User:Arbo/Sandbox.

Theorem
Every smooth $m$-dimensional manifold can be immersed in Euclidean $\left({2m-1}\right)$-space.

Proof
We will first show that any compact manifold can be embedded (immersed) in some $\R^N$, $N \gt\gt 0$. This argument can be extended to the case of arbitrary manifolds.

Statement
For every injective Function $f\colon M \rightarrow \R^n$ there exists a linear Transformation $\lambda \colon \R^n \rightarrow \R^{n-1}$, such that $\lambda\circ f$ is injective, provided that $n \gt 2m+1$. Moreover $\lambda \circ f$ is immersive, if $f$ is immersive and $m \gt 2m+1$.

Proof
For the first part, observe that $\tilde{M} = M\times M - \triangle M$ is a smooth $2m$-dimensional manifold, with $\triangle M$ being the image of the map $M \rightarrow M\times M$, $x\mapsto \left({x, x}\right)$. Define $$F\colon \tilde{M} \to S^{n-1} \\ \left(x,y\right) \mapsto \frac{f(x)-f(y)}{\left\vert f(x)-f(y) \right\vert}.$$ Verify that F is smooth and injective. Since by assumption $n \gt 2m+1 \iff n-1 \gt 2m$, a corollary from Sard's Theorem implies that $F$ is not surjective. So there exists $b \in \S^{n-1} - \operatorname{Image}(F)$. The desired Transformation $\lambda$ is the projection on $b^\perp$. It remains to check that $\lambda\circ f$ is injective. To this end suppose that $f(x)-f(y) \in \operatorname{Kernel}(\lambda)$. Then $f(x) = f(y)$ or $f(x)-f(y) = r*b$, for some $r\in\R$. The second case contradicts the injectivity of $F$, since $$F(x,y) = \frac{f(x)-f(y)}{\left\vert f(x)-f(y) \right\vert} = \frac{r*b}{\left\vert r*b \right\vert} = \frac{b}{\left\vert b \right\vert} = b$$. Therefore $f(x) = f(y)$, which implies that $x=y$ because $f$ is injective. Alltogether this shows that $f(x)-f(y) \in \operatorname{Kernel}(\lambda) \implies x=y$, where the condition of this implication is equivalent to $\lambda\circ f(x) = \lambda\circ f(y)$, since $\lambda$ is linear.

Let $\tau M$ denote the tangent bundle of $M$. Choose a riemannian metric $g$ on $\tau M$ and consider the sphere bundle $\sigma M = \{v \in \tau M ~\vert~ g(v,v)=1\}$.

Lemma
Let $...$ be a [definiend] by definition 2.

Then by definition:
 * [Definition 2 of definiend]



Thus $...$ is a [definiend] by definition 1.

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