What is a “smooth map”?

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I was trying to start reading John Milnor’s Topology from the Differentiable Viewpoint, but something on the first page really threw me.  I think I figured it out now though, so let me explain it to you too.

Milnor writes at the very beginning of the book:

FIRST let us explain some of our terms. R^k denotes the k-dimensional euclidean space; thus a point x \in R^k is an k-tuple $latex  x = (x_1, \ldots , x_k)$ of real numbers.

Let U\subset R^k and V \subset R^l be open sets. A mapping f from U to V (written f : U \rightarrow V) is called smooth if all of the partial derivatives \partial^n y/\partial x_{i_1},\cdots\partial x_{i_n} exist and are continuous.

More generally let X \subset R^k and Y \subset R^l be arbitrary subsets of euclidean spaces. A map f : X \rightarrow Y is called smooth if for each x \in X there exist an open set U \subset R^k containing x and a smooth mapping F : U \rightarrow R^l that coincides with f throughout U \cap X.

The last part is what threw me.  It really isn’t obvious what the consequences are of the general definition of a smooth map on an arbitrary subset of R^k.  There are two other ways I can think of defining this notion:

  1. A map f : X \rightarrow Y is smooth iff for there exists an open set V \subset R^k including X and a smooth mapping F : V \rightarrow R^l whose restriction to X is f.
  2. I’ll just give this second definition in the case of k=1.  It’s kind of complicated to extend it to larger k, and it’s unnecessary for what comes next.  Here it is:  For a map f : X \rightarrow Y, define g(x)=\lim_{y\in X-\{x\}, y\rightarrow x}\frac{f(y)-f(x)}{y-x}, if this limit exists.  The function g is basically the derivative of f, except that the limit is taken only over points in X.  Later we will just call it the derivative.  Now we define f to be smooth iff the limit defining g(x) exists for all x\in X and g:X\rightarrow R is continuous.

Are these two definitions equivalent to Milnor’s definition?

Well, the first definition is. It clearly implies Milnor’s if we just choose V=U for every x\in X.  To prove the converse is more tricky.  I found on a very questionable source that the way to prove this is using partitions of unity.  To understand this, I read about partitions of unity in Spivak’s Calculus on Manifolds.  Basically, partitions of unity can help you stitch together the many functions F to make one big function.  They might disagree on some points, but you can kind of take the average of the disagreeing values to get the value of the big function.  See this Math.SE question for more information.

The second definition isn’t.  Milnor’s definition implies it, but it doesn’t imply Milnor’s.  I found a counterexample which you can scroll down and read right now if you want, or you can continue reading about my thought process that led me to it.  First of all, Milnor’s definition implies it because Milnor’s definition implies my definition 1, which clearly implies it.  Now the counterexample.

How can we choose X to make a counterexample?  I want to make a stronger counterexample than necessary: I want f to be continuously differentiable on X, but have no extension to an open set containing X which is even continuous, not to mention continuously differentiable.  I think a stronger counterexample like this will be easier to find.  Well, X clearly can’t be open.  Also, X can’t be closed, because then it could be continuously extended to R^k by the Tietze extension theorem.  Furthermore, X can’t be closed in any open set V, because then it could be continuously extended to V.  Thus, it can’t be the intersection of a closed set and an open set.

But we need even more than this.  We need there to be an x\in X such that for all open neighborhoods U of x, U\cap X is not closed in U.  Otherwise, we could pick a neighborhood U_x of each x\in X such that U_x\cap X is the closed in U_x, and then we could extend f|_{U_x\cap X} to U_x using the Tietze extension theorem.  Then we would have satisfied Milnor’s definition.

I thought about how to do this.  We need it to be so no matter how small we make the neighborhood U of x, we still find limit points missing from X\cap U.  How about X=[0,1)-\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\}?  This works: in every neighborhood of 0, there are limit points \frac{1}{n} missing from X.  Now we can define a function on X.  We define it piecewise:

For \frac{1}{2}<x<1, let f(x)=\frac{1}{4}(x-\frac{1}{2}).

For \frac{1}{3}<x<\frac{1}{2}, let f(x)=\frac{1}{9}(x-\frac{1}{3}).

For \frac{1}{4}<x<\frac{1}{3}, let f(x)=\frac{1}{16}(x-\frac{1}{4}).

In general, for \frac{1}{n}<x<\frac{1}{n-1}, let f(x)=\frac{1}{n^2}(x-\frac{1}{n}).

And, let f(0)=0.

This function is continuous at x\in X, >0 because it is linear on a neighborhood of x.  It is continuous at 0 because its value gets arbitrarily close to 0 as x approaches 0.  It is differentiable at x\in (\frac{1}{n},\frac{1}{n-1}) with derivative equal to \frac{1}{n^2}.  It is differentiable at 0 because \frac{f(h)}{h} approaches 0 as h approaches  0 (while staying in X of course).  Its derivative is clearly continuous for x\in X, >0 because here it is constant on a neighborhood of x.  And 0\leq f'(x)\leq x^2, so f' is continuous at 0 as well.

However, f does not continuously extend to an function g on an open set V.  If it did, then V would have to include a neighborhood of 0, and this neighborhood would include infinitely many points \frac{1}{n}.  The function g would have to assign a value to these points, and there is no way of doing this continuously, since at such points the left- and right-hand limits are unequal.  Infinitely many jump discontinuities would be inevitable.

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