(Co)bordism invariant of Eilenberg–MacLane space becomes vanished Consider a (co)bordism invariant 
$$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$$
obtained from
$$
\Omega^5_{O}(K(\mathbb{Z}/2,2)).
$$
Here $u \in H^2(K(\mathbb{Z}/2,2),\mathbb{Z}_2)$. The $K(\mathbb{Z}/2,2)$ is Eilenberg–MacLane space. The $\mathbb{Z}/2$ is the finite group of order 2.

Question: Is it true that such a (co)bordism invariant $u_2 Sq^1 u_2+Sq^2 Sq^1 u_2$ when being pulled back from the $\Omega^5_{O}(K(\mathbb{Z}/2,2))$ to a pulldback new (co)bordism group 
  $$
\Omega^5_{\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}
$$ 
  (thus we are allowed to identify more wider classes of manifolds or more (co)bordism invariants by enlarging the cobordant structures of manifolds from
$\Omega^5_{O}(K(\mathbb{Z}/2,2))$ to $\Omega^5_{\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}$), 
the (co)bordism invariant 
  $ 
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$ becomes 0 in $\Omega^5_{\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}$? 
Namely the effective manifold generators (respect of to $ 
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$) in $\Omega^5_{O}(K(\mathbb{Z}/2,2))$ becomes trivial or vanished in $\Omega^5_{\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}$?
How to prove this?



*

*Here ${\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}$ is similar to the generalization of Pin$^c$ structure where $U(1)$ is now replaced by $SU(2)$.

 A: $\newcommand{\Pin}{\mathrm{Pin}}\newcommand{\Sq}{\mathrm{Sq}}$
To simplify notation, I'll denote $(\Pin_n^\pm\times\mathrm{SU}_2)/(\mathbb Z/2)$ by $\Pin_n^{h\pm}$, since these are
analogues of the groups $\mathrm{Spin}_n^h = (\mathrm{Spin}_n\times\mathrm{SU}_2)/(\mathbb Z/2)$. Analogously to how a
spin$c$-structure on a manifold $M$ determines a line bundle $L$ with $w_2(L) = w_2(M)$, a
pin$h\pm$-structure on $M$ determines a principal $\mathrm{SO}_3$-bundle $Q$ with $w_2(Q) = w_2(M)$ (for
pin$h+$) or $w_2(Q) = w_2(M) + w_1^2(M)$ (for pin$h-$).
This question is actually several questions in one: the argument differs for $\Pin^{h+}$ and $\Pin^{h-}$, and it's also necessary
to specify the maps $\Omega_5^{\Pin^{h\pm}}\to\Omega_5^O(K(\mathbb Z/2, 2))$. Any characteristic class of $M$ or
$Q$ in $H^2(M;\mathbb Z/2)$ determines a natural map to $K(\mathbb Z/2,2)$, and hence a map between these bordism
groups. There are several examples, such as $w_2(Q)$ and $w_2(Q) + w_1^2(M)$, and the answer could in principle differ depending
on which class one chooses.
As such, I have a partial answer: in the case of pin$h+$ 5-manifolds where the map to $K(\mathbb Z/2,2)$
is $w_2(Q)$ or $w_2(Q) + w_1^2(M)$, the answer is yes, the invariant vanishes.
Since $u_2\mathrm{Sq}^1 u_2 + \mathrm{Sq}^2\mathrm{Sq}^1u_2$ is a cobordism invariant, it suffices to check this on
a generator of $\Omega_5^{\Pin^{h+}}$. This group is isomorphic to $\mathbb Z/2$ (this is due to
Freed-Hopkins, Theorem 9.89), so we need only to find a single nonbounding
pin$h+$ 5-manifold and check there.
This generator is the Wu manifold $W := \mathrm{SU}_3/\mathrm{SO}_3$, with the principal $\mathrm{SO}_3$-bundle $Q$
given by the quotient map $\mathrm{SU}_3\twoheadrightarrow W$. It's known that $H^*(W;\mathbb Z/2) =
\mathbb Z/2[z_2,z_3]/(z_2^2, z_3^2)$ with $|z_i| = i$; its nonzero Stiefel-Whitney classes are $w_2(W) = z_2$ and $w_3(W)
= z_3$, and that $w_2(Q) = z_2$ (this is discussed, for example, at the end of section 6 of Xuan Chen's
thesis). In particular, $w_2(W) = w_2(Q)$, so $W$
admits a pin$h+$-structure whose corresponding $\mathrm{SO}_3$-bundle is $Q$, and because $\langle
w_2(Q)w_3(W), [W]\rangle = 1$, it doesn't bound as a pin$h+$ manifold. Therefore it generates
$\Omega_5^{\Pin^{h+}}$.
Using the Wu formula, $\Sq^1z_2 = z_3$ and $\Sq^2z_3 = z_2z_3$, so
$$ w_2(Q)\Sq^1w_2(Q) + \Sq^2\Sq^1w_2(Q) = 2z_2z_3 = 0,$$
so we conclude that this invariant vanishes for all pin$h+$ 5-manifolds. (Since $w_1(W)^2 = 0$, we can
also conclude this for the map to $K(\mathbb Z/2,2)$ given by $w_2(Q) + w_1(M)^2$.)

An analogous argument should be possible in the pin$h-$ case, but since
$\Omega_5^{\Pin^{h-}}\cong\mathbb Z/2\oplus\mathbb Z/2$ (this is again Freed-Hopkins, Theorem 9.89), one would have
to find two linearly independent generators and check on them. $W$ also admits a pin$h-$-structure with
corresponding bundle $Q$, but I wasn't able to figure out what the other generator is.
