Not null homotopic map from $S^3$ to $S^2 \vee S^2$ I'm asked to present a continuous function $\alpha \colon S^3 \rightarrow S^2 \vee S^2$ s.t. it is not null homotopic but taken both projections $pr \colon S^2 \vee S^2 \rightarrow S^2$ the composition $pr \circ \alpha$ is null homotopic.
I'm given the hint to use tha fact that $S^2 \times S^2$ is not homotopic equivalent to $S^2 \vee S^2 \vee S^4$ (they cohomology rings are not isomorphic).
Any help or advice is well accepted: thanks in advance.
 A: Consider the attaching map $S^3 \to S^2 \vee S^2$ of the $4$-cell in the standard CW-structure of $S^2 \times S^2$. If this is nullhmotopic, $S^2 \times S^2$ would be homotopic to $S^2 \vee S^2 \vee S^4$ as homotopic attaching map implies homotopy equivalent spaces.
But cup square $\alpha \smile \alpha$ is nontrivial in $H^*(S^2 \times S^2)$ where $\alpha$ is generator of $H^2(S^2)$, whereas it's trivial in $H^*(S^2 \vee S^2 \vee S^4) \cong H^*(S^2) \times H^*(S^2) \times H^*(S^4)$, contradiction. 
A: From the LES in homotopy groups for the pair $(S^2 \times S^2, S^2 \vee S^2)$, we have an exact sequence
$$\pi_4(S^2 \vee S^2) \to \pi_4(S^2 \times S^2) \to \pi_4(S^2 \times S^2, S^2 \vee S^2)\to \pi_3(S^2 \vee S^2) \to \pi_3(S^2 \times S^2)$$
Since we have $\pi_k(S^2 \times S^2)\cong \pi_k(S^2 ) \times \pi_k( S^2) \subset \pi_k(S^2 \vee S^2)$, the maps $\pi_k(S^2 \vee S^2) \to \pi_k(S^2 \times S^2)$ are surjective. It follows that $\pi_4(S^2 \times S^2)\to \pi_4(S^2 \times S^2, S^2 \vee S^2)$ is the zero map, giving us a short exact sequence 
$$0  \to \pi_4(S^2 \times S^2, S^2 \vee S^2)\to \pi_3(S^2 \vee S^2) \to \pi_3(S^2 \times S^2) \to 0.$$
We want an element $\alpha \in \pi_3(S^2 \vee S^2)$ such that its composition with each projection $S^2 \vee S^2 \to S^2$ is nullhomotopic. These projections factor through $S^2 \times S^2$, so $\alpha$ lies in the kernel of the composition $\pi_3(S^2 \vee S^2) \to \pi_3(S^2 \times S^2) \overset{\sim}{\to} \pi_3(S^2) \times \pi_3(S^2)$. Thus it suffices to show that $\pi_4(S^2 \times S^2, S^2 \vee S^2)$ is nontrivial, since any nonzero element in its image in $\pi_3(S^2 \vee S^2)$ will work for us. 
Here's where the hint comes in: Recall that $S^2 \times S^2$ has a cell structure with one $0$-cell, two $2$-cells, and one $4$-cell. The 3-skeleton is simply $S^2 \vee S^2$, so the boundary of the 4-cell is mapped into $S^2 \vee S^2$. Since $S^2 \vee S^2 \vee S^4$ is not homotopy equivalent to $S^2 \times S^2$, the 4-cell's attaching map is nontrivial, giving us the desired element of $\pi_3(S^2 \vee S^2)$.
