cancellation theorem in h-cobordism

Question

I'm reading milnor's book h-cobordism, in beginning of the section cancellation theorem, milnor give an example that composition of two elementary cobordism with index $0$ and $1$ may be a product cobordism.

(the definition of elementary cobordism is given as following: for the triple$(W, V_0, V_1)$ with manifold with boundary $W$ and two components of boundarys $V_0$ and $V_1$, if there exists a morse function $W\to [0,1]$ with $f^{-1}(0)=V_0$ and $f^{-1}(1)=V_1$ and $f$ has only one critical value $a$, then we say the triple $(W, V_0, V_1)$ is elementary cobordism and $index(W, V_0, V_1)=indexf^{-1}(a)$).

My question is, for two elementary cobordism $c=(W_1, V_1, V_2)$ and $\hat{c}=(W_2, V_0, V_1)$ with index $\lambda$ and $\hat{\lambda}$ ($|\lambda-\hat{\lambda}|>1$). Can the composition $c•\hat{c}$ be product cobordism?

Answer 1

It can't be a product because the relative homology won't be trivial. If $c$ is a cobordism with a single critical point of index $\lambda$ then $H_*(c,V_0)$ is nontrivial only in degree $\lambda$. In your case of composing two cobordisms with critical points of separated index, the relative homology will be nontrivial in those two degrees because it will be impossible for the boundary to algebraically cancel. This is easiest to see using the Morse chain complex. (See https://en.wikipedia.org/wiki/Morse_homology)

When the two indices differ by one, it is possible for the algebraic boundary operator to map one generator to the other, so homology is not an obstruction to canceling the critical points.