Smash and join products of spheres

Question

How to prove rigorously that $\mathbb S^n \wedge \mathbb S^m =\mathbb S^{n+m}$ and $\mathbb S^n \ast \mathbb S^m = \mathbb S^{n+m+1}$? And what intuition should i have for compute $\wedge,\ast$ for difficult spaces?

Answer 1

I will prove that $\mathbb{S}^n \wedge \mathbb{S}^m$ is homeomorphic to $\mathbb{S}^{n+m}$

(Note that throughout this post $(\cdot)^*$ refers to the one-point compactification of a topological space)

Theorem: Let $M$ be a compact manifold of positive dimension and let $p \in M$. Then $M$ is homeomorphic to the one-point compactification of $M \setminus \{p\}$.

Note that $\mathbb{S}^n$ is a compact manifold. Let $p$ be any point in $\mathbb{S}^n$, then utilizing the above theorem we get $$\mathbb{S}^n \cong \left(\mathbb{S}^n \setminus \{p\}\right)^*$$ where $\left(\mathbb{S}^n \setminus \{p\}\right)^*$ denotes the one-point compactification of $\mathbb{S}^n \setminus \{p\}$. Note also that $$\mathbb{S}^n \setminus \{p\} \cong \mathbb{R}^n.$$This then allows us to conclude that since the one-point compactification of a topological space is unique up to homeomorphism, it follows that $$\left(\mathbb{S}^n \setminus \{p\}\right)^* \cong \left(\mathbb{R}^n\right)^*$$ and thus we can conclude that $$\mathbb{S}^n \cong \left(\mathbb{R}^n\right)^*. \ \ \ \ \ \ \ \ (1)$$

But how does this help us? In order to prove the desired result, we need to make use of one further theorem.

Theorem: If $X$ and $Y$ are compact Hausdorff spaces then $$X \wedge Y \cong \left(X \setminus \{x\} \times Y \setminus \{x\}\right)^*$$ where $x$ and $y$ are chosen base points of $X$ and $Y$ respectively.

Since $\mathbb{S}^n$ and $\mathbb{S}^m$ are both compact Hausdorff spaces the above theorem implies that $$\mathbb{S}^n \wedge \mathbb{S}^m \cong \left(\mathbb{S}^n \setminus \{p\} \times \mathbb{S}^m \setminus \{q\}\right)^*$$ and since $\mathbb{S}^n \setminus \{p\} \cong \mathbb{R}^n$ and $\mathbb{S}^m \setminus \{q\} \cong \mathbb{R}^m$ and the fact that for topological spaces $X, Y, A, B$ if $X \cong A$ and $Y \cong B$ then $X \times Y \cong A \times B$ it follows that

\begin{align*} \mathbb{S}^n \wedge \mathbb{S}^m &\cong \left(\mathbb{R}^n \times \mathbb{R}^m\right)^* \ \ \text{by the above} \\ &\cong \left(\mathbb{R}^{n+m}\right)^* \ \ \ \ \text{(since } \mathbb{R}^n \times \mathbb{R}^m \cong \mathbb{R}^{n+m} \text{)} \\ &\cong \ \mathbb{S}^{n+m} \ \ \ \ \ \ \text{by $(1)$ above} \end{align*}

Answer 2

I would like to give another proof for the smash product using the definition. By the definition of the smash product, we have $$S^n\wedge S^m\cong S^n\times S^m/(S^n\vee S^m).$$

Suppose $I^n=[0,1]^n\subset \mathbb R^n$ and consider $S^n$ as $I^n/\partial I^n\cong (I^n)^\mathrm o\cup O$, where $O$ is the origin of $\mathbb R^n$. Then $$S^n\times S^m \cong (I^{n+m})^\mathrm o\cup(\partial I^n\times (I^m)^\mathrm o)\cup((I^n)^\mathrm o\times \partial I^m),$$ $$S^n\vee S^m\cong I^n\cup I^m.$$ Since $(I^{n+m})^\mathrm o/(I^n\cup I^m)\cong(I^{n+m})^\mathrm o$, $(\partial I^n\times (I^m)^\mathrm o)\cup((I^n)^\mathrm o\times \partial I^m)/(I^n\cup I^m)\cong O$, we have $S^n\wedge S^m\cong (I^{n+m})^\mathrm o\cup O\cong S^{n+m}$.

Answer 3

For join, a definition of join is that we define $X*Y=X\times Y\times I/\sim$ where the relation is given by:

$$(x,y,0)\sim(x,y',0) \text{ and } (x,y,1)\sim(x',y,1)$$

for all $x,x'\in X,y,y'\in Y$. Note that, identifying $x$ with $[(x,y,0)]$ and $y$ with $[(x,y,1)]$, we naturally have a path from $x$ to $y$.

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To prove $S^m*S^n=S^{m+n+1}$, it suffices to show that $S^m*S^0=S^{m+1}$ for all $m\ge 0$.

We have $S^m*S^0=S^m\times S^0\times I/\sim$. Say $S^0=\{x_0,x_1\}$. The relation $\sim$ "glues" two copies of $S^m\times I$ together by identifying $S^m\times \{x_0\}\times\{0\}$ with $S^m\times \{x_1\}\times\{0\}$, and then the relation $\sim$ "contracts" two copies of $S^m$ on the ends to two points (namely, it "contracts" $S^m\times \{x_0\}\times\{1\}$ to a point and $S^m\times \{x_1\}\times\{1\}$ to another). Hence, put together, $S^m\times S^0\times I/\sim$ is just $S^m\times I$ with top and bottom "contracted" to two points! Thus $S^m\times S^0\times I/\sim\,\,=S^{m+1}$.

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By the way, $X*S^0=\Sigma X$, the suspension of $X$, for any $X$.