@Mike: I spent my drive home thinking that the fact that the boundary components have no fixed points, either, meant that the fixed point we created by collapsing the boundary component couldn't be non-degenerate, but I think I decided (whilst driving) that I couldn't make sense of this smoothly.
I made this amazing program today :) You can program a group with the data structure provided and it checks it against axioms and produces a coloured cayley table :)
@TedShifrin It's tricky. I just wrote a few nonsense lines a few times before deleting them :P Clearly the image of a circle around the fixed point is something that winds $\pm 1$ times around the dot - is that all we need?
The point is that Lefschetz fixed points have to be generic, so the function shouldn't look locally like $f(x)=x$, which obviously has very non-isolated fixed points. @Mike
no, @Mike ... Lefschetz number counts (appropriately) Lefschetz fixed points of a Lefschetz map. Such a map $f\colon X\to X$ is characterized by its graph's being transverse to the diagonal in $X\times X$.
I think I could say that the simple 4-regular graph with least number of vertices is obviously $K_5$, and as that is not planar, there oughta be at least 6 vertices or more.
@Mike: The Lefschetz number is the total intersection of the graph with the diagonal. I told you how to compute the local one. The global one is the sum of the local ones.
@TedShifrin One sums over all the transverse intersections with the diagonal... and then multiplies by $\pm 1$ because of orientation or something. Anyway, like I said, don't worry about it, I'll learn it in due course.
Hmm, I'm embarrassed to realize that I've been pronouncing Sylow as if it were German. I have no idea what the proper Norwegian pronunciation is. Help @N3.
