I am struggling with the following question from Allen Hatcher's algebraic topology book.
Define the join $X*Y$ of two topological spaces $X$ and $Y$ to be the quotient of $X\times Y \times I$ under the following identifications:
$(x,y,0)\equiv (x,y',0)$ for all $y,y'$ in $Y$ $x\in X$
$(x,y,1)\equiv(x',y,1)$ for all $x,x'\in X$ and $y\in Y$
So $X*Y$ can be thought of as the union of all line segments joining points of $X$ to points of $Y$.
I am trying to prove that if $X$ is path connected and $Y$ is any space, then $X*Y$ is simply connected.
Here are two approaches I have tried:
I have read that the join $X*Y$ is homotopy equivalent to the suspension of the smash product, $\Sigma( X\wedge Y)$. So one possibility is to prove that the smash product is path-connected, and then to use the claim that the suspension of a path-connected space is simply-connected.
Another possibility is to show that the join is homeomorphic to the subspace $Cone(X)\times Y\cup X\times Cone(Y)$ of the space $Cone(X)\times Cone(Y)$. Then, if I could show that both sets in the above union are open (in the union) and simply-connected, and that their intersection is simply-connected, van Kampen's theorem would imply the result.
My problem with the first approach is that, as far as I know, the smash product depends on the choice of base points, so I am not even sure what the phrase "the join $X*Y$ is homotopy equivalent to the suspension of the smash product, $\Sigma( X\wedge Y)$" really means.
The second approach seems a bit more inviting, especially since the problem appears in Hatcher's book in the section of van Kampen's theorem; but I don't know how to prove that the sets$Cone(X)\times Y$ and $X\times Cone(Y)$ are open in their union, nor do I know how to show that they are simply-connected, unless both $X$ and $Y$ are simply-connected.
Any help (even for the case where $X$ and $Y$ are both simply connected) would be greatly appreciated.