Cone of complex and isomorphism

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?

• 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. May 10, 2020 at 19:41
• 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? May 10, 2020 at 20:18
• 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 May 11, 2020 at 10:00

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. $$i^*(\hat{\omega}) = \omega$$.