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I am doing exercise 21 of section 1.2 in Hatcher's Algebraic Topology.

Show that the join $X\ast Y$ of two nonempty spaces $X$ and $Y$ is simply connected if $X$ is path connected.

I proved that $X\ast Y$ is path connected via the existence of paths between $(x,y,t)$ and $(x,y,0)$, and then used the path connectedness of $X$. But, I do not know how to prove that fundamental group is trivial using Van Kampen's theorem.

I tried to cover $X\ast Y$ by $X \times Y \times [0,1)/ \!\sim$ and $X \times Y \times (0,1]/ \!\sim$, but their intersection is not necessarily path connected since $Y$ is not path connected. And I did see this question that proves this statement, but I would like to use Van Kampen's Theorem in the solution instead (after all, this exercise is in the section on Van Kampen's Theorem in Hatcher's book).

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Kerr
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    @StefanHamcke it is not a duplicate, see correction below – Kerr Nov 07 '15 at 20:19
  • You have posted two identical questions, this one and http://math.stackexchange.com/questions/1517838/the-join-of-nonempty-path-connected-space-x-and-any-space-y-has-a-trivial-fundam – Lee Mosher Nov 08 '15 at 18:43
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    Hmmm...you can prove it pretty easily using the basepoint-free (groupoid) van Kampen theorem, using the open cover suggested by Jane. Don't see how to do it with just the pointed van Kampen theorem though, and that's all that's in Hatcher. – Eric Wofsey Oct 20 '16 at 05:38
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    @MikePierce: In case you're interested I posted my groupoid van Kampen proof at http://math.stackexchange.com/questions/187562/proving-that-the-join-of-a-path-connected-space-with-an-arbitrary-space-is-simpl, and Kyle Miller has also posted a proof there that uses the proof (though not quite the usual statement) of ordinary van Kampen. – Eric Wofsey Oct 20 '16 at 21:45
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    But Hatcher does have the version of Seifert-van Kampen with arbitrarily many opens. Using this one, there is a way to do it if $Y$ is locally path-connected. We can reduce to this case if we replace the topology on $Y$ with the one generated by the path-components of the open subsets of $Y$. – Ben Oct 20 '16 at 21:46
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    @Ben: Right, that's basically Kyle Miller's answer that I just linked to (though you have to be careful in replacing the topology on $Y$: it's not obvious that this won't change what paths you have in $XY$ because a priori funny things might happen when you take the quotient. Kyle Miller's answer avoids having to worry about this by observing that you don't care what happens in $XY$ near $X$ since that part can just be deformation-retracted onto $X$. – Eric Wofsey Oct 20 '16 at 21:58
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    You're right, there is something more to say. Passing to this locally path-connected version induces isomorphisms on fundamental groups and I am pretty sure that passing to the locally path-connected space is compatible with the join but I agree that this is non-trivial. – Ben Oct 20 '16 at 22:33
  • This pdf contains a solution when Y is assumed to be locally path-connected: https://stanford.edu/class/math215b/Sol2.pdf – tekay-squared Jan 24 '22 at 14:41

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