= Solution
A map $g$ is <differentiable> at $a$ when a <linear map> $Dg_a:\mathbb R^n\to\mathbb R^k$ approximates its increment to first order:
$$
\boxed{g(a+h)=g(a)+Dg_a(h)+o(\|h\|).}
$$
This <Fréchet derivative> is unique, and its matrix in the standard bases is the <Jacobian matrix>. Differentiability on a domain means differentiability at each point there.
A standard sufficient condition for a zero locus to be a <submanifold> is the <regular level set theorem>: if $g$ is $C^1$ on an open domain and $Dg_a$ is <surjective> at every $a\in g^{-1}(0)$, then
$$
\boxed{M=g^{-1}(0)\text{ is an embedded }C^1\text{ submanifold of dimension }n-k.}
$$
For smooth $g$, the <submanifold> is smooth. The full-rank condition is the relevant <constraint qualification>.
Under this regularity hypothesis, its <tangent space> is
$$
\boxed{T_aM=\ker Dg_a.}
$$
Indeed, if a differentiable curve $\gamma$ lies in $M$ and passes through $a$, the <chain rule> gives $Dg_a(\gamma'(0))=0$. Thus all tangent velocities lie in the <kernel of a linear map>. The <rank-nullity theorem> makes this kernel $(n-k)$-dimensional, the same dimension as the <tangent space>, proving equality. In coordinates, solve the homogeneous linear equations given by the <Jacobian matrix>.
The regularity qualification matters: $g(x)=x^2$ has zero set $\{0\}$, a zero-dimensional <submanifold>, but $Dg_0=0$. Its <kernel> is all of $\mathbb R$, whereas its <tangent space> is zero. This is the phenomenon that <singular defining maps can overestimate tangent spaces>.
Back to article page
