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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.

Thanks!

Roy

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3 Answers 3

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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, $f((s-\delta,s+\delta))\subset A_i$.

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  • $\begingroup$ @EricWofsey That is completely true, but I didn't actually need anything beyond the ability to decompose paths into a composition of paths from the various $X*Y_i$. (This is Lemma 1.15 in Hatcher.) To do the decomposition using compactness of $[0,1]$, we just need to see that the inverse images of each piece of the decomposition are open, which I hopefully didn't make a mistake showing. $\endgroup$ Oct 20, 2016 at 8:37
  • $\begingroup$ Yep, that argument looks good now. $\endgroup$ Oct 20, 2016 at 17:55
  • $\begingroup$ Oh, except that you should also mention that the inclusion $X*Y_i\to A_i$ is a homotopy equivalence. This again takes a little bit of care to prove, since you can't just do the obvious thing of squishing $Z\setminus X*Y_i$ down to $X$ (you need to also squish down $X*Y_i$ at the same time to be sure this is continuous). $\endgroup$ Oct 20, 2016 at 22:03
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    $\begingroup$ @EricWofsey Ok, there is a homotopy $g_t:X*Y\to X*Y$ which collapses the portion associated with $[0,1/2]$ onto the $X$ side while stretching the $[1/2,1]$ portion to $[0,1]$ (say $g_t(x,y,s)=(x,y,\max(0,(1+t)s-t))$). Composing any path with $g_t$ induces a homotopy which lets us assume any piece of a loop in $X*Y$ is contained in a $X*Y_i$. $\endgroup$ Oct 21, 2016 at 4:24
  • $\begingroup$ Right, that's basically what I had in mind. $\endgroup$ Oct 21, 2016 at 5:02
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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.

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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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  • $\begingroup$ Ah, there was a typo in the title; I wish to show that the join is simply-connected. But thanks for your construction! $\endgroup$ Aug 27, 2012 at 20:20
  • $\begingroup$ @RoyBen-Abraham I've edit the answer. I believe this shows $X*Y$ is simply connected. $\endgroup$ Aug 27, 2012 at 23:49
  • $\begingroup$ Your $f$ is not well-defined: if $t(s)=1$, then $x(s)$ is not well-defined. $\endgroup$ Oct 20, 2016 at 7:08

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