# Hatcher 0.18

Hatcher, Algebraic Topology, Chapter 0

18. Show that $S^1*S^1=S^3$, and more generally $S^m*S^n=S^{m+n+1}$.

Proof. Perhaps the easiest way to prove this is to use the conclusion of Exercise 0.24 to note that $X*Y\simeq \Sigma(X\wedge Y)$ for CW-complexes, where $\Sigma$ denotes the suspension (denoted in Hatcher by $S$). This, along with Hatcher’s observations that $S^n\wedge S^m=S^{n+m}$ and that $\Sigma(S^k)=S^{k+1}$ yields that

$S^n*S^m\simeq\Sigma(S^n\wedge S^m)=\Sigma(S^{n+m})=S^{n+m+1}$.

The specific case of $X=Y=S^1$ follows directly. Of course, appealing to a later solution to prove an earlier one seems a bit cheap, so here’s an attempt at a natural solution.

As noted in Hatcher, the join $X*Y$ can be thought of as the collection of all line segments connecting points $x\in X$ to points $y\in Y$. Of course, this gives no geometric intuition as what the space $X*Y$ should look like due in part to the infinite degrees of freedom regarding placement of the spaces $X$ and $Y$, so it seems like it may be worthwhile to appeal to other sources. In Dr. Hatcher’s book, for example, it’s shown that if the spaces in question are $X=Y=I$, then $X*Y$ is a tetrahedron. Not shown, though, is how exactly this figure can be pieced together using the geometric definition involving line segments, and so it stands to reason that other information is needed. For that, consider the figure shown here and copied/pasted as Figure 1 below for completeness.

Figure 1

Worth noting in Figure 1 is the position of the (bolded) spaces $X$ and $Y$: Indeed, forming line segments between points in two copies of $[0,1]$ fails to yield anything even remotely tetrahedral if, say, $X$ and $Y$ are draw in a planar fashion (i.e., in $\mathbb{R}^2$ and are parallel to one another, say. This begs the question: How exactly should one draw the spaces $X$ and $Y$ to get the desired results?

Consider the following excerpt from Armstrong’s Basic Topology (emphasis added):

Let $X$ and $Y$ be subsets of $\mathbb{E}^n$ positioned in such a way that if $x_1,x_2$ are distinct points of $X$ and $y_1,y_2$ are distinct points in $Y$, then the line segments which join $x_1$ to $y_1$ and $x_2$ to $y_2$ do not intersect. Write $X*Y$ for the union of all the line segments which joint a point of $X$ to a point of $Y$,….(pg 199)

In particular, the above quote highlights why the case where $X=Y=[0,1]$ seems to require non-coplanar positioning in $\mathbb{R}^3$ to work, and helps to illustrate at least how the spaces $X=Y=S^1$ should look in order to deduce that result. A very crude sketch illustrating the positioning I have in mind is shown in Figure 2 below.

Figure 2

Here’s a description of what you’re looking at.

Imagine $X_i$ to denote the two copies of $S^1$, $i=1,2$. Imagine them positioned in some sort of “ambient space” in such a way that $X_1$, without loss, is completely “inside” $X_2$, i.e., $X_1$ intersects the disc $\widehat{X_2}\cong D^2$ for which $\partial\widehat{X_2}=X_2$ in precisely two points which are designated by red horizontal-ish lines in Figure 2. Note that the lightened part of the circle $X_1$ is meant to denote the portion of that circle that would be invisible if the disc $\widehat{X_2}$ were colored an opaque red. The two black, respectively green, “disks” represent points on $X_1$, respectively $X_2$, and the purple-ish segments represent the elements in $X_1*X_2$ which connect their respective black-green point pairs. Again, the lightened part of the bottom-most purple segment indicates the part that would be invisible if $\widehat{X_2}$ were opaque.

The aforementioned “ambient space” can easily be seen to require no fewer than four spatial dimensions in order to achieve the desired disjointness of the segments: In particular, note that if the space in Figure 2 were assumed to be embedded in $\mathbb{R}^3$, one couldn’t insure such disjointness. Therefore, the space formed by $X_1*X_2$ in the above figure is undoubtedly required to be a subset of $\mathbb{R}^4$. Also, one can determine that the space is homeomorphic to precisely $S^3$ by considering the CW-structure of the 1-spheres $X_i$, $i=1,2$, and how a CW-structure can be extended. To be more precise:

Both spaces $X_i$, $i=1,2$, are copies of $S^1$. From earlier results, one particular CW-structure on $S^1$ consists of two cells $e^n_i$, $i=1,2$, in each dimension $n=0,1$. For example, one could visualize the structure for $X_1$ as follows: The 0-cells $e^0_i$, $i=1,2$, consist of the red horizontal-ish lines (which are really points of intersection) mentioned above, and the two 1-cells $e^1_i$, $i=1,2$, consist of the “upper” and “lower” blue half-circles. As also mentioned above, it’s easy to imagine the existence of 2-cells in the Figure 2 space: The disc $\widehat{X_2}$ for which $\partial\widehat{X_2}=X_2$ would be one – call it $e^2_2$ – and a “filled in version” $\widehat{X_1}\cong D^2$ satisfying $\partial\widehat{X_1}=X_1$ could represent the cell $e^2_1$. This yields a 2-skeleton for the space $X_1*X_2$. Finally, attaching two 3-cells $e^3_i$, $i=1,2$, with the appropriate attachment maps yields a CW-structure on $X*Y$ consisting of two cells in each of dimensions $0,1,2,3$ and hence indicates that $X*Y=S^3$ in the case where $X=Y=S^1$.

Among the multitude of obvious drawbacks of this method is the fact that the generalized case is near impossible. For that, something else is needed. Perhaps worth considering is the method discussed here.   $\square$

## 4 thoughts on “Hatcher 0.18”

1. Aritra

I haven’t tried exercise 0.24 yet, but I think it says X*Y is homotopy equivalent to the suspension of the smash product of X and Y (wiki confirms it). But you need homeomorphism here.
Nice explanation for the second part! But I think this can be solved using the facts that join operation is associative and n-fold join of S0 gives S(n-1).

2. Hello! You can find a solution to this in Ronald Brown’s book Topology and Groupoids (Chapter 5, in the section on joins.)

It’s a pretty solid book over all too. 🙂

• Thanks for that! I’m always on the lookout for good topology books; I’ll add this one to the list!

Many thanks.

3. It can also be shown (I hesitate to use proven) rather easily with the remark at the bottom of page 9 of Hatcher. There he states that S^(n-1) is homeomorphic to the join of n copies of S^0. Then clearly it follows that the join of S^n and S^m is homeomorphic to a join of n+m+2 copies of S^0, which in turn must be homeomorphic to S^(n+m+1).