the goal of this exercise is (after getting inspired by Hatcher's proof in the even dimensional case (prop 3.22)) to compute
$$H^*(J(S^n), \mathbb{Z}), n=2k+1$$
And we are told that this should be $\bigwedge[\alpha]\otimes \Gamma[\beta]$ (all of these over $\mathbb{Z}$) with $|\alpha|=n, |\beta|=2n$.
I know this can be done material from chapter 3.C, computing instead the ponytragin product, and then dualising. I was able to use this method, I would like help with the expected direct method.
So far I have successfully gotten the abelian group structure, that is a copy of $\mathbb{Z}$ in each dimension a multiple of $n$ (Hatcher does this for all $n$ in his proof of Prop 3.22), and if we denote $\alpha$ a generator of $H^n$, then $\alpha$ generates a subalgebra of the form $\bigwedge[\alpha]$.
Let's denote $\beta$ a generator of $H^{2n}$ and $q_m: (S^n)^m\to J_m(S^n)$ the quotient map on which this proof relies on. The other very useful fact we have is that we can understand $H^i(J(S^n))$ using $H^i(J_m(S^n))$ whenever $i\leq m$ using that the inclusion induces an isomorphism.
To understand $\beta^k$ we work in $J_{2k}(S^n)$ and to understand $\alpha\beta^{k}$ we work in $J_{2k+1}(S^n)$ using that $q_m$ induces an isomorphism on $H^{nm}(J_m(S^n))$. In order to succesfully imitate the proof of Hatcher I need to understand $q^*_m(\beta)$ and this is where I am lost.
If we write $H^*((S^n)^m)=\bigwedge[x_1, ..., x_m], |x_i|=n$, my intution wants to say $q_m^*(\beta)=\sum_{(i,j)} x_ix_j$, but I think this is equal to $0$, which isn't what it should be. Hence my need for help.
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Thanks a lot for your help. Sorry if my question is a bit too rambly