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 
As usual, we calculate:
\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}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cleft%5B%5Cleft%5B%5Cleft%5Bg%5E%2A%5Ccirc+f%5E%2A%5Cright%5D%28%5Comega%29%5Cright%5D%28p%29%5Cright%5D%28v_1%2C%5Cdots%2Cv_k%29%26%3D%5Cleft%5B%5Cleft%5Bg%5E%2A%28f%5E%2A%5Comega%29%5Cright%5D%28p%29%5Cright%5D%28v_1%2C%5Cdots%2Cv_k%29%5C%5C%26%3D%5Cleft%5B%5Cleft%5Bf%5E%2A%5Comega%5Cright%5D%28g%28p%29%29%5Cright%5D%28g_%2Av_1%2C%5Cdots%2Cg_%2Av_k%29%5C%5C%26%3D%5Cleft%5B%5Comega%28f%28g%28p%29%29%29%5Cright%5D%28f_%2Ag_%2Av_1%2C%5Cdots%2Cf_%2Ag_%2Av_k%29%5C%5C%26%3D%5Cleft%5B%5Comega%28%5Cleft%5Bf%5Ccirc+g%5Cright%5D%28p%29%29%5Cright%5D%28%28f%5Ccirc+g%29_%2Av_1%2C%5Cdots%2C%28f%5Ccirc+g%29_%2Av_k%29%5C%5C%26%3D%5Cleft%5B%5Cleft%5B%28f%5Ccirc+g%29%5E%2A%5Comega%5Cright%5D%28p%29%5Cright%5D%28v_1%2C%5Cdots%2Cv_k%29%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\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 
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