Suppose X is a path-connected, locally compact, Hausdorff space and Y is its one-point 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?
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2 Answers
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More generally, if A and X0 are any finite CW complexes, and f : A → X0 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 one-point compactification of X, and the inclusion X0 → X is a homotopy equivalence. (David's example is the case A = X0 = S1, f = id.) So any map X → Y which is the mapping cone of something is homotopy equivalent to a one-point compactification.
I don't think you can realize any map of groups as the induced map on π1 of a mapping cone (0 → Z/2Z?) but you can realize (G → 0, 0 → a free group, ...)
answered Oct 27, 2009 at 17:41
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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.
answered Oct 27, 2009 at 17:03
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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). $\endgroup$ Oct 27, 2009 at 17:09
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$\begingroup$ You are right. I have edited to fix the error. $\endgroup$ Oct 27, 2009 at 17:16
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$\begingroup$ 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. $\endgroup$ Dec 27, 2009 at 10:53