I am reading Tom Dieck's page 537 and I am not sure what the vertical map that I put in the title is in the diagram in the bottom of the page. This map is labeled Thom Isomorphism. Here $MSO(k)$ is the thom space of the tautological bundle over $BSO(n)$. There is a thom isomorphism from $H_{j-1}(MSO(k))\to H_{j-k-1}(BSO(k))$ giving by capping with the $k$ dimensional cohomology class of $H^k(MSO(K))$.
The suspension isomorphism on reduced homology is from $H_j(MSO(k)) \to H_{j-k}(\Sigma BSO(k))$. I am skeptical that $H_{j-k}(\Sigma BSO(k))=H_{j-k}(BSO(k))$. For example $\Sigma BSO(2)=\Sigma BS^1=\Sigma K(\mathbb{Z},2)=\Sigma \Omega K(\mathbb{Z},3)$ which admits an evaluation map that admits isomorphisms on low dimensional homotopy groups , to $K(\mathbb{Z},3)=$ which doesn't have the same low dimensional homoptopy groups as $BSO(2)=K(\mathbb{Z},2)$
Another random idea is I know that the thom isomorphism commutes with the boundary map and hence suspension. Do you have any ideas on finding this map?
edit: I put the wrong map in the title and consequentially my post made no sense. It should be readable now.
It is just the map from $H_j(\Sigma MSO(k))=H_j(M(SO(k) \oplus SO(1))) \to H_{j-(k+1)}(B(SO(k) \oplus SO(1)))=H_{j-(k+1)}(B(SO(k)))$
But I don't know why the dimensions are not matching, for the degree of the thom isomorphism.
I personally think that tom tieck got his indices mixed up and that his book has a typo.