Cohomologous two-cocycles give equivalent extensions

Suppose $f^{\mathrm{\prime }}=f+\delta u$. Define $\mathrm{\Phi }:M{\times }_{f}G\to M{\times }_{f^{\mathrm{\prime }}}G$ by $\mathrm{\Phi }(m,g)=(m-u(g),g)$. Using the product formulas from [F1], the identity $f^{\mathrm{\prime }}=f+\delta u$, and [L1], one checks directly that $\mathrm{\Phi }((m,g)(n,h))=\mathrm{\Phi }(m,g)\mathrm{\Phi }(n,h)\mathrm{.}$ The inverse is $(m,g)\mapsto (m+u(g),g)$, so $\mathrm{\Phi }$ is a group isomorphism. It fixes the kernel and quotient maps, so [F2] makes the two extensions equivalent.

Conversely, suppose $\mathrm{\Phi }:M{\times }_{f}G\to M{\times }_{f^{\mathrm{\prime }}}G$ is an extension equivalence. Because [F2] fixes quotient and kernel, $\mathrm{\Phi }$ has the form $\mathrm{\Phi }(m,g)=(m+u(g),g)$ for a unique normalized one-cochain $u$. Comparing the image of $(0,g)(0,h)$ under $\mathrm{\Phi }$ with the product of the images gives $f^{\mathrm{\prime }}(g,h)=f(g,h)-g\cdot u(h)+u(gh)-u(g)=f(g,h)+(\delta (-u))(g,h)\mathrm{.}$ So $f^{\mathrm{\prime }}-f$ is a coboundary.

Step 1.1 proves the forward implication and step 1.2 proves the reverse implication.