Given two disjoint circles $a$ and $b$ that are projected in a way so that there's a positive and negative $a$-over-$b$ crossing, Wirtinger's presentation gives that the generators $a$ and $b$ commute in the link group, even though the group in actuality is $\mathbb{Z}*\mathbb{Z}$.
Have I made a mistake or have misinterpreted something here, or is it simply that Wirtinger's presentation only works for link diagrams with minimal crossings? I apologize for asking such a short and simple question, but I can't figure this out and it seems like the answer should be clear as day.
Edit: Here's the diagram I'm using, and the relations I got from the two crossings: two unlinked circles. The only way I see it is that the relations coming from these crossings must be trivial, e.g., $ab=ab$, or $ba=ba$, but I don't see how that could be the case.
As user8268 put it, "you should have a generator for each arc of your diagram; note that the "b"-circle has 2 arcs, so gives you 2 generators". Denoting this new generator by $b'$, we have $a,b,b'$ with a single relation $ab=b'a$, meaning $b'=aba^{-1}$ and $a$ and $b$ generate the link group freely.