Abstract Nonsense

Crushing one theorem at a time

Smooth Maps and the Category of Smooth Manifolds (Pt. II)


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

\text{ }

Of course, most of the standard results about smooth functions in the Euclidean case transfer over to this general case. In particular:

\text{ }

Theorem: Let M_1,M_2,M_3 be smooth manifolds and M_1\xrightarrow{f}M_2\xrightarrow{g}M_3 maps. Let p\in M_1, if f is smooth at p, and if g is smooth at f(p), then g\circ f is smooth at p.

Proof: Let (U_1,\varphi_1), (U_2,\varphi_2) and (U_3,\varphi_3) be charts at p,f(p),g(f(p)) respectively with f(U_1)\subseteq U_2, g(U_2)\subseteq U_3. We know that (\varphi_3\circ g\circ\varphi_2^{-1})\circ (\varphi_2\circ f\circ\varphi_1^{-1}) is smooth, being the composition of smooth maps in Euclidean spaces. But, we see that this is equal to \varphi_3\circ(g\circ f)\circ\varphi_1^{-1} and thus it’s easy to see that the charts (U_1,\varphi_1) and (U_3,\varphi_3) is a good chart at p and g(f(p)) which has smooth coordinate representation. \blacksquare

\text{ }

Theorem: Let f:M\to N be a smooth map between manifolds, then f is continuous.

Proof: Let p\in M be arbitrary, then we know that there is a neighborhood (U,\varphi) at p and (V,\psi) at f(p) with \varphi(U)\subseteq V such that \widehat{f} is smooth. That said, since smooth maps between Euclidean spaces are continuous we see that \widehat{f} is continuous. Since \widehat{f} and f differ only by homeomorphisms, we see that f is continuous on U. Since p was arbitrary the conclusion follows. \blacksquare

\text{ }

The last theorem says that (just like continuous functions [note that this was used in the last proof!]) smoothness is a local property. In particular:

\text{ }

Theorem: Let \{(U_\alpha,\varphi_\alpha)\}_{\alpha\in\mathcal{A}} be an open cover for the smooth manifold M. Suppose that for each \alpha we have a smooth map f:U_\alpha\to N (thinking about U_\alpha as an open submanifold of M) where N is some smooth manifold. If f_\alpha and f_\beta agree on U_\alpha\cap U_\beta, then there exists a unique smooth map f:M\to N such that f\mid_{U_\alpha}=f_\alpha.

\text{ }

As a last note, we discuss the notion of local diffeomorphisms. In particular, if M and n are smooth manifolds and if f:M\to N is a map, we call it a local diffeomorphism if for every point p\in M there exists neighborhoods U of p and V of f(p) so that f is a diffeomorphism of U onto V.

\text{ }

The Category of Smooth Manifolds and the Algebra of Smooth Functions

\text{ }

Now that we have defined the appropriate notion of smooth maps between smooth manifolds we can discuss the category \mathbf{Man} of smooth (finite dimensional) manifolds. In particular, we define \mathbf{Man} to be the category whose objects are smooth (finite dimensional) manifolds and whose arrows are smooth maps between these manifolds. In fact though, it is actually more convenient for us to, at this current point, consider the category \mathbf{TopMan} of topological manifolds, thought of as a full subcategory of \mathbf{Top}. We shall make the convention that \text{Hom}_{\mathbf{Man}}(M,N) shall be denoted C^\infty(M,N) and \text{Hom}_{\mathbf{TopMan}}(M,N) shall be denoted C(M,N).

\text{ }

It is easy to get rid of some of the simple properties of the category \mathbf{TopMan}. Indeed, we claim that the product of topological manifolds with the usual projections is actually the categorical product in \mathbf{TopMan}. Indeed, this is obvious since the product in \mathbf{Top} stays within in \mathbf{TopMan}. Now, we similarly claim that the product of smooth manifolds with the usual projections is the categorical product in \mathbf{Man}. Indeed, let M,N_1,N_2 be three smooth manifolds and suppose that we have smooth maps f_i:M\to N_i, we need to produce a unique smooth map f:M\to N_1\times N_2 such that \pi_i\circ f=f_i (where \pi_i is the usual projection). It’s clear that such a smooth map is necessarily unique, since there is a unique continuous map which satisfies this property (via the fact that this definition of product is a categorical product in \mathbf{TopMan} which is a supercategory of \mathbf{Man}). Thus, it suffices to show that there exists a smooth map f which satisfies these properties. Unsurprisingly, we define f(x)=(f_1(x),f_2(x)). We know this map is continuous, and thus it suffices to show it’s smooth. To do this, let for any point p\in M, (U,\varphi) and (V_i,\psi_i) (in M, and N_i and at p and f_i(p) respectively) be charts such that \varphi(U)\subseteq V_i And \psi_i\circ f\circ\varphi^{-1} is smooth. We then see that (U,\varphi) and (V_1\times V_2,\psi_1\times\psi_2) is a chart at p and f(p) respectively such that f(U)\subseteq V_1\times V_2. Moreover, we see that

\text{ }

(\psi_1\times\psi_2)\circ f\circ\varphi^{-1}=(\psi_1\circ f_1\circ\varphi^{-1})\times(\psi_2\circ f_2\circ\varphi^{-1})

\text{ }

which is clearly a smooth map between Euclidean spaces. The conclusion follows.

\text{ }

Sadly enough, the category \mathbf{Man} does not, in general, have colimits. This is somewhat of a touchy subject, and so I instead refer you to this mathoverflow thread which gives a good explanation as for why this is true.

\text{ }

\text{ }

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.

Advertisements

September 3, 2012 - Posted by | Manifold Theory, Topology | , , , , ,

1 Comment »

  1. […] Smooth Maps and the Category of Smooth Manifolds (Pt. III) Point of Post: This is a continuation of this post. […]

    Pingback by Smooth Maps and the Category of Smooth Manifolds (Pt. III) « Abstract Nonsense | September 3, 2012 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: