Can the equivalence of the covariant and contravariant "connection coefficients" be proved simply by using the flatness of the tangent plane? Do the derivations below prove the equivalence of the "connection coefficients" (5.78) and (5.78) for Riemannian and pseudo-Riemannian manifolds? Are they correct, and justified by my stated assumption that the tangent plane provides a locally rectangular Cartesian or Minkowskian reference coordinate system?
This question has its origins in Bergmann's Introduction to the Theory of Relativity.  Bergamnn begins his departure from flat spacetime by writing the equation of parallel displacement of covariant vector components in rectangular Cartesian coordinates (RCC), $0=\delta a_{\underline{i}}$, as well as in general coordinates of the same dimension.  From this he observes that the variations of general components can be expressed as bilinear functions of the coordinate displacements and the vector components.
He then promulgates the existence of such bilinear functions for both contravariant and covariant vector components in the absence of RCC.  He writes the variations as

Asserting that the gamma coefficients of these tentative laws are entirely unknown;  his next step is to derive the law of transformation for the terms in (5.79).  From there he develops various properties, and shows the two sets of coefficients are ultimately equivalent; introducing additional "postulates" as he goes.
As far as I know, there's nothing wrong with Bergmann's development, and it was apparently satisfactory to Albert Einstein, who wrote in the forward:

I do, however take issue with the assertion that

There exist "spaces" where it is not possible to introduce a Cartesian coordinate system.  Two dimensional "spaces" of this kind include the surface of a sphere.

Here we are not dealing with the distinction between proper and pseudo-Riemannian (locally Minkowskian) space.  The 'inability' Bergmann asserts has to do with intrinsic curvature.
While it is true globally, if our manifold is locally smooth (avoiding intrinsic singularities) then there will always be a local RCC or Minkowski coordinate system available. That is, there is a tangent "plane".  So everything Bergmann does under the assumption that there is a RCC available, can be done with reference to the tangent plane.  I believe this is the essence of Cartan's moving frame method.  It is also a statement of Einstein's principle of equivalence.
In the following, the under-barred indices indicate coordinates in the tangent plane.  This is an abstraction of the globally available RCC with which we began.  The $\partial^\underline{i}_{jk}=\frac{\partial^2 x^\underline{i}}{\partial x^k \partial x^j}$ notation is due to Ciufolini and Wheeler.  (Unfortunately, the underbars mixed with this notation does not render well in MathJax.)  The gammas in (5.78) are written without decoration, and those of (5.79) are decorated with an over-tilde.
The first derivation follows Bergmann (found here Why doesn't the chain rule make the second partials of the coordinate transformation functions vanish?)
\begin{align*}
0= & \left[\begin{aligned}0= & \delta a_{\underline{i}}=a_{\underline{i},\underline{j}}\delta x^{\underline{j}}=a_{\underline{i},j}\delta x^{j}\\
= & \left[a_{k}\partial_{\underline{i}}^{k}\right]_{,j}\delta x^{j}\\
= & \left(a_{k,j}\partial_{\underline{i}}^{k}+a_{k}\partial_{\underline{i}\underline{j}}^{k}\partial_{j}^{\underline{j}}\right)\delta x^{j}
\end{aligned}
\right]\partial_{i}^{\underline{i}}\\
= & \delta a_{i}+a_{k}\partial_{\underline{i}\underline{j}}^{k}\partial_{i}^{\underline{i}}\partial_{j}^{\underline{j}}\delta x^{j}\\
= & \delta a_{i}-a_{k}\tilde{\Gamma}_{ij}^{k}\delta x^{j}\\
\delta a_{i}\equiv & +a_{k}\tilde{\Gamma}_{ij}^{k}\delta x^{j}\\
\tilde{\Gamma}_{ij}^{k}= & -\partial_{\underline{i}\underline{j}}^{k}\partial_{i}^{\underline{i}}\partial_{j}^{\underline{j}}
\end{align*}
The following derivation is the same thing done for contravariant components.  It wasn't until I tried to show the two forms are equivalent that I noticed that they are effectively identical.
\begin{align*}
0= & \left[\delta_{j}^{i}\right]_{,k}= \left[\partial_{\underline{i}}^{i}\partial_{j}^{\underline{i}}\right]_{,k}=\partial_{\underline{i}\underline{k}}^{i}\partial_{j}^{\underline{i}}\partial_{k}^{\underline{k}}+\partial_{\underline{i}}^{i}\partial_{jk}^{\underline{i}}\\
0= & \partial_{\underline{i}}^{i}\left[\begin{aligned}0= & \delta a^{\underline{i}}=a^{\underline{i}}{}_{,\underline{j}}\delta x^{\underline{j}}=a^{\underline{i}}{}_{,j}\delta x^{j}\\
= & \left[\partial_{k}^{\underline{i}}a^{k}\right]_{,j}\delta x^{j}\\
= & \left(\partial_{kj}^{\underline{i}}a^{k}+\partial_{k}^{\underline{i}}a^{k}{}_{,j}\right)\delta x^{j}
\end{aligned}
\right]\\
= & -\partial_{\underline{i}\underline{k}}^{i}\partial_{j}^{\underline{i}}\partial_{k}^{\underline{k}}a^{k}\delta x^{j}+\delta a^{i}\\
= & +\Gamma_{jk}^{i}a^{k}\delta x^{j}+\delta a^{i}\\
\delta a^{i}\equiv & -\Gamma_{jk}^{i}a^{k}\delta x^{j}\\
\Gamma_{jk}^{i}= & -\partial_{\underline{i}\underline{k}}^{i}\partial_{j}^{\underline{i}}\partial_{k}^{\underline{k}}
\end{align*}

Add: I'm pretty sure this *does* work.  And it does not require our coordinate systems to be orthonormal (AKA RCC).  It only requires that coordinate bases can be determined at the ultimate point of evaluation, and at neighboring manifold points which are sufficiently close that connecting geodesics may be treated as unique.  Conceptually this is fairly simple. I am motivated by Misner, Thorne and Wheeler's

Putting it into words will require more work than I have been able to dedicate to this question.
 A: This still needs work, and would be aided greatly by illustrations.  I believe the final statement showing that the contravariant basis of $\mathcal{T}_{\parallel}$ is a covariant constant is sufficient to justify my derivation.  I will review this, and add an explanation, when I have it formulated.
As I said the concept is easy.  The explanation is not.  The concept is that $\mathcal{T}_{\parallel}$ is a covariantly constant structure near $\mathscr{P}_{\circ}$.
The first paragraphs introduce the assumptions which I am bringing
to bear on the problem under discussion.
From our physical experience with smooth 2-dimensional surfaces embedded
in 3-space we abstract concepts which we carry over to our construction
of an n-dimensional manifold.
Measurability of distance.
Measurability of angles.
Straight-on continuation (geodesic generation).
Local flatness, that is; Euclidean plane geometry in the limit as
extension decreases.
Euclidean geometry is characterized by having a constant proportion
between the diameter and circumference of a circle, called $\pi$,
and the Pythagorean metric $r=\sqrt{\delta_{ij}x^{i}x^{j}}$. On an
intrinsically curved surface we find that a circle constructed by
measuring out geodesic segments of equal length $r$ in "all"
directions from a common point will not satisfy the formula $\frac{c}{2r}=\pi.$
In the limit as $r$ tends to 0, the Euclidean ratio is obtained.
If the 2-surface embedded in physical space is curved, it is not possible
for a straight line to lie on the surface in all directions. If it
is intrinsically curved, there is no direction in which a straight
line can lie on the surface. Nonetheless, as $r$ becomes small relative
to the curvature, a straight line segment becomes indistinguishable
from a geodesic segment of the same length. That is, for all practical
purposes, a short segment of a tangent line can be treated as lying
in the surface.
It seems worth observing that the value of $\pi$ can be calculated
to arbitrary accuracy using a formula formed of only natural numbers
called Wallis's product
$$
\pi=2\prod_{i=1}^{\infty}\left(1-\left(2n\right)^{-2}\right)^{-1}.
$$
The vector space $\mathbb{R}^{n}$ can be constructed using purely
logical and algebraic methods, without appeal to geometry. This includes
endowing it with the Pythagorean formula. This mathematical construct
is call called n-dimensional Euclidean space $\mathbb{E}^{n}$. (Strictly
speaking, $\mathbb{E}^{n}$ differs from $\mathbb{R}^{n},$ in that
the placement of the origin in the former is arbitrary, as is the
orientation of the standard basis.)
Despite having no physical means of doing so, we shall assume the
same notions of measurability, straight lines and etc., apply in n-dimensional
Euclidean geometry.
Using a physical standard measuring device called a ruler we can map
a local region of physical space onto a subset of $\mathbb{E}^{3}\cong\mathbb{R}^{3}.$ This map is
called a rectangular Cartesian coordinate system. In the same sense
that a short straight line segment tangent to a curved surface effectively
coincides with a geodesic segment in the surface, we assume that a
small region of a smooth n-dimensional manifold can be effectively
mapped to a region of $\mathbb{R}^{n}$in such a way that distances
calculated using the metric$\left\{ \delta_{ij}\right\} $ agree with
measured distances. Thus we say the manifold is locally Euclidean.
Pronounce $\mathscr{P}_{\circ}$ "P naught" and $\mathscr{P}_{\bullet}$
"P spot".
$\mathcal{M}$ is an n-dimensional Riemannian manifold. By definition,
$\mathcal{M}$ is equipped with a metric, which we might think of
as a short length of $n$-dimensional string (the string itself is
1-dimensional, but has $n$ degrees of freedom). By "short", we
mean that it is effectively tangential when pulled taut anywhere on
$\mathcal{M}$ under consideration. A unit of length will then be
some multiple of the length of that reference string. A unit length
metric string will not necessarily have that tangential property.
$\mathscr{P}_{\circ}\in\mathcal{M}$ is the point at which $\left\{ \Gamma_{jk}^{i}\right\} $
is computed.
$\mathcal{N}_{\circ}\subset\mathcal{M}$ is an open neighborhood of
$\mathscr{P}_{\circ}$ small enough to treat all included geodesic
segments as unique.
$\mathcal{N}_{\bullet}\subset\mathcal{N}_{\circ}$ is an open neighborhood
of some $\mathscr{P}_{\bullet}\in\mathcal{N}_{\circ}$ small enough
to treat included geodesic segments as tangential to $\mathcal{M}.$
$d\mathscr{P}$ is the directed geodesic segment connecting $\mathscr{P}_{\circ}$
and $\mathscr{P}_{\bullet}\in\mathcal{\mathcal{N}_{\circ}}\backepsilon d\mathscr{P}\subset\mathcal{\mathcal{N}_{\circ}}.$
$\delta\mathscr{P}$ is the directed geodesic segment connecting $\mathscr{P}_{\bullet}$
and $\mathscr{P}\in\mathcal{N}_{\bullet}\backepsilon\delta\mathscr{P}\subset\mathcal{N}_{\bullet}.$
Informally we might think of this as $\mathscr{P}_{\bullet}=\mathscr{P}_{\circ}+d\mathscr{P}$
and $\mathscr{P}=\mathscr{P}_{\bullet}+\delta\mathscr{P},$ which
does not (quite) connote vector addition. Our definitions are such
that $d\mathscr{P}$ should not be considered a vector, but $\delta\mathscr{P}$
could be treated as a vector tangent to $\mathcal{M}$ at $\mathscr{P}_{\bullet}.$
That is, $\mathcal{N}_{\bullet}\subset\mathcal{N}_{\circ}$ is practically
Euclidean, but $\mathcal{N}_{\circ}\subset\mathcal{M}$ is not quite
Euclidean.
The coordinate map $\left\{ x^{i}\left[\mathscr{P}\right]\right\} $
has its origin at $\mathscr{P}_{\circ}$ and nicely covers $\mathcal{N}_{\circ}.$
That is, the coordinate map is smooth and without singularities.
For brevity, write $\left\{ x_{\circ}^{i}=x^{i}\left[\mathscr{P}_{\circ}\right]\right\} $
and $\left\{ x_{\bullet}^{i}=x^{i}\left[\mathscr{P}_{\bullet}\right]\right\} .$
Let $\left\{ \mathfrak{e}_{\underline{i}}=\frac{\partial}{\partial x^{i}}\right\} {}_{\circ}$
be the coordinate basis spanning the tangent space $\mathcal{T}_{\circ}$
at $\mathscr{P}_{\circ}$.
Let $\left\{ \mathfrak{e}_{i}=\frac{\partial}{\partial x^{i}}\right\} {}_{\bullet}$
be the local coordinate basis spanning the tangent space $\mathcal{T}_{\bullet}$
at $\mathscr{P_{\bullet}}.$
The component set of the relative position vector $\delta\mathfrak{x}\in\mathcal{T}_{\bullet}\backepsilon\delta\mathfrak{x}=\delta\mathscr{P}$
determined by $\mathscr{P}\in\mathcal{N}_{\bullet}$ is
$$
\left\{ \delta x^{i}=x^{i}\left[\mathscr{P}\right]-x^{i}\left[\mathscr{P}_{\bullet}\right]=x^{i}-x_{\bullet}^{i}\right\} .
$$
We shall assume it is meaningful to parallel transport a unit speed
tangent vector $\hat{\mathfrak{t}}$ along $d\mathscr{P}$ from $\mathscr{P}_{\circ}$
to $\mathscr{P}_{\bullet}$ and that $\left\{ \mathfrak{e}_{\underline{i}}\right\} _{\circ}$
may also be parallel transported along $d\mathscr{P}$ to $\mathscr{P}_{\bullet}$
so that $\hat{\mathfrak{t}}$ has the same component representation
on $\left\{ \mathfrak{e}_{\underline{i}}\right\} _{\parallel}=\left\{ \mathfrak{e}_{\underline{i}}\right\} _{\circ}$
after transport, as it has at$\mathscr{P}_{\circ}$.
Heuristically, consider a 2-dimensional $\mathcal{M}.$ Imagine a
geodesic $\mathscr{G}\subset\mathcal{M\backepsilon\mathscr{P}_{\circ}\in\mathscr{G}}$
represented by a 1-dimensional "monorail track". Rigidly fix a
"monorail skate" to the center of a 2-dimensional rigid window
pane $\mathcal{T}_{\parallel}$ representing a mobile copy of $\mathcal{T}_{\circ},$
such that $\mathcal{T}_{\parallel}$ moves smoothly along $\mathscr{G},$
without rotating relative to adjacent points of $\mathcal{M}.$ With
$\mathcal{T}_{\parallel}$ parked at $\mathscr{P}_{\circ}$ make a
"best trace" projecting the manifold coordinates onto $\mathcal{T}_{\parallel}.$
Note that the disk of geodesic radius $\left\Vert \delta\mathscr{P}\right\Vert $
centered on $\mathscr{P}_{\circ}$ is effectively Euclidean.
Assume the coordinate mesh is fine enough that the grid cells adjacent
to $\mathscr{P}_{\circ}$ are indistinguishable from their projections
onto $\mathcal{T}_{\circ}.$ Thus the positive unit coordinate displacements
from the origin projected onto $\mathcal{T}_{\circ}$ coincide with
the basis vectors $\left\{ \mathfrak{e}_{\underline{i}}=\partial_{i}\right\} _{\circ}.$
In $\mathcal{T}_{\circ}$ fix one end of our unit length of metric
string at the point of tangency, and describe a unit circle with the
free end. At some point on the unit circle, the string will align
with the projection of $\mathscr{G}$ as it approaches$\mathscr{P}_{\circ}.$
That point on the unit circle will be the tip of $\hat{\mathfrak{t}}$
tangent to $\mathscr{G}.$
Upon completing our tangent image of $\mathcal{M}$ surrounding $\mathscr{P}_{\circ}$
we move $\mathcal{T}_{\parallel}$ along $\mathscr{G}$ starting at
$\mathscr{P}_{\circ},$ and call the moving tangent point $\mathscr{P}_{\bullet}\in\mathscr{G}.$
After transport (when we stop moving $\mathscr{P}_{\bullet}$), $\left\{ \mathfrak{e}_{\underline{i}}\right\} _{\parallel}=\left\{ \mathfrak{e}_{\underline{i}}\right\} _{\circ}$
and $\left\{ \mathfrak{e}_{i}\right\} _{\bullet}$ will both span
$\mathcal{T}_{\bullet},$ but their elements will generally not coincide.
Nonetheless, we may express each basis as linear combinations of the
other. We therefore have two component representations of $\delta\mathfrak{x},$
and a means of transforming one component representation to and from
the other. As a result, we see that the transformation matrices are
mutually reciprocal. Thus we write
\begin{align*}
\mathfrak{e}_{i}= & \mathfrak{e}_{\underline{i}}e^{\underline{i}}{}_{i}\\
\mathfrak{e}_{\underline{i}}= & \mathfrak{e}_{i}e^{i}{}_{\underline{i}}\\
\delta\mathfrak{x}= & \mathfrak{e}_{i}\delta x^{i}\\
= & \mathfrak{e}_{\underline{i}}\delta x^{\underline{i}}\\
= & \mathfrak{e}_{\underline{i}}e^{\underline{i}}{}_{i}\delta x^{i}\\
= & \mathfrak{e}_{i}e^{i}{}_{\underline{i}}e^{\underline{i}}{}_{j}\delta x^{j}\\
= & \mathfrak{e}_{\underline{i}}e^{\underline{i}}{}_{i}e^{i}{}_{\underline{j}}\delta x^{\underline{j}}\\
\delta x^{\underline{i}}= & e^{\underline{i}}{}_{i}\delta x^{i}\\
\delta x^{i}= & e^{i}{}_{\underline{i}}\delta x^{\underline{i}}\\
\delta^{i}{}_{j}= & e^{i}{}_{\underline{i}}e^{\underline{i}}{}_{j}\\
\delta^{\underline{i}}{}_{\underline{j}}= & e^{\underline{i}}{}_{i}e^{i}{}_{\underline{j}}
\end{align*}
Prior to parallel transport of $\mathcal{T}_{\circ}$ we have (since
$\left\{ x_{\circ}^{i}=0\right\} $)
\begin{align*}
\mathscr{P}_{\bullet}=\mathscr{P}_{\circ}\implies & \left\{ \delta x^{\underline{i}}=x^{i}\mathscr{\left[\mathscr{P}\right]}-x_{\circ}^{i}=x^{i}=\delta x^{i}\right\} \\
\implies & \left\{ \mathfrak{e}_{\underline{i}}=\mathfrak{e}_{i}\right\} 
\end{align*}
To emphasize the consequences of parallel transport on the meanings
of $\delta x^{\underline{i}}$ etc., we shall make $d\mathscr{P}$
large enough that $\mathscr{P}_{\circ}\notin\mathcal{N}_{\bullet}.$
The intent of doing so is to make clear that, even though the $\delta x^{\underline{i}}$
contract with the $\mathfrak{e}_{\underline{i}},$they do not equal
the coordinate displacements from $\mathscr{P}_{\circ}$ to $\mathscr{P}.$
Our ultimate goal is to compare field vectors at $\mathscr{P}_{\circ}$
and $\mathscr{P}_{\bullet}.$ In order to accomplish this we need
a means of converting components of field vectors at $\mathscr{P}_{\bullet}$
expressed on $\left\{ \mathfrak{e}_{i}\right\} _{\bullet}$ with those
at $\mathscr{P}_{\circ}$ expressed on $\left\{ \mathfrak{e}_{i}\right\} _{\circ}.$
The omission of the underbar in the last expression was intentional,
because that's how we write it when we are not fussing about the details
of parallel transport.
The utility of $\mathcal{T}_{\parallel}$ is that we can draw the
field vector $\mathfrak{v}_{\circ}$ at $\mathscr{P}_{\circ}$ on
$\mathcal{T}_{\parallel}$, then scate $\mathcal{T}_{\parallel}$
over to $\mathscr{P}_{\bullet}$ and compare the transported copy
of $\mathfrak{v}_{\circ}$ to the field vector $\mathfrak{v}_{\bullet}.$
We can, of course, produce the same result by reversing the order
of transport, carrying $\mathfrak{v}_{\bullet}$ back to $\mathscr{P}_{\circ},$
and compare it to $\mathfrak{v}_{\circ}$ there. If $\mathfrak{v}_{\circ}=\mathfrak{v}_{\parallel}=\mathfrak{v}_{\bullet}$
we say $\mathfrak{v}$ is parallel transported along $d\mathscr{P}\subset\mathscr{G}.$
In this sense we can say $\mathfrak{v}_{\circ}$ and $\mathfrak{v}_{\bullet}$
are the same vector under the specified transport. It is possible
that parallel transporting $\mathfrak{v}_{\circ}$ along a different
path; say, first to another point $\mathscr{P}_{*}$ ("P splat")
and thence to $\mathscr{P}_{\circ}$, might produce a different result.
Lets consider what it means to compare $\underline{\mathscr{P}}=\mathscr{P}_{\circ}+\delta\mathscr{P}_{\parallel}$
with $\mathscr{P}=\mathscr{P}_{\bullet}+\delta\mathscr{P},$ where
we are still transporting along $d\mathscr{P},$ and $\delta\mathscr{P}_{\parallel}$
is the parallel transported image of $\delta\mathscr{P}=\delta\mathfrak{x}.$
Lets write the coordinates of $\underline{\mathscr{P}}$ as $x^{\underline{i}}=x^{i}\left[\underline{\mathscr{P}}\right].$
Since we have assumed $\delta\mathscr{P}$ is small enough to be treated
as a tangent vector, we conclude that $\left\{ x^{\underline{i}}=\delta x^{\underline{i}}\right\} .$
With $\mathscr{P}_{\bullet}$ fixed, we have
\begin{align*}
x^{\underline{i}}= & \delta x^{\underline{i}}\\
= & e^{\underline{i}}{}_{i}\delta x^{i}\\
= & e^{\underline{i}}{}_{i}\left(x^{i}-x_{\bullet}^{i}\right)\\
e^{\underline{i}}{}_{i}= & \frac{\partial x^{\underline{i}}}{\partial x^{i}}\left[\mathscr{P}_{\bullet}\right]\\
x^{i}= & \delta x^{i}+x_{\bullet}^{i}\\
= & e^{i}{}_{\underline{i}}\delta x^{\underline{i}}+x_{\bullet}^{i}\\
= & e^{i}{}_{\underline{i}}x^{\underline{i}}+x_{\bullet}^{i}\\
e^{i}{}_{\underline{i}}= & \frac{\partial x^{i}}{\partial x^{\underline{i}}}\left[\mathscr{P}_{\bullet}\right]
\end{align*}
Let us now define $n$ objects $\left\{ \mathfrak{e}^{i}\right\} _{\bullet}$
at $\mathscr{P}_{\bullet}$ which contract with a tangent vector giving
the vector's $i^{th}$ scalar components. That is, $\left\{ \mathfrak{e}^{i}\cdot\delta\mathfrak{x}=\delta x^{i}\right\} .$
If we apply these to the basis vectors we have
$$
\left\{ \mathfrak{e}^{i}\cdot\mathfrak{e}_{j}=e^{i}{}_{j}=\delta^{i}{}_{j}\right\} .
$$
In linear algebra, objects which contract with vectors to produce
scalars are called linear forms, and they manifest as $n$-tuples
in a way similar to vectors. Since a change of basis produces a different
$n$-tuple representation for an invariant vector, the $n$-tuple
representations of linear forms transform contragrediantly in order
to preserve the scalar value. Thus the $\mathfrak{e}^{i}$ are linear
forms which transform as $\left\{ \mathfrak{e}^{\underline{i}}=e^{\underline{i}}{}_{i}\mathfrak{e}^{i}\right\} $
and $\left\{ \mathfrak{e}^{i}=e^{i}{}_{\underline{i}}\mathfrak{e}^{\underline{i}}\right\} .$
The statement that a vector $\delta\mathfrak{x}$ is parallel transported
along a path having parameter $\lambda$ amounts to saying
$$
\frac{d\delta\mathfrak{x}}{d\lambda}=\mathfrak{0}.
$$
We will need the derivative with respect to $\lambda$ of the contraction
of a vector field with a form field. The same method as is used in
the proof of the Leibniz rule of standard calculus gives us
\begin{align*}
\frac{\Delta\left[\tilde{\sigma}\cdot\mathfrak{v}\right]}{\Delta\lambda}= & \frac{\Delta\tilde{\sigma}}{\Delta\lambda}\cdot\mathfrak{v}+\tilde{\sigma}\cdot\frac{\Delta\mathfrak{v}}{\Delta\lambda}.
\end{align*}
Taking the limit $\Delta\lambda\to0$ shows the Leibniz rule works
in this case as well. (This may be a bit hand-wavy).
Since the $\mathfrak{e}_{\underline{i}}$ are, by decree, parallel
transported along $d\mathscr{P}\subset\mathscr{G},$ we have
\begin{align*}
\mathfrak{0}= & \frac{d\mathfrak{e}_{\underline{i}}}{d\lambda}\\
0= & \frac{d}{d\lambda}\left[\mathfrak{e}^{\underline{i}}\cdot\mathfrak{e}_{\underline{j}}\right]=\frac{d}{d\lambda}\left[\delta^{\underline{i}}{}_{\underline{j}}\right]\\
= & \frac{d}{d\lambda}\left[\mathfrak{e}^{\underline{i}}\right]\cdot\mathfrak{e}_{\underline{j}}=\frac{d}{d\lambda}\left[\mathfrak{e}^{\underline{i}}\right]\cdot\mathfrak{e}_{\underline{j}}a^{\underline{j}}\\
\implies\mathfrak{0}= & \frac{d\mathfrak{e}^{\underline{i}}}{d\lambda}
\end{align*}
Because $\left\{ a^{\underline{j}}\right\} $ is the component set
of an arbitrary vector.
