Clean proof that the inverse metric tensor really is the inverse

Question

In Frederic Schuller's lectures on GR, he defines the metric tensor as a $(0,2)$-tensor field satisfying:

1. $g(X,Y)=g(Y,X) \,\,\,\forall X,Y$
2. $\flat (X)$ is a $C^\infty$-isomorphism, where $\flat(X):=g(X,\cdot)$

He then defines the inverse metric tensor as the $(2,0)$-tensor field satisfying:

1. $g^{-1}(\omega,\sigma)=\omega(\flat^{-1}(\sigma))$

He then claims without proof that $(g^{-1})^{am}g_{mb}=\delta^a_b$.

Can someone offer a clean proof of this statement, starting from Schuller's definitions?

I began with this:

$$\begin{align} (g^{-1})^{am}g_{mb} &= g^{-1}(e^a,e^m)\cdot g(e_m,e_b) \\ &= e^a(\flat^{-1}(e^m))\cdot \flat(e_m)(e_b) \end{align}$$

If I could just massage this to leave $e^a(e_b)$ that would do it, but this seems tricky as the above is really a summation over $m$, and also the $e^a$ and $e_b$ are separated by a multiplication over $\mathbb{R}$, whereas we need the $e^a$ to "act on" the $e_b$. I also tried thinking in terms of matrix multiplication, but this gets messy and confusing since $\flat$ needs to act on column vectors but produce row vectors.

Answer 1

Consider the quantity

$$ X^a{}_b = g^{-1}(e^a, \flat(e_b)) $$

Without using the definition of $g^{-1}$, this is equal to

$$(g^{-1})^{am}\flat(e_b)_m = (g^{-1})^{am} g(e_b, e_m) = (g^{-1})^{am}g_{mb} $$

Using the definition of $g^{-1}$, we have

$$X^a{}_b = e^a(\flat^{-1}( \flat(e_b))) = e^a(e_b)$$