# The Unapologetic Mathematician

## Immersions are Locally Embeddings

In both of our pathological examples last time, the problems were very isolated. They depended on two separated parts of the domain manifold interacting with each other. And since manifolds can be carved up easily, we can always localize and find patches of the domain where the immersion map is a well-behaved embedding.

More specifically, if $f:M^m\to N^n$ is an immersion, with $f_{*p}$ always an injection for every $p\in M$, then for every point $p$ there exists a neighborhood $U\subseteq M$ of $p$ and a coordinate map $(V,y)$ around $f(p)\in N$ so that $q\in f(U)\cap V$ if and only if $y^{m+1}(q)=\dots=y^n(q)=0$. Further, the restriction $f\vert_U$ is an embedding.

This is basically the actual extension of the second part of the implicit function theorem to manifolds. Appropriately, then, we’ll let $\iota_\mathbb{R}^m\to\mathbb{R}^n$ be the same inclusion into the first $m$ coordinates. We pick a coordinate map $x$ around $p$ with $x(p)=0$, and another map $\tilde{y}$ around $f(p)$ with $\tilde{y}(f(p))=0$. Then we get a map $\tilde{y}\circ f\circ x^{-1}$ from a neighborhood of $0\in\mathbb{R}^m$ to a neighborhood of $0\in\mathbb{R}^n$.

Now, the assumption on $f$ is that $f_{*p}$ is injective, meaning it has maximal rank $m$ at every point. Since $x^{-1}$ and $\tilde{y}$ are diffeomorphisms, the composite also has maximal rank $m$ at $0$. The implicit function theorem tells us there is a coordinate map $g$ in some neighborhood of $0\in\mathbb{R}^n$ and a neighborhood $W$ of $0\in\mathbb{R}^m$ such that $g\circ\tilde{y}\circ f\circ x^{-1}\vert_W=\iota\vert_W$.

We set $U=x^{-1}(W)\subseteq M$, and $y=g\circ\tilde{y}$, restricting the domain of $g$, if necessary. This establishes the first part of our assertion. Next we need to show that $f\vert_U$ is an embedding. But $f\vert_U=y^{-1}\circ\iota\circ x\vert_U$, which is a composition of embeddings, and is thus an embedding itself.

If $f$ is already an embedding at the outset, then $f(U)=f(M)\cap W$ for some open $W\in N$. In this case, with $V$ as in the theorem, we have

$\displaystyle f(M)\cap V=\{q\in V\vert y^{m+1}(q)=\dots=y^n(q)=0\}$

That is, there is always a set of local coordinates in $N$ so that the image of $M$ is locally the hyperplane spanned by the first $m$ of them.

April 19, 2011