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Whilst I was trying to think of a proof for the chain rule for Fréchet derivatives, I realized it looks very similar to the naturality axiom for functors. (Except for the need to specify points for the derivative.)

$$ D(f \circ g)_{p} = (Df)_{g(p)}\circ (Dg)_p \sim F(f \circ g) = F(f) \circ F(g) $$

Is this just a coincidence? Or is this hinting at some deeper meaning of derivatives?

user577413
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    You might find https://ncatlab.org/nlab/show/chain+rule illuminating. – Ittay Weiss Jul 20 '18 at 09:34
  • Yes, I get the gist of it, but I’m afraid I have not reached manifolds yet in my reading – user577413 Jul 20 '18 at 09:51
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    So, basically your observation is certainly onto something. That something is investigated in category theoretic circles, as the nLab page indicates, and is a good starting point if you're interested in digging in deeper. – Ittay Weiss Jul 20 '18 at 09:59
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    Let $\mathtt{Man}$ be the category of smooth manifolds and smooth maps between manifolds. Then the assignment $M \mapsto TM$ taking a manifold $M$ to its tangent bundle $TM$ can be extended to a functor $\mathtt{Man} \to \mathtt{Man}$. How does it act on arrows? It takes a smooth map $f : M \to N$ to its derivative $Df$, which naturally defines a smooth map $Df : TM \to TM$. Indeed, the chain rule really does become $D(f \circ g) = Df \circ Dg$ as maps between tangent bundles. – Branimir Ćaćić Jul 20 '18 at 11:41
  • If $D$ were a functor then the identity function $Id;x=x$ would be preserved, however $D(x) = 0$, that is $D,Id = $ constantly zero function. As such, it might not be a functor on functions, or possibly using a different notion of equality. – Musa Al-hassy Jul 20 '18 at 12:48
  • @MusaAl-hassy I think the identity is preserved, according to Branimir’s explanation. – user577413 Jul 20 '18 at 14:12
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    @MusaAl-hassy In this context, the derivative of $f : \mathbb{R}^m \to \mathbb{R}^n$ at a point $x$ is its best approximation by a linear transformation $Df(x) : \mathbb{R}^m \to \mathbb{R}^n$ at that point, in the sense that $|f(x+h)-f(x)-Df(x)(h)| = o(|h|)$; from this perspective, the derivative of the identity $I : \mathbb{R}^n \to \mathbb{R}^n$ is the identity itself, i.e., for every $x \in \mathbb{R}^n$, $DI(x) = I$, which means, in the language of vector bundles, that the derivative of the identity $I : \mathbb{R}^n \to \mathbb{R}^n$ is the identity $T\mathbb{R}^n \to T\mathbb{R}^n$. – Branimir Ćaćić Jul 20 '18 at 15:28
  • cool stuff, I didn't know that – Musa Al-hassy Jul 21 '18 at 21:40

2 Answers2

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The way I prefer to say things is this. There is a functor from, say, the category of pointed smooth manifolds to the category of vector spaces, which takes a smooth map $f : (M, m) \to (N, n)$ (so $m \in M, n \in N$ and $f(m) = n$) to the derivative $df_m : T_m(M) \to T_n(N)$. The fact that this respects composition is precisely the chain rule. There are many variations on this.

Qiaochu Yuan
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If you're not ready to think about manifolds, you can still think of the derivative as a functor on the category whose objects are $\mathbb{R}^m$ and whose morphisms are smooth maps. The (total) derivative of $f:\mathbb{R}^m\to \mathbb{R}^n$ is $Tf:\mathbb{R}^{2m}\to \mathbb{R}^{2n}$, sending $(\vec x,\vec y)$ to $(f(\vec x),Df_{\vec x}(\vec y))$. This is the specialization of the notion of "tangent bundle" mentioned in the comments to Euclidean spaces.

Kevin Carlson
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