I asked a less-well-posed version of this question here on the physics stackexchange; however I believe the answers I recieved only apply to very specific cases and not the general case I'm looking for. Please help enlighten me!
For those not aquainted, in General Relativity we consider spacetime as a 4-dimensional pseudo-Riemannian manifold with metric tensor components $g_{\mu\nu}$. There is an equivalent tetrad formulation of general relativity, wherein one utilizes (an often orthonormal) basis. In this case the metric tensor is written as:
$$g_{\mu\nu}=e_{\mu}^{a}e_{\nu}^{b}\eta_{ab}$$
Where $\eta_{ab}$ are components of the flat Minkowskian metric. In this case, the Greek and Latin indices refer to coordinate and orthonormal (generally non-coordinate) basis respectivly. One case use the tetrads to take quantities from coordinate to orthonormal indices. This formulation is extremely important, for example when using spinor fields in General relativity.
From a purely mathematical perspective, I perceive the tetrad formulation as a smooth covering map from a spacetime manifold $M$ to Minkowskian spacetime $\eta$.
$$M\rightarrow\eta$$
In which case it is rather natural that the spacetime metric tensor becomes the pullback of the metric from $\eta$. Am I right in viewing it this way? The main arguments against this viewpoint on the physics stackexhange were:
“The tetrad formalism is not related to the pullback operation. The similarity you point out has more to do with the statement that both $g_{\mu\nu}=\frac{\partial X^{a}}{\partial x^{\mu}}\frac{\partial > X^{b}}{\partial x^{\nu}}g_{ab}$ and $g_{\mu\nu}=e_{\mu}^{a}e_{\nu}^{b}\eta_{ab}$ transform covariantly under diffeomorphisms (which is a pullback of the spacetime to itself via some diffeomorphism, if you prefer that language).”
However; in reading about the pullback it would seem that this argument only applies to the pullback by a diffeomorphism (see section in link). The pullback is a much more robust operation than this.
So I'm asking the mathematicians now: Can I regard the tetrad formulation as a covering map from a spacetime manifold to Minkowskian space which implies the spacetime metric tensor is the pullback? I do understand there are issues with pulling back contravariant tensor fields, but we can just stick with differential form versions if preferred.
Also I'll add that in a great book on General relativity "Formulations of General Relativity Gravitry, Spinors and differential forms" by Krasnov. Section 3.1.3 says:
"given a tetrad, the metric on M is defined to be the pullback of the metric on E $$g_{\mu\nu}=e_{\mu}^{a}e_{\nu}^{b}\eta_{ab}$$ "
Where Krasnov is referring to the base and fiber metric, by M and E respectively.
EDIT: I didn't think I'd have to write this explicitly but here goes (in response to answer below) We are interested in a formula for the metric that is independent of coordinates of our target space. The pulled back metric will be independent of transformations which leave the quadratic form on our target space invariant. For a Minkowskian target space, such a transformation falls within the group $A\in SO\left(3,1\right)$.
$$\phi^{A}\rightarrow A_{A}^{a}\phi^{A}=\phi^{a}$$
$$\phi^{A}\phi_{A}=\phi^{A}\eta_{AB}\phi^{B}=\left(\phi^{A}A_{A}^{a}\right)A_{a}^{A}\eta_{AB}A_{b}^{B}\left(A_{B}^{b}\phi^{B}\right)=\phi^{a}\phi_{a}$$
Via the fact that $A$ is an isometry of the target space:
$$A_{a}^{A}\eta_{AB}A_{b}^{B}=\eta_{ab}=\eta_{AB}$$
Then our pulled back metric is:
$$g_{\mu\nu}=\partial_{\mu}(\phi^{A}A_{A}^{a})\eta_{AB}\partial_{\nu}(A_{B}^{b}\phi^{B})=\left(\phi_{,\mu}^{A}A_{A}^{a}+\phi^{A}A_{A,\mu}^{a}\right)\eta_{AB}\left(A_{B}^{b}\phi_{,\nu}^{B}+A_{B,\nu}^{b}\phi^{B}\right)$$
The property of orthogonal matrices $A_{a}^{A}A_{A}^{a}=1$, allows us to write the above as:
$$g_{\mu\nu}=\left(\partial_{\mu}\phi^{A}+\phi^{A}\left[\left(\partial_{\mu}A_{A}^{a}\right)A_{a}^{A}\right]\right)\eta_{AB}\left(\partial_{\nu}\phi^{B}+\left[A_{b}^{B}\left(\partial_{\nu}A_{B}^{b}\right)\right]\phi^{B}\right)$$
Reorganizing terms, we obtain:
$$g_{\mu\nu}=\left(\partial_{\mu}\phi^{A}+\phi^{A}A_{A,\mu}^{a}A_{A}^{a}\right)\left(\partial_{\nu}\phi_{A}+A_{a}^{A}A_{A,\nu}^{a}\phi_{A}\right)$$
Which contains a connection term $\mathcal{A_{\mu}=}A_{A,\mu}^{a}A_{a}^{A}$ transforming under the adjoint representation of $SO\left(3,1\right)$. Together then we have a covariant derivative $\mathcal{D_{\mu}=}\partial_{\mu}+\mathcal{A}_{\mu}$, and our embedded spacetime metric is described by:
$$g_{\mu\nu}=\left(\mathcal{D}_{\mu}\phi^{A}\right)\eta_{AB}\left(\mathcal{D}_{\nu}\phi^{B}\right)=\mathcal{D}_{\mu}\Phi^{T}\mathcal{D}_{\nu}\Phi$$
It is clear then that the basis will NOT in general Lie commute (just as we expect for tetrad basis), and we have an expression that is independent of the coordinate system of our target space. In other words, the partial derivative for a pullback seems insufficient and we need to use a covariant derivative, does this make sense?