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Sep 29, 2016 at 23:07 vote accept PVAL
Sep 29, 2016 at 10:38 comment added Danny Ruberman There are of course lots of rational homology balls in this setting, and while the map on $H_1$ is surjective as Marco says, there's no reason that the map on $\pi_1$ should be surjective in general. For a silly example, take the $W$ referred to (with $\pi_1 = Z_2$) and connect sum in the interior with a homology 4-sphere.
Sep 29, 2016 at 7:50 comment added Marco Golla Sure, feel free to drop me an email. As for the surjectivity, one can see it directly from the surgery diagram, or, less directly, from the long exact sequence for the pair $(W,\partial W)$ (and the fact that $H_1(\partial W) = \pi_1(\partial W)$).
Sep 29, 2016 at 1:36 answer added Danny Ruberman timeline score: 3
Sep 29, 2016 at 1:33 comment added PVAL @MarcoGolla Yes, X should be smooth. I've seen constructions before of $W$, but haven't went through them carefully. Is it easy to see the map $\pi_1(L(4,1)) \to \pi_1(W)$ is really surjective (maybe its forced to be and I am forgetting something silly)? If you want details to where this comes up, I'd feel more comfortable emailing you (is that okay?). I'd rather not share preliminary results on a public forum.
Sep 29, 2016 at 1:06 history edited PVAL CC BY-SA 3.0
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Sep 29, 2016 at 0:57 comment added Marco Golla Also, do you want $X$ to be smooth?
Sep 29, 2016 at 0:57 history edited Marco Golla CC BY-SA 3.0
rather picky edit on the notation for coefficients in homology.
Sep 29, 2016 at 0:53 comment added Marco Golla Nice question! Where does this come from? Anyway, here is one (maybe silly?) example where $\pi_1(X)\neq 0$: you can take the rational homology ball $W$ bounded by $L(4,1)$ and blow it up. I will try to think about more interesting examples.
Sep 28, 2016 at 23:28 history edited PVAL CC BY-SA 3.0
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Sep 28, 2016 at 23:21 history asked PVAL CC BY-SA 3.0