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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.



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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.

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Ah, there was a typo in the title; I wish to show that the join is simply-connected. But thanks for your construction! – Roy Ben-Abraham Aug 27 '12 at 20:20
@RoyBen-Abraham I've edit the answer. I believe this shows $X*Y$ is simply connected. – Alex Becker Aug 27 '12 at 23:49

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