Describe a map $f: S^2 \times S^2 \rightarrow S^4$ such that $f^{*}$ is an isomorphism.

Question

Here is the question:

Describe a map $f: S^2 \times S^2 \rightarrow S^4$ such that $f^{*}: \tilde{H^4}(S^4, \mathbb{Z}) \rightarrow \tilde{H^4}(S^2 \times S^2, \mathbb{Z})$ is an isomorphism. Does there exist a map $g: S^4 \rightarrow S^2 \times S^2$ such that $g^{*}: \tilde{H^4}(S^2 \times S^2, \mathbb{Z}) \rightarrow \tilde{H^4}( S^4 , \mathbb{Z})$ is an isomorphism? Explain your answer.

Could anyone help me in solving this question, please?

Answer 1

Assuming you know a bit about CW complexes, try to show that $S^2\times S^2$ has a cell decomposition with one 0-cell, two 2-cells which give $S^2\vee S^2$, and and 4-cell. For $S^4$ take the standard decomposition with a 0-cell and a 4-cell. Then consider the map $S^2\times S ^2\to S^4$ that collapses $S^2\vee S^2$ to the $0$-cell of $S^4$ and maps the remaining open 4-cell of $S^2\times S^2$ homeomorphically onto that of $S^4$. You should be able to study the effect of this map on the cellular complexes.

By the way, this is morally the same reasoning that Noel Lundström suggested in his comments, just without explicitly mentioning orientability.