Excision for Relative de Rham Cohomology

Question

I recently learned in Bott-Tu about the notion of relative de Rham cohomology, which is defined as follows:

If $M$ is a smooth manifold and $S\subset M$ is its (embedded) submanifold, we define the cochain complex $\Omega^*(M,S)$ by $$\Omega ^q(M,S)=\Omega ^q(M)\oplus\Omega ^{q-1}(S),\,d(\omega,\theta )=(d\omega,\omega\vert_S-d\theta).$$ The cohomology group of this cochain complex is called the relative de Rham cohomology group and are denoted by $H^q(M,S)=H^q_{dR}(M,S)$.

The cochain complex $\Omega^*(M,S)$ is just the mapping cylinder of the cochain map $\Omega^*(M)\to \Omega^*(S)$ induced by the inclusion, so there is a long-exact sequence $$\cdots\to H^q(M,S)\to H^q(M)\to H^q(S)\to H^{q+1}(M,S)\to\cdots.$$ Also, it is clear that if $M,N$ are smooth manifolds and $S\subset M$ and $N\subset R$ are their submanifolds, every smooth map $f:M\to N$ such that $f(S)\subset R$ induces a cochain map $\Omega^*(N,R)\to\Omega^*(M,S)$ by pullback, and hence linear maps between the relative cohomology groups. So the relative de Rham cohomology behaves pretty much like the relative singular cohomology. I, therefore, suspected that the two groups are actually isomorphic; but I had no clue how to prove it, so I lowered my sights and asked whether excision holds. By excision, I mean the following:

If $U,V\subset M$ are open sets and $M=U\cup V$, the inclusion $\iota:(U,U\cap V)\to (M,V)$ induces isomorphisms $$\iota^*:H^q(M,V)\xrightarrow{\cong}H^q(U,U\cap V).$$

Proving excision first looked very doable, for I could easily prove the injectivity of $\iota^*$ by using a partition of unity. But I got stuck at proving surjectivity. Does anyone know whether $\iota^*$ is surjective, and if it is surjective, how to prove it?

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Some observations

• The long exact sequence for relative de Rham cohomology arises from the short exact sequence $$0\to \Omega^{*-1}(S)\to \Omega^*(M,S)\to \Omega^*(M)\to 0,$$ whereas the singular version arises from the ses of the form $$0\to S^*(M,S;\mathbb{R})\to S^*(M;\mathbb{R}) \to S^*(S;\mathbb{R})\to 0.$$ Thus the long exact sequence of pairs is not available. (If it were the case, five lemma will immediately imply that "singular relative= relative de Rham")
• Because of this, if one is to prove that "singular relative= relative de Rham" via the five lemma, we need to construct a homomorphism $I:H^p_{dR}(M,S)\to H^p(M,S;\mathbb{R})$ (the RHS is the singular relative cohomology). One obvious candidate is $$I:[\omega,\theta]\mapsto\left([\sigma]\mapsto \int _\sigma \omega\right).$$ This, however, is not well-defined. Indeed, let $\alpha\in\Omega^1(M)$ be any 1-form such that $d\alpha\neq 0$ at some $x\in M\setminus S$. Because $[d(\alpha,0)]=0$ in $H^0_{dR} (M,S)$, one should have $I[d(\alpha,0)]=0$. But then $\int _x d\alpha=d\alpha(x)=0$, contrary to our hypothesis.

Answer 1

The relative de Rham cochain complex $Ω(M,S)$ is simply the homotopy fiber of the pullback map $Ω(M)→Ω(S)$ (computed as the mapping cocone).

Likewise, the relative singular cochain complex $S(M,S)$ is the homotopy fiber of the restriction map $S(M)→S(S)$ (computed as the kernel of a surjective map).

Thus, the natural weak equivalence of functors $Ω→S$ induces a natural weak equivalence (possibly a zigzag) $Ω(M,S)→S(M,S)$ of homotopy fibers, as desired.