Does de Rham theorem hold for manifolds with boundary?

I am following the J.Lee's book "Introduction to Smooth Manifolds", 2nd ed., page 480-486 to learn the de Rham theorem. It is proven on manifolds without boundary, which makes me curious about whether it holds for manifolds with boundary. At least in the part proving de Rham cohomology is isomorphic to the $C^\infty$ singular cohomology, I did not see a step at first glance where the manifold in question must have no boundary. ("At first glance" means I have checked it did not cite theorems like Whitney approximation theorem that specifically requires the target to be without boundary, but I did not check if all the computation/formula really works for manifolds with boundary.) I also try to search for more sources, and what I have found only talk about the de Rham theorem in no-boundary case.

Thus I wonder if de Rham theorem holds for manifolds with boundary. If it fails, is it still true that de Rham cohomology is isomorphic to the $C^\infty$ singular cohomology?

2 Answers

Yes, it holds for manifolds with boundary. One way to see this is to note that if $M$ is a smooth manifold with boundary, then the inclusion map $\iota\colon \text{Int}\ M\hookrightarrow M$ is a smooth homotopy equivalence (Thm. 9.26 in ISM), which therefore induces isomorphisms on de Rham cohomology (Thm. 17.11). Since $\iota$ also induces isomorphisms on singular cohomology, and the de Rham homomorphism (integration over chains) commutes with inclusion, the result follows.

(It would have been a good idea to include this in the book, either as a corollary or as a problem. I don't know why I didn't.)

• This is a nice way to generalize without worrying about regularity and boundary. But just to understand more, does all the steps in the proof of de Rham's theorem work as well on manifolds with boundary? Jul 27 '14 at 19:49
• I'm not sure. One problem is that my proof of the isomorphism between smooth singular & singular cohomology uses the Whitney approx theorem for a map that's already smooth on a closed subset, and this doesn't generally work when the target has nonempty boundary. But this might be fixable. Jul 27 '14 at 21:16
• I think that the existence of the homotopy operator (Lemma 11.4 in your book) does not use that the manifold has no boundary. Therefore, homotopy invariance holds for manifolds with boundary too with the same proof. So, the deRham cohomology algebra of a manifold with boundary agrees to that of its interior. Since this is the case for the singular homology too, deRham theorem follows. Correct? Nov 3 '16 at 2:36

Ray and Singer prove a version of de Rham's Theorem for manifolds with boundary (Proposition 4.2 of R-torsion and the Laplacian on Riemannian manifolds, adv. in math. 1971). They need boundary conditions to establish the Theorem. Actually the previous answer looks incorrect. The result obtained by Ray and Singer is: Let W be an h-cobordism whose boundary is M_1 U M_2. Consider relative boundary conditions on M_1 and absolute boundary conditions on M_2. Denote by H the space of harmonic forms on W. Then H is isomorphic to H^q(W,M_1) (relative cohomology). You may also want to check expression (2.7) of W. Muller, on page 735 of "Analytic Torsion and R-torsion for unimodular representations", J. Amer. Math. Soc. 1993. Muller's statement does not require the manifold to be an h-cobordism. But I am not an expert, I was only searching for an alternative proof for the Theorem, perhaps in a book, since I was having difficulty to understand these references.

• Welcome to MSE. This should be a comment, not an answer. Sep 28 '19 at 10:51
• Actually the previous answer looks incorrect. Sep 29 '19 at 11:14