Double of a manifold and cobordism

Given a manifold $$M$$, the double of $$M$$ is the boundary of $$M \times [0,1]$$. This gives doubles a special role in cobordism.

1. What is this special role?
2. Moreover, is it true that the double of a manifold $$M$$ with boundary (which should be $$\partial M \times [0,1] \cup M \times \{0,1\}$$) is the boundary of $$M \times [0,1]$$?
• 1. $M \times [0,1]$ is the identity bordism from $M$ to $M$. Equivalently, $M \times [0,1]$ is a nullbordism of the double of $M$. Apr 3 at 1:11
• Thank you very much Apr 3 at 1:21

1. Yes, $$\partial(M \times [0,1]) = \partial M \times [0,1] \cup M \times \{0, 1\}$$. Picture a standard cylinder, $$D^2 \times [0,1]$$, whose boundary is a cylindrical tube $$S^1 \times [0,1]$$ together with the disks on either end, $$D^2 \times \{0, 1\}$$
• But how can I see it is the double of $M$? Don't I need to identify the two boundaries of $M$ in the definition of double of a manifold? Apr 3 at 1:30
• @CrashBandicoot Correct, to be precise this is homotopy equivalent to the double. The cylinder part $\partial M \times [0,1]$ is like a collar neighborhood of the gluing of the two copies along the boundary. If you contracted that down ($\partial M \times [0,1] \sim \partial M \times \{0\}$) then you get the double by your definition.