Why is the boundary of an oriented manifold with its (opposite oriented) copy the empty set? Excuse the very basic question:
I'm following Milnor's Lectures on Characteristic Classes.  He defines a relation on the collection of compact, smooth, oriented manifolds of dimension $n$ by letting $M_1\sim M_2$ if $M_1\sqcup -M_2$ is a boundary (If $M$ is a manifold, then $-M$ is the same manifold with the opposite orientation).  He uses this to define cobordism classes and further goes on to define a graded ring structure on these classes.  He says clearly the relation is reflexive and symmetric.  I see symmetric, but I'm not entirely sure why this relation is reflexive.  Since we're supposed to have a group structure, and the $0$ element is the empty set, I'm assuming that the boundary of $M\sqcup -M$ is the empty set.  Why is it that $\partial (M\sqcup -M)=\emptyset $ ?  Do the two manifolds somehow cancel each other out since they have opposite orientations?  How can I picture this?

You're a bit confused about what this definition means.  First of all, the $n$-manifolds we're considering here all do not have boundary.  So $M$ should be an $n$-manifold without boundary, and you shouldn't be thinking about its boundary at all.
The relation $M_1\sim M_2$ means that $M_1\sqcup -M_2$ is the boundary of some $(n+1)$-manifold with boundary.  So to prove $M\sim -M$, you want to prove that $M\sqcup -M$ is the boundary of some $(n+1)$-manifold.  To prove this, you can just take the $(n+1)$-manifold to be $N=M\times[0,1]$.  The boundary of $N$ is $M\times \{0\}\cup M\times\{1\}$, where $M\times\{1\}$ has the given orientation of $M$ and $M\times\{0\}$ has the opposite orientation, so $\partial N\cong M\sqcup -M$ as an oriented manifold.
