$ ($c_i$ is $1$-cells, $\phi_i$ is pasting $2$-cells).
Let $Y \to X$ be the CW approximation of $X$. By inspecting the construction of $Y$ (given in proposition 4.13 in Hatcher's Algebraic Topology), we see that $Y$ has one $0$-cell and no $1$-cells. Since $X$ is a CW complex, Whitehead's theorem shows that $X$ and $Y$ are homotopy equivalent.