I want to show that $S^n \vee S^m$ is not a retract of $S^n \times S^m$, using the cup product. I am trying to do this as a special case of a more general theorem where $X \vee Y$ are arbitrary topological spaces with some nontrivial positive integral cohomology.
I want to prove this in the following fashion. Suppose such a retract, $r$, exists. Let $\alpha, \beta \in H^*(X \vee Y)$ be generators of degree $m,n$ respectively. Then $\alpha \cup \beta = 0$, but $r^*(\alpha \cup \beta) = r^*(\alpha) \cup r^*(\beta)$ should not, forcing a contradiction.
By Kunneth, we have that $H^*(X \times Y) \cong H^*(X) \otimes H^*(Y)$ by the cross product map. Assume that $r^*(\alpha) = a \times 1$ and $r^*(\beta) = 1 \times b$ for some $a \in H^*(X)$ and $b \in H^*(Y)$. Then because $r^*$ must be injective I know $a,b \neq 0$, and I know the product of these elements would be $a \times b \neq 0$, which would complete the proof.
Here is my question, how do I know that $r^*(\alpha) = a \times 1$, for some $a \in H^*(X)$. In words, how do I know that $r^*$ sends $X$-stuff to $X$-stuff. In this special case if $m \neq n$ this is clear by a degree argument, but what if $m = n$. How can I show that $r^*(\alpha) \neq (a \times 1) + (1 \times b)$ for some nontrivial $b$. I look forward to your responses!
