# A question on relative equivariant cohomology

Suppose that we have defined an extraordinary $$G$$-equivariant cohomology theory $$H$$ (say $$G$$ is a compact group). If $$X$$ is a $$G$$-space and $$A\subset X$$ is a closed $$G$$-equivarant contractible subspace, is it true that the induced map $$H(X,A)\rightarrow H(X)$$ is injective?

• I am not familiar with the term "extraordinary cohomology theory". For things I know as $G$-homology theories the easiest example is $H_G^*(X,A):=H^*(X/G,A/G)$. If we now take $G=Z/2$ and $X$ to be the mapping cone of the $G$-map $EG\rightarrow pt$ and $A$ the G-fixed point, then $H_G^*(X)$ is $H^*(pt)$, as $X$ is $G$-equivariantly contractible. By excision, the cohomology of the pair is the same as the cohomology of $X\setminus A$ which is $G$-homotopy equivalent to $EG$ and thus $H^*(X,A)=H^*(EG/G)=H^*(BZ/2)$ and so the map above cannot be injective. Aug 1, 2019 at 17:36
• The remaining question is whether my $H^*$ is extraordinary... Aug 1, 2019 at 17:38
• "extraordinary" here presumably means "generalized". I.e., including examples like equivariant k theory Aug 2, 2019 at 14:45

The answer is yes. Because $$A$$ is equivariantly contractible, the composite of the maps $$A\to X \to\ast$$ is an equivariant homotopy equivalence, thus applying any cohomology theory gives $$H(\ast)\to H(X)\to H(A)$$ so that the composite is an iso, so $$H(A)$$ is a summand of $$H(X)$$. Your claim follows by the usual long exact sequence.