I've recently encountered the following cobordism theory modulated by a class $\sigma \in H^{d+1}(B^2\mathbb{Z}/2,U(1))$.

My objects are $d$-dimensional spin manifolds with chosen spin structure. Note that a spin structure, thought of as a null-homotopy of $w_2$, gives a trivialization of $w_2^*\sigma$.

My morphisms are given by cobordisms by manifolds with $w_2^*\sigma = 0$ for which the trivializations of $w_2^*\sigma$ on the boundary components given by the chosen spin structures there all come from restricting a single trivialization of $w_2^*\sigma$ on the cobordism.

Call the resulting group $\Omega^{{\rm spin},\sigma}_d$. There is a natural surjective map $\Omega^{\rm spin}_d \to \Omega^{{\rm spin},\sigma}_d$. I'd like to calculate the kernel of this map, at least for $d=1,2,3,4,5$.

There is another map $\Omega^{{\rm spin},\sigma}_d \to \Omega^{\rm SO}_d$ forgetting all the trivialization business. The composition $\Omega^{\rm spin}_d \to \Omega^{{\rm spin},\sigma}_d \to \Omega^{\rm SO}_d$ is known to have kernel in degrees $8k+1,8k+2$ with rather explicit generators. For $k=0$, these are the circle with antiperiodic spin structure and its Cartesian square, respectively. The kernel we're after is contained in this one, so to do the calculation in the range I'm interested in, it might be possible to do things case by case.

Let's do the circle case. Trivializations of $w_2^*\sigma$ on $S^1$ form a torsor over $H^1(S^1,U(1))$. Trivializations on a surface $\Sigma$ bounding $S^1$ are a torsor over $H^1(\Sigma,U(1))$. Since the class of $\partial \Sigma$ in $\Sigma$ is zero, by universal coefficients the map $H^1(\Sigma,U(1)) \to H^1(\partial\Sigma,U(1))$ is zero. Thus, any two trivializations of $w_2^*\sigma$ on $\Sigma$ restrict to the same trivialization on the boundary. Thus, the two spin structures of the circle are distinct in $\Omega^{{\rm spin},\sigma}_d$ when $\sigma \neq 0$ since in this case the spin structures actually give different trivializations of $w_2^*\sigma$ (as can be checked from the long exact sequence coming from $\sigma:\mathbb{Z}/2 \to U(1)$).

When $\sigma = 0$, then we get the full kernel. This is the case for $d=2$ since $H^3(B^2\mathbb{Z}/2,U(1)) = 0$, but this is the last degree where this cohomology vanishes.

This finishes the calculation in the range I originally considered, so how about the general case?

The next case of interest is $d=9$, for which $H^{10}(B^2\mathbb{Z}/2,U(1))$ is nontrivial but indescribable to me.

A first subproblem for the general attack might be ``when is a cohomology operation injective?" I don't know the answer to this either, except in the case $d=1$ when $\sigma:H^2(-,\mathbb{Z}/2) \to H^2(-,U(1))$ is given by a homomorphism $\mathbb{Z}/2 \to U(1)$ and we have a long exact sequence where the kernel of $\sigma$ is the image of $H^1(-,U(1))$.

Any help or references would be much appreciated.