(I mean that the complement of X in P should be a ball after X has been compressed.)

Looks good! I want to use a homotopy to make X as big as possible in P, so the complement is either empty or a ball whose boundary contains one distinguished disc. If we do so then I think we get that, after a homotopy, the map factors as: glue on some 2-handles; glue on some 1-handles; and then a finite-sheeted covering. (Note that, in the question, I mentioned that by "inclusion", I meant gluing on 1-handles.) Does that sound right?

That's exactly the order I had in mind!

Sam, I see what you mean - I'm a little confused about what the correct statement about the boundary should be. I realise now that my original comment didn't make sense. Well, as I said, I'd be happy for any factorization to start with, and to worry about the boundary later.

I'd be happy to start with any factorization, and worry about whether the homotopy can be made nice later.

At each "time", the homotopy should map the boundary to itself.

OK. But I suspect that social scientists can answer the question "What is the impact of Mathematics in social science today?" better than mathematicians can.

I assumed it was a deliberate joke. After all, these things aren't knots. Though it now seems to have been changed...

I'm reminded of an old joke. Q: What's the physicist's definition of a group? A: A Lie group without the manifold structure.

Of course, the existence of a non-principal ultrafilter uses the axiom of choice. So there's a very real sense in which you can prove "more theorems" using these ideas.

I assume that we're talking about continuous automorphisms of pi hat?

Surely the precise statement of the questions should be: "Is the closure of the image of Mod(S) equal to the preimage of {+-1} under the map you define Out(pihat) - > Zhat^*?"