In the text book of Weinberg, there is a proof to show that path integral is independent of gauge fixing functional $f_a[\phi; x]$. $\phi_\Lambda$ is the result of gauge transformation on $\phi$ by an arbitrary gauge $\Lambda^a(x)$, then \begin{equation} I = \int \left [\prod_{n,x} d\phi_{\Lambda,n}(x)\right] G[\phi_\Lambda]B[f[\phi_\Lambda]]\ \mathrm{Det}\, F[\phi_\Lambda]\tag1 \end{equation} After a few steps we encounter the matrix, whose determinant we're interested in.
\begin{equation} F_{xa,yb}[\phi_\Lambda] = \left. \frac{\delta f_a[\phi_\Lambda;x]}{\delta \lambda^b(y)} \right\rvert_{\lambda=0} \tag2 \end{equation}
Since gauge transformations form a group, gauge transformation with $\Lambda^a(x)$ followed by $\lambda^a(x)$ is a single transformation $\tilde \Lambda^a(x)$, i.e $(\phi_\Lambda)_\lambda = \phi_{\tilde \Lambda(\Lambda, \lambda)}$. Using chain rule,
\begin{equation} F_{xa,yb}[\phi_\Lambda] = \int J_{xa, zc} [\phi, \Lambda] R^{zc}_{yb}[\Lambda] d^4z\tag3 \end{equation} where \begin{align} J_{xa, zc} [\phi, \Lambda] &\equiv \left. \frac{\delta f_a[\phi_{\tilde \Lambda};x ] }{\delta \tilde \Lambda^c(z)} \right \rvert_{\tilde \Lambda=\Lambda} = \left. \frac{\delta f_a[\phi_\Lambda;x]}{\delta \lambda^c(z)} \right\rvert_{\lambda=0} \tag4\\ R^{zc}_{yb} &\equiv \left. \frac{\delta \tilde \Lambda_c[z; \Lambda , \lambda ]}{\delta \lambda^b(y)} \right\rvert_{\lambda=0}\tag5 \end{align}
Now I don't understand how (6) follows from (3)? What happens to the integral on spacetime coordinates when we take determinants on both sides? If I understand correctly, determinant is only on gauge group indices.
$$\mathrm{Det}\, F[\phi_\Lambda] = \mathrm{Det}\, J[\phi, \Lambda]\ \mathrm{Det}\, R[\Lambda] \tag6$$
