For any cobordism invariant (or simply bordism invariant) quantity $\omega$ that satisfy the conditions:

$\omega$ can be fully decomposed from the cup product of

**characteristic classes**(such as Stiefel Whitney class, Chern class, etc.).$\omega$ can be defined on an

*unorientable smooth*manifold.

then question:

Could we show that this $\omega$ has to be $\mathbb{Z}/2\mathbb{Z} : = \mathbb{Z}_2$ valued (a mod 2 class)? Is this true or false?

From the recent post: Cobordism invariants: topological v.s. geometric, I recall that:

(1). The 2d $Pin^-$ bordism invariant (defined on unorientable manifolds) from $\Omega_2^{Pin^-}=\mathbb{Z}_8$, where any 2-manifold $M$ always admits a $Pin^-$ structure. The $\mathbb{Z}_8$ is a Arf-Brown-Kervaire (ABK) invariant. $Pin^-$ structures are in one-to-one correspondence with quadratic enhancements
$q: H^1(M,\mathbb{Z}_2)\to\mathbb{Z}_4$ such that $x,y \in \mathbb{Z}_2$, we have
$q(x+y)-q(x)-q(y)=2\int_M x\cup y\mod4.$ In particular,
$q(x)=\int_M x\cup x\mod2.$ The $x,y$ counts as a *cohomology class*, but the ABK invariant seems not be decomposed from decomposed from a cup product of **characteristic classes** only? *But does this count as a counter example of the above statement?*

(2) The 4d $\Omega_4^{Pin^+}(B^2 \mathbb{Z}_2)=\mathbb{Z}_4 \times \mathbb{Z}_{16}$ contains an $\eta$ invariant of $\mathbb{Z}_{16}$, and a quadratic refinement of $\mathbb{Z}_4$ (from $B \in H^2(M,\mathbb{Z}_2)$ to $H^4(M,\mathbb{Z}_4)$. (It is probably similar to Pontryagin square $P(B)$ of $B \in H^2(M,\mathbb{Z}_2)$; but Pontryagin square $P(B)$ is defined on an orientable $SO$ structure, here I am looking at the unorientable $Pin^+$ structure. *But does this count as a counter-example of the above statement?*

If the above statement is false, can one give counter examples?

Many thanks for comments/answers.