Are measurements using the first fundamental form preserved by re-parameterization? I may be conflating a number of different concepts, but I am confused about measurements using the first fundamental form.
Here is what I think I know:


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*The first fundamental form can be used to compute distance on a parametric surface

*The coefficients of the first fundamental form can be re-parameterized through substitution, as with the relation given in Sec. 1.3 here.


Here is what confuses me:


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*Re-parameterizing the sphere through a map projection (e.g. gnomonic projection, mercator projection, etc.) results in different distances on the mapped surface. That is, $ds_{map}^2 \ne ds_{sphere}^2$. I can walk through this derivation and it makes logical sense.

*Yet, "arclength is invariant to coordinate system transformations," at least according to the proof on the Metric Tensor Wikipedia page (screenshot pasted below) and a handful of other Google search results on the topic. But isn't a spherical projection a coordinate transformation $M : (\lambda, \phi) \rightarrow (x, y)$?



I believe that this text clears it up a bit (bottom of page 8):

A mapping of a portion of a manifold M to a portion of a manifold N is called isometric, if the length of any curve on N is the same as the length of its pre-image on M.

In other words, arclength is preserved by isometric transformations. Spherical map projections are provably not isometric, so, sure, that's why  $ds_{map}^2 \ne ds_{sphere}^2$. I think I follow this.
Why I'm confused is because I thought I followed the logic of proof on the Wikipedia page, and I don't understand why it wouldn't generalize to non-isometric mappings. All the same derivatives can be calculated through a spherical projection, for example.
So does arclength hold or not?
[Note: when answering this question, feel free to use the example of spherical projections, because it's something I think I understand fairly well. Also, I'm not a mathematician, so formal mathematical language, while likely the most accurate way to answer this question, is probably going to be lost on me.]
 A: Let $\dot S:=S^2\setminus\{(0,0,1)\}$ be the $2$-sphere with the north pole removed. There is the stereographic projection 
$$\sigma: \quad \dot S\to{\mathbb R}^2,\qquad (x,y,z)\mapsto(\xi,\eta)$$
according to well known formulas. This $\sigma$ is a map of one surface onto another one. It is not an isometry when we adopt on $\dot S$ and ${\mathbb R}^2$ their "usual" standard metrics, which come about as follows:  
$\dot S$ acquires its standard metric from the embedding of $S^2$ into the euclidean ${\mathbb R}^3$. If $(\phi,\theta)$ are geographical coordinates on $\dot S$ you compute $ds_{\rm sphere}^2=E(\phi,\theta)d\phi^2+G(\phi, \theta)d\theta^2$, whereby $E$ and $G$ in fact depend only on the geographical latitude. On ${\mathbb R}^2$ you have the standard euclidean $ds^2_{\rm plane}=d\xi^2+d\eta^2$.
On $\dot S$ one might think of other coordinate systems, say $(x,y)$ in the vicinity of the south pole, or the $(\xi,\eta)$ furnished by $\sigma$. The latter is to be understood in the following way: Each point $P=(x,y,z)\in\dot S$ so far had geographical coordinates $(\phi,\theta)$, but $P$ obtains now new coordinates $(\xi,\eta):=\sigma(P)$. But we do not want to change the length measurement on $\dot S$. This means it is forbidden to write $ds^2_{\rm sphere}=d\xi^2+d\eta^2$. Instead we have to do some computation as quoted by you from Wikipedia, and we would then obtain
$$ds^2_{\rm sphere}=G(\xi,\eta)(d\xi^2+d\eta^2)\ .$$
(In this example the resulting transformation formula is so simple because the stereographic projection $\sigma$ is conformal.) Note that in this second thought experiment we have not mapped one surface onto another, but we have (for whatever reasons) changed the local coordinate system on the given surface $\dot S$.
