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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? – Qiaochu Yuan Sep 22 '11 at 4:21
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. – Adam Smith Sep 22 '11 at 4:25
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 :) – Dylan Moreland Sep 22 '11 at 4:35
@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. – Qiaochu Yuan Sep 22 '11 at 4:49
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? – Adam Smith Sep 22 '11 at 5:02

Let $M = N = \mathbb R$, and consider $f: M \to N$, so $f$ is a real-valued function of a real variable. Take $v = \partial_x$, based at $x_0$. Then $Tf(v)(a) = \partial_x(a\circ f)(x_0),$ for any $a:N \to \mathbb R$.

To identify this, we apply the chain rule to compute that $\partial_x(a\circ f)(x_0) = (\partial_x a)(f(x_0))\cdot f'(x_0),$ to conclude that $Tf(v)$ is equal to $f'(x_0)\partial_x$, based at $f(x_0)$.

It seems to me that you need this calculation in order to derive the classical chain rule from the the functorial version, and of course I used the classical chain rule in its derivation. Am I missing something?

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I'm not certain that I need to know how to break up the Tf(v)(a) into derivatives of f and a, in order to prove functoriality of T. – ziggurism Sep 23 '11 at 0:30
@ziggurism: Dear ziggurism, I agree. But to relate the general functoriality statement to the classical chain rule, I think you do need to so break things up. (In your post, you stated that the general functorial chain rule implies the classical chain rule, but you didn't say how. The suggestion of my answer is that if you try to carry this derivation through, you will find yourself having to use the classical chain rule to do so.) Regards, – Matt E Sep 23 '11 at 2:30
So if I'm understanding you right, you're saying that I don't get the chain rule until I express my function in local coordinates, and I can't do so without invoking the standard chain rule. Is that right? – ziggurism Oct 8 '11 at 15:06
Dear ziggurism, Yes, more or less. (It's not so much about expressing your function in local coordinates, as that the classical chain rule is about differentiation with respect to the global coordinate $x$ on $\mathbb R$.) Rather than trying to think about my answer, you could just try yourself to derive the classical chain rule from the functorial statement, and check whether my claim is correct, or whether I'm blundering. Regards, – Matt E Oct 9 '11 at 2:40

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