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Let $X$ be a $G$-space, for a compact Lie group $G$. If $U\subset X$ is a $G$-invariant open subspace, is it true that the restriction map on equivariant $K$-theory $$K_G(X)\rightarrow K_G(U)$$ is always surjective?

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    $\begingroup$ Let $G$ be the trivial group, $X=\mathbb{C}$ and $U=\mathbb{C}\smallsetminus\{0\}$. Then $K(X)=\mathbb{Z}$ but $K(U)=\mathbb{Z}\oplus\mathbb{Z}/2$ and the map is just the inclusion of the first factor. $\endgroup$
    – Denis Nardin
    Aug 6, 2019 at 20:30

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