# Computing relative de Rham cohomology?

Let $f: N \to M$ be a smooth map. Then in Bott and Tu they define a relative de Rham complex $\Omega^{\bullet}(f)$ where $\Omega^k(f) = \Omega^k(M) \oplus \Omega^{k-1}(N)$ with coboundary map given in terms of the ordinary exterior derivative $d$ by $(\omega, \theta) \mapsto (d\omega, f^*\omega - d\theta)$. The cohomology of this complex is denoted $H^q(f)$.

I am wondering if there are any examples (other than $f$ being the identity) where it is possible to compute $H^k(f)$ directly from the definition, without using the accompanying long exact sequence in cohomology? $$\ldots \to H^{k-1}(f) \to H^{k-1}(M) \to H^{k-1}(N) \to H^k(f) \to H^k(M) \to H^k(N) \to \ldots$$

As far as I can tell, Bott and Tu don't have any examples where they do this.

In particular, I am interested in using this for the embedding $i: \partial M \hookrightarrow M$ of the boundary into a manifold with boundary.

I am playing around with a simple toy example, with $M$ being the closed unit ball in $\mathbb{R}^n$, and $\partial M$ being the unit sphere, but unfortunately my algebraic topology background is a bit weak so I can't see how to compute without resorting to this long exact sequence.

• I don't think it's really possible. In general, computing (co)homology straight from the definitions is very hard, unless you're in very special cases. It's almost always necessary to go through "abstract" arguments (long exact sequences, homotopy invariance, Mayer-Vietoris...). – Najib Idrissi Apr 28 '16 at 8:21
• You could probably do the case you want ($H^n(i)$) by hand in much the same way one calculations $H^n(M)$ by hand. In general, whenever you can compute cohomology by hand, you can probably compute relative cohomology by hand. But there are not many cases in which this is true. – user98602 Apr 28 '16 at 14:06
• Have you tried the case of the projection map $M\times S^1\to M$, say? Maybe even with a very specific, simple $M$? – Ted Shifrin Apr 28 '16 at 16:44
• The first rule of the cohomology club is that one simply does not compute cohomology directly. – Mariano Suárez-Álvarez Apr 28 '16 at 19:10