[proofplan]
The strategy is to substitute the [Singular Value Decomposition](/theorems/3071) of $JL$ into the Gram matrix $JL^\top JL$ and exploit orthogonal invariance of the determinant. Writing $JL = P\Sigma Q^\top$ with $P \in \mathbb{R}^{n\times n}$, $Q \in \mathbb{R}^{m\times m}$ orthogonal and $\Sigma \in \mathbb{R}^{n\times m}$ rectangular diagonal with diagonal entries $\sigma_1 \ge \cdots \ge \sigma_m \ge 0$, the Gram matrix becomes $JL^\top JL = Q(\Sigma^\top\Sigma)Q^\top$, where $\Sigma^\top\Sigma = \operatorname{diag}(\sigma_1^2, \dots, \sigma_m^2)$. The determinant of this $m\times m$ matrix is $\sigma_1^2\cdots\sigma_m^2$, and taking the square root gives $J_m L = \sigma_1\cdots\sigma_m$. The injectivity criterion follows by relating $\sigma_m = 0$ to the existence of a non-zero kernel vector via the SVD.
[/proofplan]
[step:Recall the definition of the $m$-dimensional Jacobian and substitute the SVD]
The theorem concerns a linear map $L: \mathbb{R}^m \to \mathbb{R}^n$ with $m \le n$. Let $A := JL \in \mathbb{R}^{n \times m}$ denote the matrix of $L$ in the standard bases (the Jacobian matrix of the linear map, which is independent of the base point). The $m$-dimensional Jacobian of $L$ is defined by
\begin{align*}
J_m L := \sqrt{\det(A^\top A)},
\end{align*}
where $A^\top A \in \mathbb{R}^{m \times m}$ is the Gram matrix. The argument under the square root is non-negative: for every $x \in \mathbb{R}^m$, $x^\top (A^\top A) x = |Ax|^2 \ge 0$, so $A^\top A$ is positive semidefinite and its determinant (the product of its eigenvalues) is non-negative.
By the [Singular Value Decomposition](/theorems/3071) (whose hypothesis $m \le n$ is exactly the hypothesis of the present theorem), there exist orthogonal matrices $P \in \mathbb{R}^{n\times n}$ and $Q \in \mathbb{R}^{m\times m}$ and a rectangular diagonal matrix $\Sigma \in \mathbb{R}^{n\times m}$ with diagonal entries $\sigma_1 \ge \sigma_2 \ge \cdots \ge \sigma_m \ge 0$ (the singular values of $L$) and zeros elsewhere, such that
\begin{align*}
A = P \Sigma Q^\top.
\end{align*}
[/step]
[step:Compute the Gram matrix in terms of the singular values]
Substituting $A = P \Sigma Q^\top$ into $A^\top A$:
\begin{align*}
A^\top A &= (P \Sigma Q^\top)^\top (P \Sigma Q^\top) \\
&= Q \Sigma^\top P^\top P \Sigma Q^\top \\
&= Q \Sigma^\top \Sigma Q^\top,
\end{align*}
using $P^\top P = I_n$ since $P$ is orthogonal.
We compute $\Sigma^\top \Sigma \in \mathbb{R}^{m \times m}$ explicitly. The matrix $\Sigma$ has entries $\Sigma_{ij} = \sigma_i \delta_{ij}$ for $i \in \{1, \dots, m\}$ (and $\Sigma_{ij} = 0$ for $i > m$). Hence for $j, k \in \{1, \dots, m\}$,
\begin{align*}
(\Sigma^\top \Sigma)_{jk} = \sum_{i=1}^n \Sigma_{ij} \Sigma_{ik} = \sum_{i=1}^m \sigma_i \delta_{ij} \cdot \sigma_i \delta_{ik} = \sigma_j^2 \delta_{jk},
\end{align*}
where the sum collapses to a single term when $j = k$ (giving $\sigma_j^2$) and vanishes otherwise. Thus
\begin{align*}
\Sigma^\top \Sigma = \operatorname{diag}(\sigma_1^2, \sigma_2^2, \dots, \sigma_m^2) \in \mathbb{R}^{m \times m}.
\end{align*}
[/step]
[step:Take the determinant and apply orthogonal invariance]
We compute $\det(A^\top A)$ using the multiplicativity of the determinant:
\begin{align*}
\det(A^\top A) = \det(Q \Sigma^\top \Sigma Q^\top) = \det(Q) \det(\Sigma^\top \Sigma) \det(Q^\top).
\end{align*}
Since $Q$ is orthogonal, $\det(Q) \in \{+1, -1\}$ and $\det(Q^\top) = \det(Q)$, so $\det(Q) \det(Q^\top) = \det(Q)^2 = 1$. Therefore
\begin{align*}
\det(A^\top A) = \det(\Sigma^\top \Sigma) = \prod_{i=1}^m \sigma_i^2,
\end{align*}
since the determinant of a diagonal matrix is the product of its diagonal entries.
Taking the square root (and recalling $\sigma_i \ge 0$ for all $i$),
\begin{align*}
J_m L = \sqrt{\det(A^\top A)} = \sqrt{\prod_{i=1}^m \sigma_i^2} = \prod_{i=1}^m \sigma_i = \sigma_1 \sigma_2 \cdots \sigma_m.
\end{align*}
[/step]
[step:Identify the injectivity criterion via the smallest singular value]
We show $J_m L = 0$ if and only if $L$ fails to be injective.
The product $\sigma_1 \sigma_2 \cdots \sigma_m$ vanishes if and only if at least one $\sigma_i = 0$, and since the singular values are ordered non-increasingly, $\sigma_i = 0$ for some $i$ is equivalent to $\sigma_m = 0$ (the smallest one).
Suppose $\sigma_m = 0$. Let $q_m \in \mathbb{R}^m$ be the $m$-th column of $Q$. Then $q_m \neq 0$ (it has unit norm since $Q$ is orthogonal) and
\begin{align*}
A q_m = P \Sigma Q^\top q_m = P \Sigma e_m^{(m)} = P (\sigma_m e_m^{(n)}) = 0,
\end{align*}
where $e_m^{(m)}$ is the $m$-th standard basis vector of $\mathbb{R}^m$ (we used $Q^\top q_m = e_m^{(m)}$ because $Q$ is orthogonal and $q_m$ is its $m$-th column). Since $A q_m = 0$ with $q_m \neq 0$, the linear map $L$ has non-trivial kernel and is not injective.
Conversely, suppose $L$ is not injective. Then there exists $v \in \mathbb{R}^m \setminus \{0\}$ with $Av = 0$. Then
\begin{align*}
0 = |Av|^2 = v^\top A^\top A v = v^\top Q (\Sigma^\top \Sigma) Q^\top v = w^\top (\Sigma^\top \Sigma) w
\end{align*}
where $w := Q^\top v \in \mathbb{R}^m$ has $|w| = |v| > 0$ (since $Q$ is orthogonal, it preserves norms). Writing $w = (w_1, \dots, w_m)^\top$, the right-hand side equals $\sum_{i=1}^m \sigma_i^2 w_i^2$. This sum vanishes with all $\sigma_i^2 \ge 0$ and $w_i^2 \ge 0$, so $\sigma_i w_i = 0$ for every $i$. Since $w \neq 0$, some $w_i \neq 0$, forcing the corresponding $\sigma_i = 0$, hence $\sigma_m = 0$ (as $\sigma_m \le \sigma_i$).
This completes the proof: $J_m L = \prod_{i=1}^m \sigma_i$ and $J_m L = 0 \iff \sigma_m = 0 \iff L$ is not injective.
[/step]
