Orientation on the boundary

Question

If $M$ is an oriented without boundary manifold, and $\mu$ is it volume form, is true that the boundary of $M\times [0,1]$ is $ M \cup M$, right? It is true also that the orientantion on the boundary is $\mu$ for the superior part, If I can say this way, and $-\mu$ for the inferior part?

Answer 1

Yes, so proper way to formulate is to pick outward pointing vector fields on $\partial M\times I$. Let $\omega=\lambda\times\mu$ be the orientation on $M\times I$.

Then the orientation on $\partial M$ is defined to be $\omega(V_{out},-)$. In one case $\lambda(V_{out})$ is positeve and so we get back $\mu$ in the other is negative and we get $-\mu$ (if you use the standard orientation of $I$ it will be as you said $\mu$ on $M\times\{1\}$.