How to show the isomorphism between the following Spin and the Pin- bordism group, known as the Smith isomorphism:
$$ \Omega^{Spin}_{d}(B\mathbb{Z}_2)' \to \Omega^{Pin-}_{d-1}(pt) $$
in particular, the $\Omega^{Spin}_{d}(B\mathbb{Z}_2)'$ is not exactly the the usual Spin bordism group $\Omega^{Spin}_{d}(B\mathbb{Z}_2)'$, but the reduction.
Hint: One need to first show a relation: $$ \Omega^{Spin}_{d}(BG)=\Omega^{Spin}_{d}(BG)' \oplus \Omega^{Spin}_{d}(pt) $$ where the reduction of the spin bordism group $\Omega^{Spin}_{d}(BG)$ to $\Omega^{Spin}_{d}(BG)'$ gets rid of the $\Omega^{Spin}_{d}(pt)$. This part has something to do with the kernel of the forgetful map to $\Omega^{Spin}_{d}(pt)$.
I like to understand what role does this forgetful map play, and how does the reduction help us to achieve the Smith isomorphism?