Weak Whitney Immersion Theorem

Theorem
Every $k$-dimensional manifold $X$ admits a one-to-one immersion in $\R^{2 k + 1}$.

Proof
Let $M > 2 k + 1$ be a natural number such that $f: X \to \R^M$ is an injective immersion.

Define a map $h: X \times X \times \R \to \R^M$ by:
 * $h \left({x, y, t}\right) = t \left({(f \left({x}\right) - f \left({y}\right)}\right)$

Define a map $g: T \left({X}\right) \to \R^M$ by:
 * $g \left({x, v}\right) = \mathrm d f_x \left({v}\right)$

where $T \left({X}\right)$ is the tangent bundle of $X$.

Since $M > 2 k + 1$, the Morse-Sard Theorem implies $\exists a \in \R^M$ such that $a$ is in the image of neither function.

Let $\pi$ be the projection of $\R^M$ onto the orthogonal complement of $a, H$.

Suppose:
 * $\left({\pi \circ f}\right) \left({x}\right) = \left({\pi \circ f}\right) \left({y}\right)$

for some $x, y$.

Then:
 * $f \left({x}\right) - f \left({y}\right) = t a$

for some scalar $t$.

Aiming for a contradiction, suppose $x \ne y$.

Then as $f$ is injective:
 * $t \ne 0$

But then $h \left({x, y, 1/t}\right) = a$, contradicting the choice of $a$.

By Proof by Contradiction:
 * $x = y$

and so $\pi \circ f$ is injective.

Let $v$ be a nonzero vector in $T_x \left({X}\right)$ (the tangent space of $X$ at $x$) for which:
 * $\mathrm d \left({\pi \circ f}\right)_x \left({v}\right) = 0$

Since $\pi$ is linear:
 * $\mathrm d \left({\pi \circ f}\right)_x = \pi \circ \mathrm d f_x$

Thus:
 * $\pi \circ \mathrm d f_x \left({v}\right) = 0$

so:
 * $\mathrm d f_x \left({v}\right) = t a$

for some scalar $t$.

Because $f$ is an immersion, $t \ne 0$.

Hence:
 * $g \left({x, 1/t}\right) = a$

contradicting the choice of $a$.

Hence $\pi \circ f: X \to H$ is an immersion.

$H$ is obviously isomorphic to $\R^{M-1}$.

Thus whenever $M > 2 k + 1$ and $X$ admits of a one-to-one immersion in $\R^M$, it follows that $X$ also admits of a one-to-one immersion in $\R^{M-1}$.

Also see

 * Whitney Immersion Theorem, a strictly stronger result: every $k$-manifold admits of a one-to-one immersion in $\R^{2 k}$, and an immersion (not necessarily one-to-one) in $\R^{2k-1}$.