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Let $G$ be a compact group and $H$ be a closed subgroup of $G$. I know that if $G\rightarrow G/H$ is a principal $H$-bundle, then we can choose $E_{H}$ as $% E_{G}$, where $E_{G}$ is the Milnor space of $G$. I know that it is not always possible to choose $E_{H}$ as $E_{G}$.

What I'm wondering is that: the inclusion $E_{H}\subset E_{G}$ induces cohomology monomorphism $H^{\ast }\left( E_{G}/H;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( E_{H}/H;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) =H^{\ast }\left( B_{H};% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) $?

Here $E_{G}/G$ denotes the orbit space of the action $G$ on the contractible space $E_{G}$.

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  • $\begingroup$ At least for Lie groups $EG/H$ is a $BH$: it is the quotient space of a contractible free right $H$-space. So the map you mention is an isomorphism, even with integral coefficients. Did you mean the map $H^*(BG,\mathbb{Q})\to H^*(BH,\mathbb{Q})$ instead? $\endgroup$
    – algori
    Commented Jul 28 at 20:35
  • $\begingroup$ @algori For Lie groups and for groups of finite-dimensional, it is clear. I am interested in the most general situation. $\endgroup$
    – Mehmet Onat
    Commented Jul 28 at 21:18

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