Euler Characteristic of a boundary of a Manifold

Question

I need some guidance in understanding a specific passage of the following result taken from [tom Dieck Algebraic Topology, page 456]

Proposition 18.6.2. Let $B$ be a compact $(n+1)$-manifold with boundary $\partial B=M$. Then $\chi(M)=(1+(-1)^n)\chi(B)$. In particular $\chi (M)$ is always even.

Proof Let $W=B \cup (\partial B \times [0,1)\ )$. Then $B$ is a compact deformation retract of $W$ and $W\setminus B = \partial B \times (0,1) \simeq \partial B$. Hence $$\chi(B)=\chi(W)=\chi(W\setminus B)+(-1)^{n+1}\chi(B)$$ therefore we get $\chi(B)=\chi(\partial B)-(-1)^n\chi(B)$, which implies the result we are looking for.

My question: I don't have the slightest idea about how to justify the identity $$ \chi(B)=\chi(W)=\chi(W\setminus B)+(-1)^{n+1}\chi(B) $$ in particular, how do I get the $(-1)^{n+1}$ sign in front of $\chi(B)$? Clearly $W \neq W\setminus B \sqcup B$ from a topological viewpoint, and therefore cannot be the simple addition between the two characteristics. The only other result I know is the inclusion-exclusion principle, but I don't know if it can be applied here, and moreover I don't see how it would give us the right factor in front of $\chi(B)$.

Addendum It was suggested to me to use M-V, and I considered the l.e.s. $$ \to H_{i+1}(W) \to H_i(\partial B) \to H_i(W\setminus B)\oplus H_i(B) \to H_i(W) \to $$ which with the use of the retraction, can be rewritten as $$ \to H_{i+1}(B) \to H_i(\partial B) \to H_i(\partial B)\oplus H_i(B) \to H_i(B) \to $$ and then using this exact sequence to relate the Euler characteristic leads, according to me to the triviality $0=0$ due to the fact that everything cancels out right hand side. Is there something wrong in this reasoning?

Any advice or idea would be helpful

Answer 1

Don't let the fancy formula fool you. It's only a compact way to say what is a common result about Euler Characteristic: Let $B$ a $n+1$-dim manifold with boundary. $\partial B$ is a $n$-dim.manifold.

Now consider the M-V sequence for the triad $(2B,B,\partial B)$ , where $2B$ is a $n+1$ manifold without boundary obtained by glueing two copies of $B$ along $\partial B$. After some standard computations, we have the following $$\chi(2B)=2\chi(B)-\chi(\partial B)$$

• If $n+1$ is odd, $\chi(2B)=0$ (by P-D), and hence as your formula says $$2\chi(B)=\chi(\partial B)$$
• if $n+1$ is even, then apply the first point to $\partial B$ which is a odd dimensional manifold.

NB i just hide under the carpet all orientations-related problem. if $2B$ happens to be non orientable, just work with the orientation cover of it! (and recall the relation between coverings and Euler characteristic)