[[Differential geometry MOC]]
# Immersion and submersion

A $C^k$ [[Differentiability|differentiable]] function $f : X \to Y$ between $C^k$ [[Differentiable manifold|differentiable manifolds]] $X,Y$ of dimensions $n,m$ respectively 
is **(im/sub)mersive** at $x \in X$ iff the [[Differential pushforward]] $T_{x}f : T_{x}X \to T_{f(x)}Y$ at $x$ is a linear ([[linear monomorphism|mono]]/[[Linear epimorphism|epi]])morphism.
Such function is said to be an **(im/sub)mersion** iff it is an (im/sub)mersion everywhere. #m/def/geo/diff

An immersion may be thought of as a map which _locally_ resembles the **canonical immersion** defined for $n \leq m$ as
$$
\begin{align*}
i : \mathbb{R}^n &\to \mathbb{R}^n \times \mathbb{R}^{m-n} \\
x &\mapsto(x;0)
\end{align*}
$$
whereas submersion may be thought of as a map which _locally_ resembles the **canonical submersion** defined for $n \geq m$ as
$$
\begin{align*}
j : \mathbb{R}^m \times \mathbb{R}^{n-m} &\to \mathbb{R}^m \\
(x;y) &\mapsto x
\end{align*}
$$
This point of view is justified by the [[#Local (im/sub)mersion theorem]].

## Local (im/sub)mersion theorem

Let $f : X \to Y$ be a $C^\infty$ [[Differentiability|differentiable]] map between $C^\infty$ [[Differentiable manifold|differentiable manifolds]] $X,Y$ of dimension $n,m$ respectively,
and let $f : x \mapsto y$.
Then $f$ is an (im/sub)mersion at $x$
iff there exist $C^\infty$ [[Coördinate chart|charts]] $\varphi : U \to V$ on $X$ about $x$ and $\tilde{\varphi} : \tilde{U} \to \tilde{V}$ on $Y$ about $y$ 
with $f(U) \sube \tilde{U}$ 
such that $\psi f \varphi^{-1}$ is a restriction of the canonical (im/sub)mersion, #m/thm/geo/diff 
i.e. in the immersion case
$$
\begin{align*}
\tilde{\varphi} f \varphi^{-1}(v) = i(v) = (v;\vab 0)
\end{align*}
$$
and in the submersion case
$$
\begin{align*}
\tilde{\varphi} f\varphi^{-1}(v;w) = j(v;w) = v
\end{align*}
$$

> [!check]- Proof
> Without loss of generality, we can consider $X$ and $Y$ to be open subsets of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively,
> since we are only interested in local properties and locally these are diffeomorphic.
> We also assume $n \neq m$, since otherwise this is a special case of the [[Inverse function theorem]].
> 
> Assume $f$ is an immersion at $x$.
> Let $W = Df(x)(\mathbb{R}^n) \leq \mathbb{R}^m$ where $\dim W = n$.
> Choose some [[complement subspace]] $W^c$ where $\dim W^c = m-n$.
> We can then define
> $$
> \begin{align*}
> F : X\times W^c &\to \mathbb{R}^m \\
> (\xi; w) &\mapsto f(\xi) + w
> \end{align*}
> $$
> which has the [[total derivative]]
> $$
> \begin{align*}
> DF(x,0) : \mathbb{R}^n \times W^c &\to \mathbb{R}^m \\
> (\vab a, \vab b) &\mapsto Df(x_{0})(\vab a)+\vab b
> \end{align*}
> $$
> which is a [[Linear isomorphism]],
> so by the [[Inverse function theorem]] $F$ is locally a diffeomorphism.
> Thus taking the canonical immersion $i : X \to X \times W^c = \mathbb{R}^m$, we have $Fi = f$, as required.
> 
> Now assume $f$ is a submersion at $x$.
> Let $K = \ker Df(x)$ where $\dim K = n-m$ by the [[Rank-nullity theorem]]
> and let $P : \mathbb{R}^n \twoheadrightarrow K$ be a [[projection operator]] onto $K$
> (where we make the natural identification of $\mathbb{R}^n$ with $T_{x}X$).
> We can then define
> $$
> \begin{align*}
> F : X &\to \mathbb{R}^m \times K \\
> \xi &\mapsto (f(\xi), P(\xi))
> \end{align*}
> $$
> which has the total derivative
> $$
> \begin{align*}
> DF(x) : \mathbb{R}^n &\to \mathbb{R}^m \times K \\
> \vab a &\mapsto (Df(x)\,\vab a, P\,\vab a)
> \end{align*}
> $$
> which is a [[Linear isomorphism]],
> so by the [[Inverse function theorem]] $F$ is locally a diffeomorphism.
> Thus taking the canonical submersion $j : \mathbb{R}^m \times K \to \mathbb{R}^m$,
> we have $jF = f$, as required.
> 
> For the converses note, that the composition of (im/sub)mersions is an (im/sub)mersion,
> and coördinate charts are diffeomorphisms and hence both immersions and submersions. <span class="QED"/>

## Properties

- Iff $f$ is immersive at $x$, then $\rank(T_{x}f)=n \leq m$
- Iff $f$ is submersive at $x$, then $\rank(T_{x}f)=m \leq n$


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