This itself does not produce a solution of (1), but an additional hypothesis such as the Palais-Smale condition does provide such a solution.
We have to show that $M$ itself is an algebra. [Or: $M$ is itself an algebra.]
But $H$ itself can equally well be a member of $S$.
Now $f$ is independent of the choice of $\gamma$ (although the integral itself is not).
Thus in $G$ itself there can be no such equivalences with $f(r)=2$.
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