# The Unapologetic Mathematician

## 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 »

1. […] We know that a map induces a chain map , which induces a map on the de Rham cohomology. This is what we mean when we say that de Rham cohomology is functorial. […]

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