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The pushforward of maps between smooth manifolds is defined as follows:

If $f: M \to N$ and $a \in C^\infty(N)$, then $Tf: TM \to TN$ takes $v \mapsto Tf(v)$ which operates on functions on $N$ as $Tf(v)(a) = v(f^*a) = v(a \circ f)$.

It can be shown that $T(g \circ f) = T(g) \circ T(f)$, where $f: M \to N$ and $g: N \to P$. This is one of the necessary criteria for $T$ to be a functor from the category of smooth manifolds to the category of smooth tangent bundles. The proof requires nothing more than moving some symbols around according to the definition given above. No further assumptions about $M$ or $C^\infty(M)$ are required.

But the functoriality of $T$ is just an abstract generalized statement of the chain rule of calculus, which states that the derivative of a composition of functions is the composition of the derivatives, evaluated at the appropriate points. In particular, by setting $M = N = P = \mathbf{R}$, then $f$ and $g$ become real functions of one variable, and the functoriality becomes $f(g(x))' = f'(g(x))g'(x)$. So it contains the standard chain rule of calculus as a special case.

Is this a valid proof of the chain rule? It seems to me that the proof of the chain rule requires an analytic argument resting on analytic properties of the real numbers. There should be no way around this requirement. The generalization of the chain rule should not be able to escape the requirements and give you the proof of the chain rule "for free".

Is there an assumption about the underlying spaces in the proof of functoriality that I've missed? Or does category theory really make such an important calculus trivial?

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    You're assuming a lot of background about the construction of the tangent bundle. Probably all of the analytic arguments are there?2011-09-22
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    Let me state what Qiaochu said in a different way. In a situation like this, all category theory gives is a definition (namely, functorality). You still have to prove that the objects in question satisfy this definition. In the case of the tangent bundle, any such verification is going to use the chain rule. However, it is worth noting that the chain rule is a pretty trivial result -- if you think of a derivative as a linear approximation, all it is saying is that the linear approximations compose like linear maps. And the proof is basically following your nose.2011-09-22
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    @AdamSmith Yes, that's exactly what I thought! The proof of functoriality must somehow get dirty with the calculus and invoke the chain rule. But it doesn't do so!2011-09-22
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    But there's a certain amount of work that goes into showing that derivations at a point of $\mathbf R^n$ and the old directional derivative are the same, right? No such thing as a free lunch :)2011-09-22
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    @DylanMoreland, yes, that work is there if you're interested. But it's not required (you can define your vectors as derivations and never mention directional derivatives). And anyway, is that work supposed to somehow bake the chain rule into the smooth map of manifolds? I can't see how...2011-09-22
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    @ziggurism: okay, but if you define vectors as derivations and plan to actually _do anything_ with them, at some point you have to exhibit nontrivial derivations. This requires actual work in that there exist $\mathbb{R}$-algebras with no nontrivial derivations. You can't get away with abstract nonsense forever.2011-09-22
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    It's not just that you have to exhibit some nontrivial derivations, you have to actually show that this agrees with the obvious tangent bundle on $\mathbb{R}^n$. Otherwise, you can't prove any of the theorems of local differential topology or geometry, and if you can't do that, what's the point of defining the tangent bundle?2011-09-22
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    @ziggurism: oh, and it also occurred to me that I don't know how to show that the composition of smooth functions is smooth (which you assume) without the chain rule.2011-09-22
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    It sounds like everybody's saying that the analytic information content of the chain rule is somehow included in some level of the definition of manifolds. Either it's in the directional derivative, or the composition of maps, or somewhere else. I find this answer plausible, but I'm somehow not entirely persuaded.2011-09-23

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