Domenico Fiorenza Dijkgraaf-Witten model

Let $\mathbf{c} : \mathbf{B}G \to \mathbf{B}^n U(1)$ be a cocycle and assume for simplicity that $G$ is discrere (Dijkgraaf-Witten type of case), so that for $\Sigma$ a manifold we have

etc. We had shown that $\tau_{n-dim \Sigma} Top(\Sigma, B^n U(1)_{disc}) \simeq B^{n-dim \Sigma} U(1)_{disc}$. Following Freed-Hopkins-Lurie-Teleman, what we need to show is the following:

for $\Sigma_{in} \to \Sigma \leftarrow \Sigma_{out}$ a cobordism we need to produce a diagram

Here the lower two morphisms should be the ones we already have, and the task is to show that we have the natural transformation filling this diagram.

Now, here is how this should work, and my question would be if you can see the remaining missing step needed to show that it does work:

let’s hom the entire cobordism diagram into the entire background field morphism: this yields

You see, if we could show that the bottom middle term here, the “relative cohomology term” is also equivalent to $B^{n-dim \Sigma} U(1)$, then we’d be done. More concretely, imagine the above diagram formed concretely in some 1-categorical model, then we should show that it is equivalent to the following 1-categorical diagram

where now in the bottom middle is the path object. The middle morphism into that would exhibit the right homotopy/natural transformation that we are after.

So, it seems all to point in the right direction. But can we show the crucial statement that

?

This might just follow from a slight variation of the argument for

But I am not sure yet how to do it.