[Math] Euler Characteristic of a boundary of a Manifold

algebraic-topologymanifolds-with-boundaryproof-verification

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

Best Answer

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)

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