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This is a question about the definition of a smooth function in Guillemin and Pollack's "Differential Topology". G&P define all manifolds as objects embedded in $\mathbb{R}^N$ for some $N$, which (as a zillion people have reminded me) is not how mathematicians usually think of manifolds - nevertheless, this is a question about how things are done in G&P:

Right on the bottom of page 2 and top of page 3, G&P say: "a mapping $f$ is smooth if it has continuous partial derivatives of all orders. However, when the domain of $f$ is not open, on usually cannot speak of partial derivatives...So we adapt the open situation to more general spaces. A map $f:X \to \mathbb{R}^m$ defined on an arbitrary subset $X$ of $\mathbb{R}^n$ is called smooth if it may be locally extended to a smooth map on open sets; that is, if around each point $x$ in $X$ there is an open set $U$ in $\mathbb{R}^n$ and a smooth map $F: U\to\mathbb{R}^m$ such that $F$ equals $f$ on $U \cap X$".

So consider the map $f$ from the $x$-axis in $\mathbb{R}^2$ to $\mathbb{R}$ defined by $f(x,0) = 0$. Since the $x$-axis is not open in $\mathbb{R}^2$, we need to extend $f$ to a smooth function $F$ around (say) $(0,0)$. There are TWO ways to do this (at least): $F(x,y) = y$ and $F(x,y) = 0$ for all points in the plane. These two different extensions give different derivatives around $(0,0)$, so it would seem that when specifying an extension we also need to specify which onee, if we want to talk about "THE" derivative at a point in the domain of $f$.

Clearly I'm misunderstanding something basic - what is it?

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    $\begingroup$ Your observation about there being many extensions is correct. However, the authors do not attempt to define "THE" derivative of a function in that way, but rather whether or not the function is smooth. $\endgroup$ May 2, 2012 at 20:05
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    $\begingroup$ There is only a problem if G&P claim that a smooth function on a non-open set has well-defined derivatives. Can you point to a place where they do that? (I haven't looked at the book myself.) If $X$ is an embedded submanifold without boundary, the correct statement would be that for any two extensions, the derivatives (considered as linear maps from $\mathbb{R}^n$ to $\mathbb{R}^m$) have the same restriction to the tangent space of $X$. If $X$ is more complicated then it is harder to make a clean statement, but perhaps G&P do not need one. $\endgroup$ May 2, 2012 at 20:08
  • $\begingroup$ Thanks for the comment, Neil. Well, later in the book "the" derivative of a smooth function from one manifold to another is an object of study, and the authors talk about f'(x) all the time. For example, if x is a point in a manifold X (embedded, since that's all they consider) and f is a smooth function from X to another manifold Y, f'(x) is a map that takes the tangent space of X at x into the tangent space of Y at f(x), so they seem to be claiming that there's one natural derivative to consider. $\endgroup$ May 2, 2012 at 20:30
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    $\begingroup$ @Bruce: you've missed a key detail to the definition in G&P. The derivative is a map from the tangent bundle of one manifold to the tangent bundle of the other. i.e. at each point it maps the tangent space of one to the tangent space of the other. G&P only use extendability as a convienient technical tool to set up the derivative. But the derivative itself does not depend on the choice of extension. This follows from the definition of a manifold and the chain rule. $\endgroup$ May 2, 2012 at 20:39
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    $\begingroup$ Also, on page 86, figure 2-19, there are a giraffe, a moose, and an elephant. On page 125, in a further development of Moose Theory, they discuss hot fudge topping and the oozing trajectory. $\endgroup$
    – Will Jagy
    May 2, 2012 at 20:58

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when you read a little further, page 8 or so, you will find that the derivative of a smooth map is a linear map defined only on vectors tangent to the manifold. it will turn out that all extensions of the smooth function act the same on vectors tangent to the manifold. In your example both extensions y and 0 act as the zero map on vectors tangent to the axis. so it works out ok. – roy smith 2 days ago

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