let $M$ be a smooth manifold with boundary $\partial M= X\times F$, where $X$ and $F$ are smooth manifolds, $F$ compact. The embedding $i : \partial M\longrightarrow M$ induces the restriction $$i^{*} : \Omega^{*}(M)\longrightarrow \Omega^{*}(\partial M) $$ The projection $\pi : \partial M \longrightarrow X$ induces the mapping $$\pi_{*} : \Omega^{*}(\partial M)\longrightarrow) \Omega^{*-\nu}(X), \quad dimF=\nu $$ which takes the forms to their integrals along the fiber $F$. Consider the morphism $$\alpha: (\Omega^{*}(M),d)\longrightarrow (\Omega^{*-\nu}(X),d)$$ of de Rham complexes $M$ and $X$ and denote the cone of this morphism by $$\Omega^{k}(M,\pi) = \Omega^{k}(M)\oplus \Omega^{k-\nu-1}(X),\quad \partial = \begin{pmatrix} d & 0 \\ -\alpha & -d \end{pmatrix}.$$ let $\quad \Omega^{*}_{0}(M) = \ker \alpha$. How to prove that this inclusion $\quad r : \Omega^{*}_{0}(M)\longrightarrow \Omega^{k}(M,\pi)\quad $ induces isomorphism in cohomology groups?

  • $\begingroup$ This question would be more appropriately tagged [algebraic-topology] instead of [algebraic-geometry]. I have adjusted the tags on this question, and I would ask you to look out for this going forwards. $\endgroup$
    – KReiser
    May 10, 2020 at 19:41
  • $\begingroup$ There are couple of problems in your question. First, you may want to assume $F$ being compact to allow integration along the fiber. Second, in the boundary morphism of mapping cone, you may want to specify $\alpha=\pi_*i^*$. Lastly, it does not seems to me that your claim is true in general. Can you provide a reference? $\endgroup$
    – AG learner
    May 10, 2020 at 20:18
  • $\begingroup$ AG, Hello! I found a lemma saying that: if a morphism of chain (cochain) complex $f$ is surjective, then $cone(f)$ and $\ker f$ have the same cohomology groups $\endgroup$
    – Ady Fall
    May 11, 2020 at 10:00

1 Answer 1


As you noted in the comments, if a morphism is surjective, then its kernel is quasi-isomorphic to what you call its "cone" (I would call it the homotopy kernel). It's not too hard to prove. Suppose $f : X \to Y$ is a surjective cochain map and consider the inclusion $\iota : \ker f \to \operatorname{cone} f$.

  • $\iota_*$ is injective. Let $x \in \ker f$ be a cocycle ($dx = 0$), such that $\iota_*[x] = 0$, i.e. $\iota(x) = (x,0) = d\alpha$ for some $\alpha = (x',y') \in \operatorname{cone} f$. But $d\alpha = (dx', f(x') + dy')$ so $(x,0) = d\alpha$ implies that $x = dx'$ is therefore a coboundary, i.e. $[x] = 0$.
  • $\iota_*$ is surjective. Let $(x,y) \in \operatorname{cone} f$ be a cocycle, i.e. $d(x,y) = (dx, f(x) + dy) = (0,0)$. Since $f$ is surjective, we can find $x' \in X$ such that $f(x') = y$. Therefore $$(x,y) = (x - dx',0) + d(x',0) \implies [(x,y)] = \iota_*[x-dx'].$$

So now the question is why your map $\alpha = \pi_* i^*$ is surjective: it's because both $\pi_*$ and $i^*$ are surjective.

  • $\pi_*$ is surjective. Suppose $\omega \in \Omega^*(X)$ is some form. Let $\mathrm{vol}_F$ be a volume form for $F$. Then $\omega = \pi_*(\omega \times \mathrm{vol}_F)$.
  • $i^*$ is surjective. You can find a collar around $\partial M$, i.e. a smooth embedding $g : \partial M \times [0,1) \to M$ such that $g(\partial M \times \{0\}) = \partial M$. Let $U \cong \partial M \times [0,1)$ be the image of $g$ which is open in $M$. Find some open set $\partial M \subset V \subset U$ such that $\bar{V} \subset U$. Define a bump function $\rho : M \to [0,1]$ such that $\rho_{| \partial M} = 1$ and $\rho$ vanishes outside $V$.
    Now let $\omega \in \Omega^*(\partial M)$ be some form. Consider $\rho \cdot \omega \times 1 \in \Omega^*(\partial M \times [0,1)) = \Omega^*(U)$. Its support is contained in $V$, so you can extend it by zero outside $U$ to get a well-defined form $\hat{\omega} \in \Omega^*(M)$ which restricts to $\rho \cdot \omega \times 1$ on $U$. In particular, it restricts to $\omega$ on $\partial M$, i.e. $i^*(\hat{\omega}) = \omega$.

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