Let $X$ be a CW complex satisfying $\pi_0(X) = \pi_1(X) = 0$, $H_2(X) \cong \mathbb Z^2$, and $H_j(X) = 0$ for $j \ne 2$.
I am trying to prove that $X$ is homotopic equivalent to $S^2 \vee S^2$.
By Hurewicz's theorem, $\pi_2(X) \cong \mathbb Z^2$. Let $[f], [g] \in \pi_2(X)$ be two generators of $\pi_2(X)$ where $f, g : S^2 \rightarrow X$.
Define $F : S^2\vee S^2 \rightarrow X$ by mapping one of the sphere to $X$ by $f$, and the other sphere by $g$.
The proof is done by Whitehead's theorem if I can show that $F_* : H_2(S^2\vee S^2) \rightarrow H_2(X)$ is an isomorphism, which is the place I am stuck.
I would appreciate any help, hint or reference.