Relative de Rham Cohomology is Homotopy Invariant

Question

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.

Answer 1

• 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.

Answer 2

I'm a little late to answer this, but I'm working through Bott and Tu myself now and had the exact same issue, so I figure why not post.

As you say, we want to use the five lemma on this diagram: $$\require{AMScd}\begin{CD} \rightarrow H^{q-1}(M) @>>f^*> H^{q-1}(N) @>>> H^{q}(f) @>>> H^{q}(M) @>>f^*> H^{q}(N) \rightarrow\\ @| @| @VV?V @| @| \\ \rightarrow H^{q-1}(M) @>>g^*> H^{q-1}(N) @>>> H^{q}(g) @>>> H^{q}(M) @>>g^*> H^{q}(N) \rightarrow\\ \end{CD}$$ We know that $f^* = g^*$, but we need a well defined map $H^{q}(f) \to H^{q}(g)$. As you point out in the comments, the chain level identity map does not send closed forms to closed forms. But we can create such a map using a fixed homotopy $F: N \times I \to M$. Let $i_0, i_1: N \to N \times I$ be the inclusions to $N \times \{0\}$ and $N \times \{1\}$ respectively, so that $F \circ i_0 = f$ and $F \circ i_1 = g$. Also let $p_0: N \times I \to N$ be the projection onto $N \times \{0\}$.

Now suppose $(\omega, \theta) \subset \Omega^q(M) \oplus \Omega^{q-1}(N) = \Omega^q(f)$ is closed, so $i_0^* F^* \omega = f^* \omega = d\theta$. Since $i_0$ is a homotopy equivalence, $F^*\omega$ is exact as well; write $F^* \omega = d\Theta$ for some $\Theta \in \Omega^{q-1}(N \times I)$. Next, define $\Theta' = \Theta - p_0^*(i_0^*\Theta - \theta).$ Note that $$d\Theta' = d\Theta - p_0^*(i_0^* d\Theta - d\theta) = d\Theta - p_0^*(i_0^*F^*\omega - f^*\omega) = d\Theta = F^*\omega \tag{1}$$ and $$i_0^*\Theta' = i_0^*\Theta - i_0^*p_0^*i_0^*\Theta + i_0^*p_0^*\theta = \theta, \tag{2}$$ since $p_0 \circ i_0$ is the identity. Modulo exact forms, $\Theta'$ is the unique $(q-1)$-form on $N \times I$ satisfying (1) and (2). That's because (1) determines $\Theta'$ up to closed forms, and (2) fixes the cohomology class of $i_0^*\Theta' - \theta$. Finally we can define our map $H^q(f) \to H^q(g)$ by $[(\omega, \theta)] \mapsto [(\omega, i_1^*\Theta')]$.

This is well defined. Trivially the square to the right of the arrow in the diagram above commutes. For the box on the left, note that if $\omega = 0$ and $\theta$ is closed, then we may take $\Theta' = p_0^*\theta$, and then our map becomes $[(0,\theta)] \mapsto [(0,\theta)]$. The five lemma gives the result.

Answer 3

There is no need to do anything weird here (like the five lemma)

It's a simple consequence of prop6.49's long exact sequence:

$$...\rightarrow H^q(f)\rightarrow H^q(M)\rightarrow H^q(S)\rightarrow H^{q+1}(f)\rightarrow ...$$

Remember long exact sequence can be split into short exact sequences which must split because we are dealing with vector spaces here. We have $H^q(S)=coker f^\ast\oplus ker f^\ast$, here $f^\ast$ is the pullback map $H^\ast(M)\rightarrow H^\ast(S)$. Same with $g^\ast$.