# Abstract Nonsense

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

Point of Post: This is a continuation of this post.

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Now, what we’d like to discuss is the set $C^{\infty}(M)$ of all smooth functions (for us, when we remember, we shall obey history’s tough hand, and call smooth maps into the real line “functions” and reserve this word for this case) $M\to\mathbb{R}$. It’s easy to see that this is an $\mathbb{R}$-algebra. Since the set $\mathbf{Map}(M,\mathbb{R})$ of all functions (not necessarily smooth) $M\to\mathbb{R}$ is an algebra, it really only needs to be checked that constant functions are in $C^\infty(M)$, and that it is closed under addition and multiplication. Clearly constant functions are contained in $C^\infty(M)$, and multiplication and addition are fairly easy. Indeed, it’s trivial that the maps $+,\cdot:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ given by $\cdot(x,y)=xy$ and $+(x,y)=x+y$ are smooth. From this, we see that if $f,g:M\to \mathbb{R}$ are two smooth maps, then their sum $f+g$ and their product $fg$ (defined pointwise) are smooth since $f+g=+\circ(f\times g)$ and $fg=\cdot\circ(f\times g)$, and the composition of smooth maps is smooth.

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Now, let $M$ be a topological manifold. We can define $C(M)$ to be the set of all continuous maps $M\to\mathbb{R}$, and similar to the case of smooth maps, it’s easy to show that this is a subalgebra of $\text{Map}(M,\mathbb{R})$.

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Now, suppose that $M$ and $N$ are smooth manifolds and $f:M\to N$ a smooth map. There is an obvious way to define a map $f^\ast:C^\infty(N)\to C^\infty(M)$ given by $g\mapsto g\circ f$. This is clearly an algebra map since

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$((g+g')\circ f)(x)=(g+g')(f(x))=g(f(x))+g'(f(x))=(g\circ f)(x)+(g'\circ f)(x)$

$((gg')\circ f)(x)=(gg')(f(x))=g(f(x))g'(f(x))=(g\circ f)(x)(g'\circ f)(x)$

$(c\circ f)(x)=c(f(x))=c$

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where $g,g'\in C^\infty(M)$ and $c$ is a constant map.

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Following the exact same process, we see that if we have a continuous map $f:M\to N$ this induces an algebra map $f^\ast:C(N)\to C(M)$.

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It’s obvious that these association of spaces and maps are  contravariant functors $\mathbf{Man}\to\mathbb{R}\text{-}\mathbf{Alg}$ and $\mathbf{TopMan}\to\mathbb{R}\text{-}\mathbf{Alg}$. Indeed, it’s easy to see that these are precisely the contravariant Hom functors $C^\infty(\bullet,\mathbb{R})$ and $C(\bullet,\mathbb{R})$ respectively.

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Now, we can ask a very interesting question about how $C^\infty$ and $C$ react with one another. In particular, suppose that we have smooth manifolds $M$ and $N$ and a continuous map $f:M\to N$. We know that this induces an algebra map $f^\ast:C(N)\to C(M)$. Does this map $f^\ast$ restrict down to an algebra map $C^\infty(N)\to C^\infty(M)$? Of course, this will restrict down to an algebra map $C^\infty(N)\to C(M)$ and so the operative part of this question is whether $f^\ast(C^\infty(N))\subseteq C^\infty(M)$. This is not always true, and in fact, we actually have the following:

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Theorem: Let $M$ and $N$ be smooth manifolds and $f:M\to N$ a continuous map. Then, $f^\ast(C^\infty(N))\subseteq C^\infty(M)$ if and only if $f$ is smooth.

Proof: We have already discussed the fact that if $f$ is smooth, then necessarily the containment condition holds.

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Conversely, suppose that the containment condition holds and let $p\in M$ and let $(U,\varphi)$ and $(V,\psi)$ be charts around $p,f(p)$ respectively such that $f(U)\subseteq V$. We need to show that $\psi\circ f\circ\varphi^{-1}$ is a smooth map. That said, note that

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$\psi\circ f=(\psi_1\circ f)\times\cdots\times(\psi_n\circ f)$

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if $N$ is an $n$-dimensional manifold and $\psi_i=\pi_i\circ f$. Now, we know that each $\psi_i$ is a smooth map $N\to\mathbb{R}$ and thus we know that $\psi_i\circ f$ is a smooth map $M\to\mathbb{R}$. Thus, we see that $\psi\circ f$ is the product of smooth maps $M\to\mathbb{R}$ and thus a smooth map $M\to\mathbb{R}^n$. We see then that $\psi\circ f\circ\varphi^{-1}$, being equal to $(\psi\circ f)\circ\varphi^{-1}$ is a composition of smooth maps, and thus smooth. $\blacksquare$

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In fact, it’s easy to see from this that we have the following:

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Theorem: Let $M$ and $N$ be smooth manifolds and $f:M\to N$ a homeomorphism. Then, $f^\ast$ restricts to an isomorphism $C^\infty(N)\to C^\infty(M)$ if and only if $f$ is a diffeomorphism.

Proof: Since the assocation $M\mapsto C^\infty(M)$ and $f\mapsto f^\ast$ is a functor, it follows from first principles that if $f$ is a diffeomorphism then $f^\ast$ is an isomorphism of algebras.

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Conversely, suppose that $f^\ast$ restricts to an isomorphism $C^\infty(N)\to C^\infty(M)$. The fact that $f^\ast(C^\infty(N))\subseteq C^\infty(M)$ tells us that $f^\ast$ is smooth. That said, it’s easy to see $(f^{-1})^\ast=(f^\ast)^{-1}$ and so $(f^{-1})^\ast(C^\infty(M))\subseteq C^\infty(N)$ and so $f^{-1}$ is smooth. $\blacksquare$

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We thus see that, in a very precise sense, the notion of a differentiable structure on a topological manifold is characterized by its algebra of smooth functions. In fact, some authors define a smooth manifold as being a topological manifold equipped with a particular type of subalgebra of the continuous real-valued functions on that manifold (c.f. [4]).

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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.