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Suppose $ f:N\rightarrow M$ is a smooth map between two manifolds. Relative de Rham cohomology is defined through the complex $ \Omega^{q}(f)=\Omega^{q}(M)\oplus\Omega^{q-1}(N)$ with $d(\omega,\theta)=(d\omega,f^{*}\omega-d\theta)$.

I'm trying to show that this relative cohomology is homotopy-invariant as in if $ f $ and $ g $ are homotopic maps from $ N $ to $ M $, the induced relative cohomologies are isomorphic algebras.

I've been trying to define a chain map between the following exact sequence $$0\rightarrow\Omega^{q-1}(N)\rightarrow\Omega^{q}(f)\rightarrow\Omega^{q}(M)\rightarrow 0$$ and the corresponding one for $g$, and after that use naturality and five lemma. The maps in the short exact sequences are inclusion and projection. I haven't been able to get this to work so far. Could you please help?

Edit: If you have another method to prove the same result, I'm also interested in that. The book I'm reading (Bott - Tu) hasn't introduced singular cohomology at this point, so I cannot use it.

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It will be true that $f^*\omega=g^*\omega$ as cohomology classes, since homotopic maps induce the same maps on de Rham cohomology to begin with. Are you trying to prove this as an independent result? – Connor Jun 3 '14 at 15:19
You can look at the homotopy $F:N \times [0,1] \rightarrow M$ and consider the mapping cone complex corresponding to this map $\Omega(F)$ and show that it has the same homology as the mapping cone complex for $f$ and $g$. You won't get a map between $\Omega(f)$ and $\Omega(g)$. To avoid having to deal with manifolds with boundary you can do this for singular homology instead. – apurv Jun 3 '14 at 17:22
@Connor I know that $ f $ and $ g $ induce the same map on cohomology. How can I use this to prove the isomorphism? – PeterM Jun 4 '14 at 0:44
@PeterM Sorry, I may have missed that you wanted an actual isomorphism of algebras. If $f^*=g^*$, then you'll get the same boundary map and hence the same chain complex and cohomologies. As in the approach that apurv suggested, this also won't give you an actual map between the two. – Connor Jun 4 '14 at 0:52
@Connor I'm sorry if I'm not making myself clear. I'm just interested in proving the existence of an algebra isomorphism using what the book has introduced so far. Constructing an explicit isomorphism is a plus but not necessary. $ f $ and $ g $ are the same on cohomology but not chains. Which boundary map do you mean? Would you mind spelling out the details? – PeterM Jun 4 '14 at 9:28
  • Suppose given a diagram of cochain complexes $$\begin{array} AA & =& A \\ \downarrow{f} & & \downarrow{g} \\ B & \stackrel{h}{\longrightarrow} & C. \end{array} $$
    If $h$ is a homotopy equivalence, with homotopy inverse $h'$ also commuting with $f$ and $g$, then the induced map on mapping cones $$\mathrm{cone}(f) \xrightarrow{h_*} \mathrm{cone}(g)$$ is a homotopy equivalence with homotopy inverse $h'_*$. (Use a homotopy $1_B \simeq hh'$ to construct a homotopy $1_{\mathrm{cone}(f)} \simeq h_*h'_*$.)
  • Now consider homotopic maps $f, g : N \to M$ of smooth manifolds, with a homotopy $$K: f \to g : N \times I \to M.$$ Let $i_0$ and $i_1$ be the inclusions $N \to N \times I$ of the fibres over $0$ and $1$ respectively, so that $i_0^*K = f$ and $i_1^*K = g$. By the homotopy invariance of usual de Rham cohomology, the maps $i_0^*$ and $i_1^*$ in the diagram below $$\begin{array} A \Omega_M & =& \Omega_M & = & \Omega_M\\ \downarrow{f^*} & & \downarrow{K^*} & & \downarrow{g^*} \\ \Omega_N & \stackrel{i_0^*}{\longleftarrow} & \Omega_{N \times I} & \stackrel{i_1^*}{\longrightarrow} & \Omega_N \end{array} $$ are homotopy equivalences of cochain complexes. Therefore, using a homotopy inverse of $i_0^*$, we obtain a diagram

    $$\begin{array} A\Omega_M & =& \Omega_M \\ \downarrow{f^*} & & \downarrow{g^*} \\ \Omega_N & \stackrel{i}{\longrightarrow} & \Omega_N. \end{array} $$
    in which $i$ is a homotopy equivalence. By the previous bullet, this implies that we have a homotopy equivalence $\mathrm{cone}(f^*) \simeq \mathrm{cone}(g^*)$, and taking cohomology we have the result.

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