Let $M$ be a manifold, $(N,g)$ be a Riemannian manifold and $f:M \to N$ be a diffeomorphism.
Does the following equallity holds? $$(f^*g^{-1}) = (f^*g)^{-1} $$ (here $g^{-1}$ - inverse metric to g, $(f^*g)^{-1}$ - inverse metric to pullback metric $f^*g$)
If it does, how to prove it? Can the following identities help? $$C(C(g^{-1}\otimes g))=dim N = dimM \\ C \circ f^* = f^*\circ C$$ (here $C$ is a contraction map for tensor fields)
Hint Since contractions depend only on the smooth structure of a manifold, they commute with pullbacks by diffeomorphisms. In particular, $$C(f^* (g^{-1} \otimes g)) = f^* C(g^{-1} \otimes g).$$ (As in the question, $C$ is the contraction map $T^{2, 0} M \otimes T^{0, 2} M \to T^{1, 1} \cong \operatorname{End} TM$.)
For a more concrete argument, for any point $p \in M$, pick local coordinates $(x^a)$ around $f(p)$; these pull back to coordinates $(f^* x^a) = (x^a \circ f)$ on $M$, and by construction the coordinate representations of $g, f^* g$ coincide, hence so do those of their inverses, whence the inverses agree.