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Mar 26, 2014 at 13:49 comment added Alex Degtyarev Yes, exactly. Plus, use excision.
Mar 26, 2014 at 13:37 comment added Peter Crooks Are you are referring to the Thom class of the normal bundle $N$ of $Y_G$ in $X_G$? In this case, we have a map $\pi:H^{*-d}(Y_G)\rightarrow H^*(N,N-\text{zero-section})$. The Thom isomorphism is then $\varphi:H^{*-d}(Y_G)\rightarrow H^*(N,N-\text{zero-section})$ $$x\mapsto\pi(x)\cup T,$$ where $T$ is the Thom class. Is this generally what you are describing?
Mar 25, 2014 at 22:44 history answered Alex Degtyarev CC BY-SA 3.0