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.
Thanks!
Roy
Best Answer
Let $\{Y_i\}_i$ be the collection of path components of $Y$, so then $X*Y=\bigcup_i X*Y_i$. Let $Z$ be the portion of $X*Y$ associated with $[0,1/2)$, and let $A_i=Z\cup (X*Y_i)$. The intersection of any two of these $A_i$ is $Z$, which is path connected (deformation retracting onto $X$), so if we prove that $X*Y_i$ is simply connected, then $X*Y$ is simply connected by the van Kampen theorem. While it is true that the $A_i$ are not open in $X*Y$, this is fine because the surjectivity part of the theorem only relies on $f^{-1}(A_i)$ being open in $[0,1]$ for any path $f:[0,1]\to X*Y$. We defer showing openness to the appendix.
Because of this, let us assume $Y$ is path connected as well. The space $X*Y$ decomposes into two open subsets, one associated with $[0,2/3)$ and the other associated with $(1/3,1]$. Call these $Z_1$ and $Z_2$. The first deformation retracts onto $X$, the second deformation retracts onto $Y$, and their intersection $Z_1\cap Z_2$ deformation retracts onto $X\times Y$. The inclusion maps $X\times Y\to Z_1$ and $X\times Y\to Z_2$ induce projections on $\pi_1$, so the van Kampen theorem gives $\pi_1(X*Y)$ to be the free product $\pi_1(X)*\pi_1(Y)$ modulo the subgroup generated by all $i_{1*}(x,y)i_{2*}(x,y)^{-1}$, which are all $xy^{-1}$. Since $x,y$ may be arbitrary, the resulting $\pi_1(X*Y)$ is trivial, so $X*Y$ is simply connected.
Appendix. First, $Z$ is open in $X*Y$, so $f^{-1}(Z)$ is open. Take a point $f(s)=(x,y,t)$ in $X*Y_i$ with $t>0$. By continuity of $f$ there is a $\delta$ such that $f((s-\delta,s+\delta))\subset X\times Y\times (0,1]$. Since the image of this interval must lie in a path component, $f((s-\delta,s+\delta))\subset A_i$.