I've been going through Nakahara's Topology, Geometry and Physics from 2003 and I'm struggling to fully understand this derivation from page 410:
Say there exists two different fields, $\mathcal{F}_i = d\mathcal{A}_i + \mathcal{A}_i\land \mathcal{A}_i$ and $\mathcal{F}_j = d\mathcal{A}_j + \mathcal{A}_j\land \mathcal{A}_j$, then the compatibility function between the two is computed as
\begin{equation} \mathcal{F}_j = d(t^{-1}_{ij}\mathcal{A}_it_{ij} +t^{-1}_{ij}dt_{ij}) + (t^{-1}_{ij}\mathcal{A}_it_{ij} +t^{-1}_{ij}dt_{ij})\land (t^{-1}_{ij}\mathcal{A}_it_{ij} +t^{-1}_{ij}dt_{ij})\\ = (-t^{-1}_{ij}dt_{ij}\land t^{-1}_{ij}\mathcal{A}_it_{ij} + t^{-1}_{ij}d\mathcal{A}_it_{ij} -t^{-1}_{ij}\mathcal{A}_i\land dt_{ij} -t^{-1}_{ij}dt_{ij}t^{-1}_{ij}\land dt_{ij})+(t^{-1}_{ij}\mathcal{A}_i\land \mathcal{A}_it_{ij} +t^{-1}_{ij}\mathcal{A}_i\land dt_{ij} + t^{-1}_{ij}dt_{ij}t^{-1}_{ij}\land \mathcal{A}_it_{ij} + t^{-1}_{ij}dt_{ij}\land t^{-1}_{ij}dt_{ij})\\ = t^{-1}_{ij}(d\mathcal{A}_i + \mathcal{A}_i\land \mathcal{A}_i)t_{ij} = t^{-1}_{ij}\mathcal{F}_it_{ij} \end{equation}
where the compatibility function for gauge potentials has been substituted in and $t_{ij}$ is the transition function. My problems mostly lie with the first term as I'm not entirely sure what the exterior derivative $d$ does in this context. Any help is much appreciated.
The exterior derivative here works the same as it would on regular forms, except here the forms are valued in the endomorphism bundle. I think what might be a bit confusing is how they simplified $dt_{ij}^{-1}$. Try to to think about how you can simplify that. Hint: $t_{ij}^{-1}t_{ij}=1$.