# The Unapologetic Mathematician

## Submersions

Another quick definition: we say that a smooth map $f:M^m\to N^n$ is a “submersion” if it is surjective, and if every point $p\in M$ is a regular point of $f$. Despite the similarity of the terms “immersion” and “submersion”, these are very different concepts, so be careful to keep them separate.

The nice thing about submersions is that every value of $N$ is a regular value, and every one has a nonempty preimage. Thus our extension of the implicit function theorem applies to show that $f^{-1}(q)$ is an $m-n$-dimensional submanifold of $M$.

One obvious example of submersion is a projection from a product manifold. As we’ve seen, the determinant of this projection is always a surjection. In fact, it’s a projection itself.

Another example is the projection from the tangent bundle $\mathcal{T}M$ down to its base manifold $M$. Indeed, given any tangent vector $v$ at $p$ we can pick a coordinate patch $U\subseteq M$ around $p$ and the corresponding patch $\pi^{-1}(U)$ of $\mathcal{T}M$. Within these coordinates we can easily calculate the derivative of $\pi$ and see that it’s just a projection onto the first $\dim(M)$ components, which is surjective. In this case, the preimages $\pi^{-1}(p)$ are the stalks of the tangent bundle $\mathcal{T}_pM$.

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May 2, 2011 - Posted by | Differential Topology, Topology

## 2 Comments »

1. I am not used to the requirement that f be itself surjective, only its derivative. Do you just include it to avoid empty preimages of regular values?

Comment by Landau | May 2, 2011 | Reply

2. Well, even if $f$ isn’t surjective it’s surjective onto its image, which is then a submanifold, and this restriction of the domain is a submersion. I’d say it’s more to make submersion more fully the dual of immersion.

Comment by John Armstrong | May 2, 2011 | Reply