# Abstract Nonsense

## Smooth Maps and the Category of Smooth Manifolds (Pt. I)

Point of Post: In this post we define what it means for a map between two manifolds to be smooth.

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Motivation

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We literally defined smooth manifolds to be the topological spaces where we will have a relatively sound meaning of what a “smooth map” is. Thus, it would seem that the first order of business is to fully define and explore this notion of smooth map. The basic idea though is precisely what we have said before. A map between smooth manifold will be smooth if it is smooth locally around each point and its image–when we think about the space locally as Euclidean space.

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The interesting part is that once we define smooth map we will then be able to define the category of (finite dimensional) smooth manifolds. We will then be able to discuss the functor which takes a smooth manifold to it’s algebra of smooth functions (for us, function will mean a map into $\mathbb{R}$). We will then be able to make sense of the following statement: the smooth structure of a manifold  is largely encoded in its algebra of smooth functions.

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Smooth Maps

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Let $M$ and $N$ be smooth manifolds of dimensions $m$ and $n$ respectively. Suppose then that we have a map $f:M\to N$. We call $f$ smooth at $p\in M$ if for every pair of charts $(U,\varphi)$ and $(V,\psi)$ at $p$ and $f(p)$ respectively, such that $f(U)\subseteq V$, one has that the map $\psi\circ f\circ\varphi^{-1}:\varphi(U)\to\psi(V)$ is a smooth map at $\varphi(p)$ in the usual Euclidean sense. We call the map $\psi\circ f\circ\varphi^{-1}$ the coordinate representation of $f$ with respect to $(U,\varphi)$ and $(V,\psi)$ and denote it $\widehat{f}_{(U,\varphi),(V,\psi)}$. It shall often be clear from context, or irrelevant which charts we are talking about, in which case we shall merely refer to the coordinate representation of $f$ and write $\widehat{f}$.

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We see that, of course, the statement that a map $f:M\to N$ is smooth at $p$ is equivalent to the statement that every coordinate representation $\widetilde{f}$ is smooth at $\varphi(p)$. This very nicely highlights the fact this definition mirrors exactly the intuition we had for what a smooth map on a topological manifold should be–one which, when we pretend that it’s locally defined on Euclidean space (this is the coordinate representation), that it’s smooth. That said, we defined smooth manifolds exactly so that we really only have to care about one coordinate representation–not all of them. This is formalized by the following:

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Theorem: Let $M$ and $N$ be smooth manifolds and $f:M\to N$ a map. Then, $f$ is smooth at $p\in M$ if and only if there exists a single coordinate representation of $f$ that is smooth at $\varphi(p)$.

Proof: Suppose that $(U,\varphi)$ and $(V,\psi)$ is the pair of charts for which $f$‘s coordinate representation is smooth. Now, let $(U_1,\varphi_1)$ and $(V_1,\psi_1)$ be another pair of charts with $\varphi(U_1)\subseteq V_1$. To prove that $\psi_1\circ f\circ\varphi_1^{-1}$ is smooth at $\varphi(p)$, it suffices to show that it is smooth in some neighborhood of $\in\varphi(U_1)$. That said, we know that on $\varphi_1(U\cap U_1)$ and $\psi(V\cap V_1)$ we have that $\psi_1\circ\psi^{-1}$ and $\varphi\circ\varphi_1^{-1}$ are smooth. Noting then that

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$\psi_1\circ f\circ\varphi_1^{-1}=(\psi_1\circ\psi^{-1})\circ(\psi\circ f\circ\varphi^{-1})\circ(\varphi\circ\varphi_1^{-1})$

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we see that $\psi_1\circ f\circ\varphi_1^{-1}$ is the composition of smooth maps $\varphi_1(U\cap U_1)\to\psi_1(V\cap V_1)$ and thus is smooth. But, since $\varphi_1(p)\in\varphi_1(U\cap U_1)$ we may conclude that $\psi_1\circ f\circ\varphi_1^{-1}$ is smooth at $\varphi_1(p)$ as desired. $\blacksquare$

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We call a map between smooth manifolds smooth if it is smooth at every point. A bijective smooth map which smooth inverse is called a diffemorphism, if there exists a diffeomorphism between smooth manifolds we say that they are diffeomorphic. It’s obvious that “is diffeomorphic to” is an equivalence relation on smooth manifolds.

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Let’s now look at some examples of smooth maps between manifolds.

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Let $\mathbb{S}^1$ and $\mathbb{R}$ have their standard smooth structures and define $f:\mathbb{R}\to\mathbb{S}^1$ by $f(t)=e^{it}$. To see this is smooth, choose $t\in\mathbb{R}$. If $t\notin \frac{\pi}{2}+2\pi \mathbb{Z}$ we know that $f(p)\ne (0,1)$ and so we know a chart for $p$ is $\mathbb{S}^1-\{(1,0)\}$ and $\displaystyle \varphi(x,y)=\frac{x}{1-y}$. Thus, taking any neighborhood $U$ of $t$ not intersecting $\displaystyle \frac{\pi}{2}+2\pi\mathbb{Z}$ we see that $\varphi\circ f:U\to\mathbb{R}$ is given by $\displaystyle s\mapsto \frac{\cos(s)}{1-\sin(s)}$ which is evidently smooth. If $t\in \frac{\pi}{2}+2\pi\mathbb{Z}$ we know a chart of $f(t)$ is $(\mathbb{S}-\{(0,-1)\},\psi)$ where $\displaystyle \psi(x,y)=\frac{x}{1+y}$. Now,  it’s easy then to see that if we take a neighborhood $V$ of $t$ not intersecting $\frac{3}{2}\pi+2\pi\mathbb{Z}$ we can see clearly see that $\psi\circ f:U\to\mathbb{R}$ is given by $\displaystyle s\mapsto \frac{\cos(s)}{1+\sin(s)}$ which is evidently smooth. Since $t$ was arbitrary we see that $f$ is indeed smooth.

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Now, let $U\subseteq\mathbb{R}^n$ be open. It’s easy to see that a map $f:U\to\mathbb{R}^m$ is smooth in the calculus sense if and only if it’s smooth in the manifold sense.

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We claim that for any $\delta>0$ one has that $\mathbb{R}^n$ is diffeomorphic to $B_\delta(0)\subseteq\mathbb{R}^n$. Indeed, define $f:B_\delta(0)\to\mathbb{R}^n$ by $\displaystyle x\mapsto\frac{x}{\delta^2-|x|^2}$. Evidently this map is smooth (since it is smooth in the usual calculus) sense, is bijective, and has smooth inverse.

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It’s easy to see that any smooth chart $(U,\varphi)$ on an $m$-dimensional manifold $M$ is a smooth map $U\to\mathbb{R}^n$, and is, in fact, a diffeomorphism $U\to\varphi(U)$.

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As a last example, we come back to an example we’ve discussed before. In particular, we’ve discussed that the smooth structures induced by the atlases $\{(\mathbb{R},\varphi)\}$ and $\{(\mathbb{R},\text{id})\}$ (where $\varphi(x)=x^3$) are different. That said, the manifolds aren’t that different since the map $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3$ is a diffeomorphism (where the domain $\mathbb{R}$ is the one with the non-standard structure). Indeed, it really suffices to note that $f\circ\varphi^{-1}=\text{id}$ and since this a coordinate representation of $f$ whose domain covers $\mathbb{R}$, we’re done.

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References:

[1] Lee, John M. Introduction to Smooth Manifolds. New York: Springer, 2003. Print.

[2] Lee, John M. Introduction to Topological Manifolds. New York: Springer, 2000. Print.

[3] Milnor, John Willard, and David W. . Weaver. Topology from the Differentiable Viewpoint. Charlottesville (Va.): University of Virginia, 1969. Print.

[4] Bredon, Glen E. Topology and Geometry. New York: Springer-Verlag, 1993. Print.