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Related to the question, the $\Omega_4^{Pin^+}(pt)=\mathbb{Z}_{16}$. However, the topological $Pin^+$ bordism group is $\mathbb{Z}_{8}$ rather than $\mathbb{Z}_{16}$. (This fact I don't quite understand well.)

There is a manifold $M'$ homeomorphic to the smooth generator $\mathbb{RP}^4$ but not smoothly $Pin^+$ cobordant to it.

The $M'$ has a $\mathbb{Z}_{16}$ invariant equal to $(k \mod 16)$ as opposed to $\mathbb{RP}^4$’s $(k \mod 8)$.

question: What is the manifold $M'$? That homeomorphic to the smooth generator $\mathbb{RP}^4$ but not smoothly $Pin^+$ cobordant to it?

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    What is the "topological Pin-bordism group"? – 2017-01-08
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    Check the statement here: "It is interesting to note that the topological Pin+ bordism group in 4d is Z8 rather thanZ16. There is a manifold homeomorphic to the smooth generator RP4 but not smoothly Pin+ cobordant to it which has a Z16 invariant equal to 9 as opposed to RP4’s 1 (these numbers are equal mod 8)." – 2017-01-08
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    See page 10 of [this article download PDF on the right](http://link.springer.com/article/10.1007/JHEP12(2015)052) – 2017-01-08

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