# The Unapologetic Mathematician

## Smooth Maps

As usual, we’re going to want our objects of study — smooth (or differentiable) manifolds — to be objects in a category. And a category means we need morphisms. The morphisms between smooth manifolds are smooth maps.

Given two smooth manifolds, $M^m$ and $N^n$, a continuous map $f:M\to N$ is smooth at a point $p\in M$ if we can find a coordinate patch $U\subseteq M$ containing $p$ and a coordinate patch $V\subseteq N$ containing the image $f(U)$ so that the composite function $\phi_V\circ f\circ\phi_U^{-1}$ is smooth as a function from $\phi_U(U)\subseteq\mathbb{R}^m$ into $\phi_V(V)\subseteq\mathbb{R}^n$. We say it’s smooth if it’s smooth at all points.

But wait, maybe we just got lucky when we picked these coordinate patches. Well, it actually doesn’t matter. If $f$ is smooth according to one pair of coordinate patches, it’s smooth according to any other pair. Indeed, if we take another set of coordinates $(U',\phi_{U'})$ around $p$ then the compatibility condition says that the transition function $\phi_U\circ\phi_{U'}^{-1}$ is smooth. And then so is the composite:

$\displaystyle\left(\phi_V\circ f\circ\phi_U^{-1}\right)\circ\left(\phi_U\circ\phi_{U'}^{-1}\right)=\phi_V\circ f\circ\phi_{U'}^{-1}$

But this is just the smoothness condition in terms of $U'$ and $V$. Similarly, if we change coordinates in the target from $V$ to $V'$, the compatibility condition says that $\phi_{V'}\circ\phi_V^{-1}$ is smooth, and so the composite

$\displaystyle\left(\phi_{V'}\circ\phi_V^{-1}\right)\circ\left(\phi_V\circ f\circ\phi_U^{-1}\right)=\phi_{V'}\circ f\circ\phi_U^{-1}$

is smooth as well. The condition for smoothness at a point, therefore, really only depends on the behavior of the function near that point, and not on what particular coordinates we use to attest to its smoothness.

In particular, let’s consider what it means for a function to be smooth from a manifold $M^m$ to the real space $\mathbb{R}^n$. In this case, we can choose the entire space with the identity function as the coordinate patch on the target manifold. Thus a function $f:M\to\mathbb{R}^n$ is smooth at a point $p$ if there is some $U\subseteq M$ containing $p$ with coordinate map $\phi_U:U\to\mathbb{R}^m$ such that the composite $f\circ\phi_U^{-1}:\phi_U(U)\to\mathbb{R}^n$ is smooth.

Finally, just like we have the fancy word “homeomorphism” for isomorphisms of topological spaces, we have the fancy word “diffeomorphism” for isomorphisms of differentiable manifolds.

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

1. John, I am so happy that you have decided to write about differential geometry. I have been studying Diff. Geom. for a year now. I have been using Do Cormo’s Diff. Geom of Curves and surfaces and O’Neill’s Intro to D. Geom.
I look forward to your perspective on this subject.

Comment by Jonathan | March 7, 2011 | Reply

2. […] specifically, a partition of unity is a collection of nonnegative smooth functions indexed by some set , subject to two conditions. First: the collection of supports is a locally […]

Pingback by Partitions of Unity « The Unapologetic Mathematician | March 7, 2011 | Reply

3. […] from the additive group of real numbers to the diffeomorphism group of . Indeed, each is a diffeomorphism of — a differentiable isomorphism of the manifold to itself — with […]

Pingback by One-Parameter Groups « The Unapologetic Mathematician | May 31, 2011 | Reply

4. […] say we have a diffeomorphism from one -dimensional manifold to another. Since is both smooth and has a smooth inverse, we must […]

Pingback by Integrals and Diffeomorphisms « The Unapologetic Mathematician | September 12, 2011 | Reply

5. […] been using up until now don’t really match up to this picture. Homeomorphism — or diffeomorphism, for differentiable manifolds — is about having continuous maps in either direction, but […]

Pingback by Homotopy « The Unapologetic Mathematician | November 29, 2011 | Reply