I heard in class that the trace operator (Sobolev context!) is neither surjective nor injective, and other properties that are of more importance I guess. The definitions were precisely as on Wikipedia.
I get that it is not surjective, but why is it not injective?
(Bonus question: Are these properties of importance when solving BVPs?)
The trace operator is linear, so if it vanishes any non-zero function it won't be injective. This is indeed so, with the trace-zero functions comprising an interesting vector space of functions. (Roughly speaking, these are the "nice" functions that vanish in the region to which our operator restricts attention.) It's interesting because it's useful; your suspicion they'd help solve BVPs is correct.