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View S^3 as the sphere in C^2 (with coordinates, let's say, x and y). Let P and Q be the subsets of S^3 given by the equations x = 0 and y = 0...

View S^3 as the sphere in C^2 (with coordinates, let's say, x and y). Let P and Q be the subsets of S^3 given by the equations x = 0 and y = 0 respectively. What is the fundamental group of S^3 − A − B?


P and Q are obviously disjoint circles imbedded in S^3. I am trying to show that the deformation retract of S^3 - P - Q is the torus T(a), where for some a in (0, 1), T(a) = {(x, y) in S^3 : |x|^2 = a} (let's take a = 1/2). After that, it is trivial that the group we are looking for is the group of a torus, i.e. Z^2.

So to get the retract, I see that every point of S^3 - Q has a unique nearest point in P and a unique minimizing geodesic in the usual metric. I know I have to use that, but I cannot use that geodesic to retract onto |x|^2 = |y|^2 = 1/2.

What is the retraction? Any help is appreciated. Many thanks.

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