Suppose $X$ and $Y$ are topological spaces. Let's define the join $X\ast Y$ as the quotient space $X\times Y\times [0,1]/\sim$, where $\sim$ is the equivalence relation generated by $(x,y,0)\sim(x,y',0)$ and $(x,y,1)\sim(x',y,1)$. In particular, define the cone over $X$, $Cone(X)$, as the join of $X$ with a point. Is it true that $Cone(X\ast Y)$ is homeomorphic to $Cone(X)\times Cone(Y)$? If not, when does this happen?
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3$\begingroup$ is there any reason to expect this to be true? Is there a non-trivial example where it is true? $\endgroup$– Will SawinCommented Mar 21, 2012 at 4:09
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$\begingroup$ Hint: If you forget all the coordinates from X and Y, the first space maps to a triangle and the second maps to a square; try to find a homeomorphism between those that preserves the type of preimage. $\endgroup$– Tyler LawsonCommented Mar 21, 2012 at 5:07
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$\begingroup$ (This may be a compactly-generated hint.) $\endgroup$– Tyler LawsonCommented Mar 21, 2012 at 5:27
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$\begingroup$ Isn't this homework ? $\endgroup$– BS.Commented Mar 21, 2012 at 10:09
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$\begingroup$ It´s not a homework I came uo with this problem when I was traying to understand the proof of a Proposition. $\endgroup$– AntonioCommented Mar 21, 2012 at 13:29
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2 Answers
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If you use initial topologies to define the join, as in Section 5.7 of my book "Topology and groupoids", then the result you want is exactly 5.7.4 on p. 174, and the picture for it is as suggested by Tyler (Fig. 5.7).
Of course it can't be true generally with quotient topologies, as products don't preserve quotients (this is well known and is an example on p. 111). I have never worked out a proof that the two versions of the join are equivalent in the compactly generated case (as defined in Section 5.9 of the book), so I'd be grateful if this can be supplied.
answered Apr 8, 2012 at 17:02
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Take X=Poincare 3-sphere and Y=S^2, then your first space is homeomorphic to D^6, because double suspension of X is sphere, but second doesn't have topological manifold structure.
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3$\begingroup$ No, this is not a counterexample. $\endgroup$ Commented Apr 8, 2012 at 14:00