Proving that the join of a path-connected space with an arbitrary space is simply-connected 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 
 A: 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, $f((s-\delta,s+\delta))\subset A_i$.
A: Here is a proof using van Kampen's theorem for the fundamental groupoid (see http://www.math.harvard.edu/~hirolee/pdfs/231a-35-van-kampen.pdf, for instance).  To start, it is easy to see that $X*Y$ is path-connected, since $X$ is path-connected, and every point of $X*Y$ by definition is on a path to a point of $X$ (thinking of $X*Y$ as $X\coprod Y$ together with a path from each point of $X$ to each point of $Y$).
Now let $U\subset X*Y$ be the image of $X\times Y\times[0,1)$ under the quotient map and $V\subset X*Y$ be the image of $X\times Y\times (0,1]$ under the quotient map.  By van Kampen, we can compute the fundamental groupoid $\pi_{\leq 1}(X*Y)$ as the pushout in groupoids of $\pi_{\leq 1}(U)\leftarrow\pi_{\leq 1}(U\cap V)\to\pi_{\leq 1}(V)$.
Let's now take a closer look at $U$, $V$, and $U\cap V$.  Note that $U$ deformation-retracts to $X$, by sending $(x,y,s)$ to $(x,y,st)$ as $t$ goes from $0$ to $1$ (this respects the equivalence relation that defines $X*Y$ since $U$ contains no points with $s=1$).  Similarly, $V$ deformation-retracts to $Y$, by linearly moving the $s$ coordinate to $1$.  Finally, $U\cap V$ is just $X\times Y\times (0,1)$ which is homotopy-equivalent to $X\times Y$.
In particular, it follows any path $\gamma$ in $V$ between points $(x,y,s)$ and $(x',y',s')$ in $U\cap V$ is homotopic to a path in $U\cap V$.  Namely, on the $X$-coordinate choose any path from $x$ to $x'$ in $X$ (here we use path-connectedness of $X$), on the $Y$-coordinate take the path given by $\gamma$, and on the $I$-coordinate take the linear path from $s$ to $s'$.  This path has the same projection onto $Y$ as $\gamma$, and hence is homotopic to $\gamma$ since the projection to $Y$ is our homotopy equivalence from $V$ to $Y$.
Since $\pi_{\leq 1}(X*Y)$ is the pushout of $\pi_{\leq 1}(V)$ and $\pi_{\leq 1}(U)$ over $\pi_{\leq 1}(U\cap V)$, we conclude that any path in $X*Y$ between two points of $U$ is homotopic to a path contained in $U$ (since all the parts that are in $V$ can be homotoped to be in $U\cap V$).  In particular, for any basepoint $p\in U$, the inclusion map $\pi_1(U,p)\to \pi_1(X*Y,p)$ is surjective.  But the inclusion $X\to U$ is a homotopy equivalence, and the inclusion $X\to X*Y$ is nullhomotopic since any point of $Y$ gives a cone over $X$ in $X*Y$.  It follows that the inclusion $U\to X*Y$ is nullhomotopic, and in particular induces a trivial map on $\pi_1$.  Since the inclusion map $\pi_1(U,p)\to \pi_1(X*Y,p)$ is both surjective and trivial, $\pi_1(X*Y,p)$ is trivial.  Thus $X*Y$ is simply connected.
A: Let $(x,y,t),(x',y',t')$ be two points in $X*Y$, and $\alpha:[0,1]\to X$ a path connecting $x$ and $x'$. Then define 
$$\gamma(s)=\begin{cases}
(x,y,(1-3s)t) &\text{if } 0\leq s\leq 1/3\\
(x,y',(3s-1)t') &\text{if } 1/3\leq s\leq 2/3\\
(\alpha(3s-2),y',t') &\text{if } 2/3\leq s\leq 1\\
\end{cases}$$
to get a path joining $(x,y,t)$ and $(x',y',t')$. Thus $X*Y$ is path-connected.
To see that $X*Y$ is simply-connected, let $\gamma:[0,1]\to X*Y$ be a path with $\gamma(0)=\gamma(1)$. Write $\gamma(s)=(x(s),y(s),t(s))$. We want to show that $\gamma$ is null-homotopic. WLOG we can assume $\gamma(0)=\gamma(1)=(x,y,0)$. Let $f(s,r)=(x(s),y(s),rt(s))$ for $r\in [0,1]$. Then $f$ is a homotopy between $\gamma$ and $\gamma'$ defined by $\gamma'(s)=(x(s),y,0)$. We can then define a homotopy 
$$g(s,r)=\begin{cases}
(x,y,2s) &\text{if } s\leq r/2\\
\left(x\left(\frac{s-r/2}{1-r}\right),y,r\right) &\text{if } r/2< s< 1-r/2\\
(x,y,2-2s) &\text{if } s\geq 1-r/2
\end{cases}$$
between $\gamma'$ and $\gamma''$ defined by 
$$\gamma''(s)=\begin{cases}
(x,y,2s)&\text{if } s\leq 1/2\\
(x,y,2-2s)&\text{if } s\geq 1/2
\end{cases}$$
which is clearly null-homotopic. Composing homotopies shows that $\gamma$ is null-homotopic, hence $X*Y$ is simply-connected.
