The Unapologetic Mathematician

Mathematics for the interested outsider

Functoriality of the Derivative

We’ve said that the tangent bundle construction is a functor with the derivative as the action on morphisms. But we haven’t actually verified that it obeys the conditions of functoriality.

First off, if f=1_M:M\to M is the identity map on a smooth manifold M, then f_*:\mathcal{T}M\to\mathcal{T}M should be the identity map between the tangent bundles. That is, at each point p we should have f_{*p}:\mathcal{T}_pM\to\mathcal{T}_pM the identity map on this vector space. And indeed, if we let (U,x) be any coordinate patch around p we know that the matrix of f_{*p} with respect to these local coordinates is the Jacobian of the coordinate function x\circ1_M\circ x^{-1}=1_{x(U)}. But this Jacobian is clearly the identity matrix, proving our claim.

More importantly, if f:L\to M and g:M\to N are two smooth maps, then their composition g\circ f is also smooth. Given a point p\in L we can define the derivatives f_{*p}:\mathcal{T}_pL\to\mathcal{T}_{f(p)}M, g_{*f(p)}:\mathcal{T}_{f(p)}M\to\mathcal{T}_{g(f(p))}N, and (g\circ f)_{*p}\mathcal{T}_pL\to\mathcal{T}_{g(f(p))}N. I say that g_{*f(p)}\circ f_{*p}=(g\circ f)_{*p}. And since this holds at every point we can write (g\circ f)_*=g_*\circ f_*, proving functoriality.

So, let’s take a vector v\in\mathcal{T}_pL and see what happens. Taking a test function \phi:N\to\mathbb{R} we calculate

\displaystyle\begin{aligned}\left[(g\circ f)_{*p}(v)\right](\phi)&=v\left(\phi\circ(g\circ f)\right)\\&=v\left((\phi\circ g)\circ f\right)\\&=\left[f_{*p}(v)\right](\phi\circ g)\\&=\left[g_{*f(p)}\left(f_{*p}(v)\right)\right](\phi)\\&=\left[\left[g_{*f(p)}\circ f_{*p}\right](v)\right](\phi)\end{aligned}

And so (g\circ f)_{*p}=g_{*f(p)}\circ f_{*p}, just as we claimed.

We should note, here, how this recalls the Newtonian notation for the chain rule, where we wrote \left[g\circ f\right]'(p)=g'(f(p))f'(p). Of course, multiplication is changed into composition of linear maps, but that little detail will be cleared up soon (if you don’t already see it).

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April 7, 2011 - Posted by | Differential Topology, Topology

2 Comments »

  1. [...] the last two are the constant maps with the given values. We can thus use the chain rule to calculate the derivatives of these [...]

    Pingback by The Tangent Space of a Product « The Unapologetic Mathematician | April 27, 2011 | Reply

  2. [...] chain rule lets us combine these two outer derivatives into [...]

    Pingback by Integral Curves and Local Flows « The Unapologetic Mathematician | May 28, 2011 | Reply


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