The Unapologetic Mathematician

Mathematics for the interested outsider

De Rham Cohomology is Functorial

It turns out that the de Rham cohomology spaces are all contravariant functors on the category of smooth manifolds. We’ve even seen how it acts on smooth maps. All we really need to do is check that it plays nice with compositions.

So let’s say we have smooth maps f:M_1\to M_2 and g:M_2\to M_3, which give rise to pullbacks f^*:\Omega(M_2)\to\Omega(M_1) and g^*:\Omega(M_3)\to\Omega(M_2). All we really have to do is show that g^*\circ f^*=(f\circ g)^*, because we already know that passing from chain maps to the induced maps on homology is functorial.

As usual, we calculate:

\displaystyle\begin{aligned}\left[\left[\left[g^*\circ f^*\right](\omega)\right](p)\right](v_1,\dots,v_k)&=\left[\left[g^*(f^*\omega)\right](p)\right](v_1,\dots,v_k)\\&=\left[\left[f^*\omega\right](g(p))\right](g_*v_1,\dots,g_*v_k)\\&=\left[\omega(f(g(p)))\right](f_*g_*v_1,\dots,f_*g_*v_k)\\&=\left[\omega(\left[f\circ g\right](p))\right]((f\circ g)_*v_1,\dots,(f\circ g)_*v_k)\\&=\left[\left[(f\circ g)^*\omega\right](p)\right](v_1,\dots,v_k)\end{aligned}

as asserted. And so we get maps f^*=H^k(f):H^k(M_2)\to H^k(M_1) and g^*=H^k(f):H^k(M_3)\to H^k(M_2) which compose appropriately: H^k(g)\circ H^k(f)\to H^k(f\circ g).

July 23, 2011 Posted by | Differential Topology, Topology | 1 Comment

   

Follow

Get every new post delivered to your Inbox.

Join 388 other followers