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?
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$\begingroup$ 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 Gfixed 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. $\endgroup$– HenrikRüpingAug 1, 2019 at 17:36

$\begingroup$ The remaining question is whether my $H^*$ is extraordinary... $\endgroup$– HenrikRüpingAug 1, 2019 at 17:38

2$\begingroup$ "extraordinary" here presumably means "generalized". I.e., including examples like equivariant k theory $\endgroup$– Charles RezkAug 2, 2019 at 14:45
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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.