J. Milnor: smooth manifold triad vs cobordism

cobordismdiffeomorphismdifferential-topologysmooth-manifolds

I started studying Milnor's "Lectures on the H-cobordism theorem". On page 2 he provides the following definitions

Definition 1.3. $(W;V_0,V_1)$ is a smooth manifold triad if $W$ is a compact smooth manifold and $\partial W$ is the disjoint
union of two open and closed submanifolds $V_0$ and $V_1$.

My first question is: what exactly is meant by saying "open and closed submanifolds $V_0$ and $V_1$"?

My second question: Everywhere i looked up the definition of a cobordism it was equivalent to Milnor's definition of smooth manifold triads? What am i missing?

Wikipedia says:

A cobordism between manifolds M and N is a compact manifold W whose
boundary is the disjoint union of M and N, $\partial W = M \sqcup N$

Isn't this exactly what milnor defines as a smooth manifold triad?

Milnor then continues with

Definition 1.5 Given two closed smooth $n$-manifolds $M_0$ and $M_1$ (i.e. $M_0,M_1$ compact, $\partial M_0 = \partial M_1 =
\emptyset$
) a cobordism from $M_0$ to $M_1$ is a $5$-tuple
$(W;V_0,V_1;h_0,h_1)$ where $(W;V_0,V_1)$ is a smooth manifold triad
and $h_i:V_i \to M_i$ is a diffeomorphism, $i=0,1$.

Am i correct if i conclude "a smooth manifold triad corresponds up to diffeomorphism to a cobordism"? Does this explain the equivalent definitions of smooth manifold triads and cobordisms?

Thanks for any help.

Best Answer

To answer your first question, each of $V_0,V_1$ is required to be a submanifold of $\partial W$, and an open subset of $\partial W$, and a closed subset of $\partial W$. This is equivalent to the statement that each is a union of components of $\partial W$ (because each component of $\partial W$ is itself open and closed in $\partial W$).

As for your other questions, you are not missing anything in particular. Milnor is being more careful to insist that $V_0$ and $V_1$ actually "be" part of a triad, but not that they actually be part of the cobordism. To say that the two definitions are exactly the same is not exactly correct.

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