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

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

Now, what we’d like to discuss is the set 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) . It’s easy to see that this is an -algebra. Since the set of all functions (not necessarily smooth) is an algebra, it really only needs to be checked that constant functions are in , and that it is closed under addition and multiplication. Clearly constant functions are contained in , and multiplication and addition are fairly easy. Indeed, it’s trivial that the maps given by and are smooth. From this, we see that if are two smooth maps, then their sum and their product (defined pointwise) are smooth since and , and the composition of smooth maps is smooth.

Now, let be a topological manifold. We can define to be the set of all continuous maps , and similar to the case of smooth maps, it’s easy to show that this is a subalgebra of .

Now, suppose that and are smooth manifolds and a smooth map. There is an obvious way to define a map given by . This is clearly an algebra map since

where and is a constant map.

Following the exact same process, we see that if we have a continuous map this induces an algebra map .

It’s obvious that these association of spaces and maps are contravariant functors and . Indeed, it’s easy to see that these are precisely the contravariant Hom functors and respectively.

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

**Theorem: ***Let and be smooth manifolds and a continuous map. Then, if and only if is smooth.*

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

Conversely, suppose that the containment condition holds and let and let and be charts around respectively such that . We need to show that is a smooth map. That said, note that

if is an -dimensional manifold and . Now, we know that each is a smooth map and thus we know that is a smooth map . Thus, we see that is the product of smooth maps and thus a smooth map . We see then that , being equal to is a composition of smooth maps, and thus smooth.

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

**Theorem: ***Let and be smooth manifolds and a homeomorphism. Then, restricts to an isomorphism if and only if is a diffeomorphism.*

**Proof: **Since the assocation and is a functor, it follows from first principles that if is a diffeomorphism then is an isomorphism of algebras.

Conversely, suppose that restricts to an isomorphism . The fact that tells us that is smooth. That said, it’s easy to see and so and so is smooth.

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]).

**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.

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