Lyusternik's theorem on tangent spaces

Remarks

The converse inclusion is easy: if $x_0 + th + r(t) \in M$ with $r(t) = o(t)$, then differentiating $g(x_0 + th + r(t)) = 0$ at $t = 0$ gives $Dg(x_0) h = 0$. The main inclusion is proved by a Newton-type iteration that corrects $x_0 + th$ back onto $M$. The open mapping theorem gives a constant $C$ such that each $v \in Y$ equals $Dg(x_0) w$ for some $w$ with $\norm{w} \le C \norm{v}$, and this makes the corrections converge. When $\ker Dg(x_0)$ has a closed complement, for example when $X$ is a Hilbert space, the result also follows from the implicit function theorem.

Surjectivity cannot be dropped. For $g \colon \R^2 \to \R$, $g(x_1, x_2) = x_1^2 + x_2^2$, one has $M = \set{0}$ and $Dg(0) = 0$, so $\ker Dg(0) = \R^2$, while the only tangent vector to the single point $M$ is $0$.

Going deeper

For a student

This is the key step in the proof of Lagrange multipliers on Banach spaces. In the calculus of variations, $X$ is a function space and $g$ encodes constraints such as a fixed integral, so the theorem describes the admissible variations of a constrained problem. The condition that $Dg(x_0)$ is onto plays the role of the constraint qualification.