most recent 30 from http://mathoverflow.net 2013-05-19T13:18:17Z http://mathoverflow.net/feeds/question/91790 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91790/cone-over-the-join-of-two-topological-spaces Antonio 2012-03-21T04:05:27Z 2012-04-08T17:02:39Z

<p>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?</p>

http://mathoverflow.net/questions/91790/cone-over-the-join-of-two-topological-spaces/92142#92142 Bad English 2012-03-25T08:36:34Z 2012-03-25T08:36:34Z

<p>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.</p>

http://mathoverflow.net/questions/91790/cone-over-the-join-of-two-topological-spaces/93507#93507 Ronnie Brown 2012-04-08T17:02:39Z 2012-04-08T17:02:39Z

<p>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). </p> <p>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. </p>