I have been asking quite a few questions lately on this same topic, but I am really trying to get a grasp of a lot of the content of my Riemannian Geometry course, so I appreciate the help!
on p. 14 of do Carmo's Riemannian Geometry, he starts through a proof that, given a differential immersion $\varphi:M_{1}^{n} \rightarrow M_{2}^{m}$, then for every point $p \in M_1$, $\exists$ a neighborhood $V\subset M_1$ of $p$ such that the restriction $\varphi: V \rightarrow M_2$ is an embedding.
During his proof, he paramaterizes $M_1$ and $M_2$ by maps $\textbf{x}_1: U_{1} \subset \mathbb{R}^n \rightarrow M_1$ and $\textbf{x}_2: U_{2} \subset \mathbb{R}^m \rightarrow M_m$, respectively, and defines $\widetilde{\varphi} = \mathbf{x}_{2}^{-1} \circ \varphi \circ \mathbf{x}_1$. Then by denoting the coordinates of $\mathbf{R}^n$ and $\mathbf{R}^m$ by $(x_1, ..., x_n)$ and $(y_1, ..., y_m)$, respectively, he writes: $$\widetilde{\varphi} = (y_1(x_1, ... , x_n), ... , y_m(x_1, ... , x_n)).$$ Up until here, this all makes perfect sense to me. But the next thing that he does I find confusing. He says:
Let $q = \textbf{x}_{1}^{-1}(p)$. Since $\varphi$ is an immersion, we can suppose, renumbering the coordinates for both $\mathbb{R}^n$ and $\mathbb{R}^m$, if necessary, that $$\frac{\partial (y_1, ... , y_n)}{\partial (x_1, ... , x_n)}(q) \neq 0\text{.}$$
There are two things I do not understand here:
How can we simply decide to renumber the coordinates for $\mathbb{R}^n$ and $\mathbb{R}^m$? I know that the determinant of the Jacobian is only defined for the case that $n = m$, but our assumption was that the dimensions of the manifolds were not the same, so how exactly do we just decide to adjust the dimensions so we can take the determinant?
Why does the fact that $\varphi$ is an immersion (its differential is an injective linear transformation) imply that the determinant of the differential is non-zero? It has been a few years since I studied linear algebra, but I do not recall that an injective linear transformation implies non-zero determinant. Is this true in general, or is there another piece of the puzzle I am missing here?
My first idea was to look at the differential of $\widetilde{\varphi}$, i.e. $D\widetilde{\varphi} = D(\textbf{x}_{2}^{-1} \circ \varphi \circ \textbf{x}_1)$ and applying the chain rule. which (I think) should be: $$D(\textbf{x}_{2}^{-1} \circ \varphi \circ \textbf{x}_1) = D\textbf{x}_{2}^{-1} (\varphi \circ \textbf{x}_1) D\varphi(\textbf{x}_1)D\textbf{x}_1.$$
And $D\textbf{x}_{2}^{-1}(\varphi \circ \textbf{x}_1)$ is a vector in $\mathbb{R}^m$, $D\varphi(\textbf{x}_1)$ is a ($m$-dimensional) vector in $T_{\varphi(p)}M_2$, and $D\textbf{x}_1$ is a (injective, $n\times n$) linear transformation from $\mathbb{R}^n$ to $T_{p}M_1$. So It starts to make sense that $D\widetilde{\varphi}$ should be an $n\times n$ matrix, but I don't get how to treat the multiplication of two $m$-dimensional vectors. Is one of them a dual vector, and I somehow missed that fact or is my reasoning so far just completely off?
Your chain rule is basically right, but the interpretation is a bit off. Differentials take vectors to vectors, and are hence matrices, so you can straighten things out by looking at the domain and range of each:
The matrix $Dx_1$ takes an $n$-dimensional vector $v\in T_{{\bf x}}\mathbb{R}^n\cong \mathbb{R}^n$ and outputs an $n$-dimensional vector $(x_{1*}v)\in T_{x_1({\bf x})}M_1^n$, hence is an $n\times n$ matrix.
The matrix $D\varphi$ takes this vector and outputs a $m$-dimensional vector $\varphi_*(x_{1*}v) \in T_{(\varphi\circ x_1)({\bf x})}M_2^m$, hence is an $m\times n$ matrix.
The matrix $Dx_2^{-1}$ takes this vector and outputs an $m$-dimensional vector ${x_{2}^{-1}}_*(\varphi_*(x_{1*}v))\in T_{(x_2^{-1}\circ \varphi \circ x_1)({\bf x})}\mathbb{R}^m\cong \mathbb{R}^m$, hence is an $m\times m$ matrix.
So the differential $D\tilde{\varphi}=Dx_2^{-1}\times D_\varphi\times Dx_1$ is an $m\times n$ matrix.