Suppose X is a pathconnected, locally compact, Hausdorff space and Y is its onepoint compactification. Let G be the fundamental group of X and H be the fundamental group of Y. Is it true that the embedding of X into Y always induces a monomorphism G>H? More generally, what is the relationship between G and H?
More generally, if A and X_{0} are any finite CW complexes, and f : A → X_{0} is any map, let Y be the mapping cone of f, and let X be Y with the cone point removed; then Y is the onepoint compactification of X, and the inclusion X_{0} → X is a homotopy equivalence. (David's example is the case A = X_{0} = S^{1}, f = id.) So any map X → Y which is the mapping cone of something is homotopy equivalent to a onepoint compactification.
I don't think you can realize any map of groups as the induced map on π_{1} of a mapping cone (0 → Z/2Z?) but you can realize (G → 0, 0 → a free group, ...)
It is not always a monomorphism. Let X be R^2 \setminus { (x,y) : x^2 + y^2 < 1 }. Then X has fundamental group the integers, but Y is contractible.

3$\begingroup$ I think you mean { (x,y) : x^2 + y^2 < 1}. i.e. you want to leave the boundary circle there. Then Y is contractible. Otherwise Y is a pinched torus and also has fundamental group the integers (although the induced map is zero). $\endgroup$ – Chris SchommerPries Oct 27 '09 at 17:09

$\begingroup$ You are right. I have edited to fix the error. $\endgroup$ – David E Speyer Oct 27 '09 at 17:16

$\begingroup$ I think the good old inversion z > 1/z makes this easier to see: i.e., just take the closed unit disk in the plane with the origin removed. $\endgroup$ – Pete L. Clark Dec 27 '09 at 10:53