So I'm studying the treatments of vectors and covectors across multiple resources and this is an area where different authors take wildly different approaches; but all of them seem to lead to the same results. For example, in A Students Guide to Tensors and Vectors; in section 4.5 the author defines dual basis vectors $e^1, e^2, ...$; where he only mentions the idea of a co-vector or a 1-form in section 5.9; where he states that these definitions are in some way equivalent to his definitions.
However, I have read the common multilinear map introduction from many other authors, such as the one presented in Manifolds, Tensors and Forms. Where a dual basis $V^*$ is populated by 1-forms; and in this context, one forms are functions that take vectors as inputs and return real numbers. Usually, 1-forms are pictured as stacked planes.
I was trying to show that these two approaches are equivalent, but I'd like to make sure that I've done everything correctly.
Equivalence
So I define a basis $e_i \in V$ and then a dual basis $\epsilon^i \in V^* $, I then define another vector in the original space $ e^i \in V $. I have the following picture in mind throughout my working:
So I'm asking, is the statement that $e_i \cdot e^j = \delta_i^j$ equivalent to stating that $\epsilon_i(e^j) = \delta_i^j$. Or is it incorrect to work entirely within $V$?
So the basic issue I have is that we have the musical isomorphism between $e_i$ and $\epsilon^i$, we define it as follows (I left out the $\otimes$ symbol):
$$ \flat\left( e_i \right) = g(e_i, \cdot ) = g_{km}\epsilon^k\epsilon^m e_i = g_{km}\epsilon^k \delta^m_i\\ \flat\left( e_i \right) = g_{ki}\epsilon^k $$
Equivalently and similarly, we can define the sharp operator to state:
$$ \sharp \left(\epsilon^k\right) = g^{ik}e_i $$
Now this is fine, but then it should be okay to also have a linear map, which asserts that:
$$ e_i \cdot e^j = \delta_i^j $$
Of course I'm only using superscripts for convenience here. But if this is the case, then by that logic; we can use the metric tensor in a different way to figure out what the map should be. First, lets assume that some linear map can take $e_i$ to $e^j$. I've put the j on the bottom because $e^j$ is a vector:
$$ e^j = L_j^m e_m $$
Now from the definition above:
$$ e_i \cdot e^j = g(e_i, e^j) = g(e_i, L_j^m e_m) = g_{np} \epsilon^n \epsilon^p e_i L_j^m e_m = g_{np} \delta^p_i \delta_m^n L_j^m = g_{ni} L_j^n\\ e_i \cdot e^j = g_{ni} L_j^n $$
Now if we want $e_i\cdot e^j = \delta_i^j$, then we can show that:
$$ \delta_i^j = g_{ni} L_j^n $$
So now if we multiply both sides by the components of the inverse metric tensor, then:
$$ \delta_i^j g^{ni} = g^{ni} g_{ni} L_j^n = L_j^n \\ g^{nj} = L_j^n $$
So then immediately we see that this linear map has the exact same components as the inverse metric tensor:
$$ e^j = L_j^m e_m = g^{mj} e_m $$
So:
$$ e^k = g^{ik} e_i $$
Now when you compare these two equations that I've highlighted; it would seem that they communicate the same ideas, except for the fact that $e^i$ lives in $V$ and $\epsilon^i$ lives in $V^*$:
So why should we use an isomorphism to a whole new vector space, rather than just using a linear map to a different element in the same space?
I appreciate the usefulness of visualising a covector as a set of stacked planes. But is it necessary to conjure a whole new vector space for this, rather than just imagining the dual vector to behave in this way.