TITLE: Not null homotopic map from $S^3$ to $S^2 \vee S^2$
QUESTION [7 upvotes]: 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.

REPLY [8 votes]: 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.