Motivation
I am trying to derive the Riemannian Metric of the Inner Disc model as presented by by cannon et al. in the special case of 1 dimension. My understanding is that this is the induced metric as described by wikipedia, but I get different results, which hints that I am missing some underlying understanding.
Pushforward function
Let $U$ be the open Subset in $\mathbb{R}$ which denotes our inner disk model, so it is limited to $(-1,1)$. Let $\varphi$ be a continuous differential function mapping $U$ to $\mathbb{R}^2$. Lets fix the basis of $\mathbb{R}^2$ to $e_1 = \begin{pmatrix} 1 \\\\ 0 \end{pmatrix}$ and $e_2 = \begin{pmatrix} 0 \\\\ 1 \end{pmatrix}$. In this case $\varphi$ is given by the hyperbolic stereographic projection for any $x \in U$, so in our case is: $$ \varphi : U \to \mathbb{R}^2 := \varphi(x) = \frac{2x}{1-x^2} \cdot e_1 + \frac{1+x^2}{1-x^2} e_2 $$ The derivative of this function is $$ \frac{\partial \varphi(x)}{\partial x} = \frac{\partial \varphi^1}{\partial x} \cdot e_1 + \frac{\partial \varphi^2}{\partial x} \cdot e_2 \text{ where } \frac{\partial \varphi^1}{\partial x} = \dfrac{2\left(x^2+1\right)}{\left(1-x^2\right)^2} \text{ and } \frac{\partial \varphi^2}{\partial x} = \dfrac{4x}{\left(1-x^2\right)^2} $$ as can be seen in this image and as as a gif here. Let $v,w \in T_xU$ be vectors at the tangent space at $x$, which since $U$ is 1-dimensional are just given by scalars. Now the pushforward $\varphi_{\star}(v)$ maps $v$ from $T_xU$ to it's mapped tangent space in $\mathbb{R}^2$. In our case this reduces to: $$ \varphi_*(v) = \sum_{i=1}^{n} \sum_{a=1}^{m} v^i \frac{\partial \varphi^a}{\partial x^i} e_a = v \frac{\partial \varphi^1}{\partial x} e_1 + v \frac{\partial \varphi^2}{\partial x} e_2 = v \frac{2\left(x^2+1\right)}{\left(1-x^2\right)^2} e_1 + v \frac{4x}{\left(1-x^2\right)^2} e_2 = \begin{pmatrix} v \dfrac{2\left(x^2+1\right)}{\left(1-x^2\right)^2} \\\\ v \dfrac{4x}{\left(1-x^2\right)^2} \end{pmatrix} $$ which can be seen in this image for v = (0.5)
Induced Metric
Now if my understanding is correct we should be able to define the riemannian metrix just by the standard scalar product in $\mathbb{R}^2$ using $g_{x}(v,w) = \langle\varphi_{\star}(v), \varphi_{\star}(w)\rangle$ which in our case is $$ g(v,w) = \langle\varphi_\star(v), \varphi_\star(w)\rangle = \frac{ 4 v w (x^2+1)^2}{(1-x^2)^4} + \frac{ 16v w x^2}{(1-x^2)^4} $$
This in turn gives rise of the 1 x 1 matrix form of the riemannian metric, which should be: $$ \begin{bmatrix} \frac{ 4 dx^2 (x^2+1)^2}{(1-x^2)^4} + \frac{ 16 dx^2 x^2}{(1-x^2)^4} \end{bmatrix} $$ which in turn is just the line element $$ ds^2 = \frac{ 4 dx^2 (x^2+1)^2}{(1-x^2)^4} + \frac{ 16 dx^2 x^2}{(1-x^2)^4} $$
And now I'm stumbled, because cannon et al. tell me that it is instead: $$ ds{}_I^2 = 4 \frac{dx^2}{(1-x^2)^2} $$ Which is similar, but different enough as can be seen here. Why is this the case and where did I make a mistake?