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The accepted answer to this question contains the following statement:

If $M$ is a compact $d$-dimensional manifold with nonempty boundary, then $H^d(M)=0$.

This does not look obvious to me, so I thought it was worth asking. Can you give a proof of this statement (or, at least, a reference)?

Alessio Di Lorenzo
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  • https://math.stackexchange.com/questions/604044/top-homology-of-an-oriented-compact-connected-smooth-manifold-with-boundary –  Sep 21 '17 at 20:50
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    It would be nice if one could give an informal proof of this statement without the use of Poincaré-Lefschetz Duality, like Lavinias answer to the question posted by John on the same statement for homology. – H1ghfiv3 Sep 21 '17 at 21:08
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    Okay, I am giving it a try. First, take a finite simplicial structure on $M$ and use the fact that de-rham cohomology is isomorphic to real simplicial cohomology. The top dimensional simplicial cohomology is the $\mathbb R$-vectorspace generated by the the maps $\alpha_i$, where $i$ runs over the number of all $n$-simplices and $\alpha_i$ assigns the $i$-th $n$-simplex $\omega_i$ the number $1$ and every other $n$-simplex the number $0$. – H1ghfiv3 Sep 21 '17 at 21:21
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    To show that $H^d(M) = 0$, it there suffices to show that each of the $\alpha_i$ is a coboundary, i.e, $\alpha_i$ can be described as a map on the $n-1$ simplices- In order to do that, try to carefully "invert" the argument used by Lavina in her proof for the vanishing of top homology, using the fact that collection of $n-1$ simplices on $\partial M$ give you enough freedom to do so. – H1ghfiv3 Sep 21 '17 at 21:22
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    Use the following idea to define a cochain $\beta_i$ with $\delta(\beta_i) = \alpha_i$. First, assign all the (oriented) $n-1$ simplices on $\omega_i$ numbers that add up to $1$ in total. The coboundary of the map defined this way is not necessarily $\alpha_i$, the problem is some $n$-simplices adjacent to $\omega_i$ will be assigned a non-zero value. Now how do you proceed ? :) – H1ghfiv3 Sep 21 '17 at 21:30

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As $M$ is compact and smooth, there exists a finite triangulation $T$ of $M$. The de Rham Theorem now tells you now that the de Rham cohomology of differential forms is isomorphic to the real simplicial cohomology on $M$ induced by $T$. Denote by $C^*_T(M,\mathbb R)$ the corresponding real simplicial cochain complex.

For a fixed (oriented) $d$-simplex $\omega$ of $T$, consider the associated dual cochain $\alpha \in C_T^d(M,\mathbb R)$ that evaluates to $1$ at $\omega$ and to $0$ at every other $d$-simplex. As $C_T^d(M,\mathbb R)$ is generated by the collection of all such dual cochains, showing that $\alpha$ is a coboundary will suffice to prove that $H^d(M) = 0$.

As $M$ can WLoG assumed to be path-connected, we can choose a sequence $\omega = \omega_1,\omega_2,\omega_3,\dots,\omega_n$ of distinct wandering oriented $d$ simplices, such that for $k=1,\dots,n$, $\omega_k$ doesn't share a face with any other member of this sequence other than with $\omega_{k-1}$ and $\omega_{k+1}$, such that two consecutive members $\omega_k$ and $\omega_{k+1}$ have precisley one d-1 face $\sigma_k$ in common, and such that $\omega_n$ has a face $\sigma_n$ lying in $\partial M \neq \emptyset$.

Define a cochain $\beta \in C^{d-1}_T(M,\mathbb R)$ by assigning the value $\pm 1$ to the (oriented) simplicies $\sigma_k$ for $k=1,\dots,n$ and $0$ to any other $d-1$ simplex. By choosing the right signs each time, one can guarantee that $\delta \beta (\omega_k) = 0$ for every $k \geq 2$ and $\delta \beta(\omega) = \delta \beta (\omega_1) = 1$. Moreover, from the definition of $\beta$ and the choice of our sequence $\omega_1,\dots,\omega_n$, we see that $\delta \beta = 0$ on any $d$-simplex other than $\omega$. Thus, $\delta \beta = \alpha$, so $\alpha$ indeed is a coboundary.

H1ghfiv3
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    The de Rham theorem and the existence of triangulations are results considerably more complex than Poincaré duality... – Mariano Suárez-Álvarez Sep 21 '17 at 23:06
  • You also need the fact that singular cohomology coincides with the simplicial cohomology for a specific triangulation. – Mariano Suárez-Álvarez Sep 21 '17 at 23:07
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    In order to even apply Poincaré-Lefschetz duality, you need to pass from de Rham cohomology to singular cohomology. Thus, you certainly need the de Rham theorem. Secondly, although the existence of a triangulation might be a deeper result than Poincaré-Lefschetz duality, simply applying the latter result without knowing all the machinery behind it doesn't give you any real understanding on why this result is true. My goal was to give a visual, combinatorial proof of this fact, and this is, as far as I know, best achieved when passing to the world of simplicial complexes. – H1ghfiv3 Sep 22 '17 at 06:13
  • Also, my proof works in the case when $M$ is not oriented. In this case, you cannot even apply Poincaré-Lefschetz duality. – H1ghfiv3 Sep 22 '17 at 06:50