I have just come across the definition of a cobordism between two closed oriented $n$-dimensional manifolds $M,M'$ as an oriented $(n+1)$-dimensional manifold $W$ with boundary such that $\partial W = M \sqcup -M'$.
If $M$ is a closed oriented $n$-dimensional manifold then $(M \times [0,1]; M \times \{0\}, M \times \{1\})$ should be a cobordism between $M$ and itself.
What orientation is $[0,1]$ given here? Thinking of the orientation as an arrow, I thought the arrow should point from $\{0\}$ to $\{1\}$ so that the induced orientation on the boundary of $[0,1]$ will be $+1$ at $\{1\}$ and $-1$ at $\{0\}$. However this seems to produce a cobordism $(M \times [0,1]; M \times \{1\}, M \times \{0\})$ and not $(M \times [0,1]; M \times \{0\}, M \times \{1\})$. Should the arrow in fact point from $\{1\}$ to $\{0\}$ to produce $(M \times [0,1]; M \times \{0\}, M \times \{1\})$ or am I thinking about something incorrectly?
Gosh. I'd never considered this, but it seems as if you're correct. But since $M \times [0,1]'$ is a cobordism in the other direction (where $[0,1]'$ just means "reverse orientation"), it's kinda moot. (And $[0, 1] \times M$ would work too).
By the way, the part where we write $$ \partial W = M \sqcup -M' $$ is usually taken to mean that $\partial M$ is homoe/diffeomorphic to the union of the two manifolds, not literally equal. So in your example, you don't need to write $M \times \{0\}$, since there's an obvious diffeomorphism to just $M$.