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Let $S^{n}$ be an $n$-sphere. I'd like to compute the reduced homology (with $\mathbb{Z}$-coefficients) of the space $\bigvee^{r}_{i = 1} S^{n_{i}} * \bigvee^{s}_{j = 1} S^{m_{j}}$, where $r, s, n_i, m_j \geq 0$ and $*, \vee$ denote the join and wedge sum operations, respectively. I can compute these homology groups of pure wedges sums of spheres or of the pure topological joins of spheres, but I'm not sure how to proceed for this general mixed case.


Does the join operation factor over wedge sum, i.e., $$S^{n} *(S^{m_1} \vee S^{m_2}) \simeq (S^{n}*S^{m_1}) \vee (S^{n} * S^{m_2}) \simeq S^{n + m_1 + 1} \vee S^{n + m_2 + 1}?$$

Can one reduce $\bigvee^{r}_{i = 1} S^{n_{i}} * \bigvee^{s}_{j = 1} S^{m_{j}}$ in a similar fashion?

I do know that this statement is true for suspensions $$\Sigma(\bigvee^{s}_{j = 1} S^{m_{j}} ) \simeq \bigvee^{s}_{j = 1} S^{m_{j} + 1},$$ which can be viewed as the join with $S^{0}$ from the left.

Update: Yes, I believe that I can iterate the suspension from the left to prove the identity $$S^{0} * \cdots * S^{0} *(S^{m_1} \vee S^{m_2}) \simeq S^{n + m_1 + 1} \vee S^{n + m_2 + 1},\;\; \text {where}\;\; S^{0} * \cdots * S^{0} \simeq S^{n}.$$

I'm still working on the general case.

Update 2: I have a hunch that $$ \bigvee^{r}_{i = 1} S^{n_{i}} * \bigvee^{s}_{j = 1} S^{m_{j}} \simeq \bigvee^{r}_{i = 1} \bigvee_{j = 1}^{s} S^{n_{i}} * S^{m_{j}} \simeq \bigvee^{r}_{i = 1} \bigvee_{j = 1}^{s} S^{n_{i} + m_{j} + 1}. $$ If $n_{i} = n$ and $m_{j} = m$ for $1 \leq i \leq r$ and $1 \leq j \leq s$, then $\bigvee^{r}_{i = 1} S^{n_i} * \bigvee^{s}_{j = 1} S^{m_j} \simeq \bigvee^{rs}_{i = 1} S^{n + m +1}$.

Does any of this generalize to CW complexes or to topological spaces of finite homological type?


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In general, the join in the suspension of the smash product... Then, with field coefficients for example, $\tilde{H}^*(A \ast B) \cong \tilde{H}^{\ast -1}(A)\otimes \tilde{H}^{\ast-1}(B)$. – Dylan Wilson Jun 13 '11 at 3:51
Also, using the fact $A \ast B \cong \Sigma(A \vee B)$ you could probably prove your hunch – Dylan Wilson Jun 13 '11 at 3:57
@Dylan: Is that smash $\wedge$ or wedge $\vee$ in the last identity? – user02138 Jun 13 '11 at 4:08
up vote 6 down vote accepted

The answer to your questions is yes. Specifically, define the reduced join of pointed spaces $(X,x_0)$ and $(Y,y_0)$ by the formula $$ X {\;\ast_\text{red}\;} Y \;=\; (X \ast Y) / \bigl(\{x_0\} \ast Y \;\cup\; X \ast \{y_0\}\bigr). $$ If $X$ and $Y$ are CW complexes and $x_0,y_0$ are vertices, then the reduced join $X {\;\ast_\text{red}\;} Y$ is homotopy equivalent to $X \ast Y$ (see the argument below). However, it is easy to see that $$ (X\lor Y){\;\ast_\text{red}\;}Z \;\;\cong\; (X{\;\ast_\text{red}\;}Z)\lor(Y{\;\ast_\text{red}\;}Z) $$ where $\cong$ denotes homeomorphism. It follows that $$ (X \lor Y)\ast Z \;\simeq\; (X\lor Y){\;\ast_\text{red}\;}Z \;\;\cong\; (X{\;\ast_\text{red}\;}Z)\lor(Y{\;\ast_\text{red}\;}Z) \;\simeq\; (X\ast Z) \lor (Y\ast Z). $$ where $\simeq$ denotes homotopy equivalence.

Why the reduced join is homotopy equivalent to the join: Assume that $X$ and $Y$ are CW complexes, and the basepoints $x_0$, $y_0$ are vertices. Then the join $X\ast Y$ has a natural cell structure, and the subspaces $$ A \;=\; \{x_0\} \ast Y \qquad\text{and}\qquad B \;=\; X \ast \{y_0\} $$ are subcomplexes. But $A$ is homeomorphic to the cone of $Y$ and $B$ is homeomorphic to the cone of $X$, so both $A$ and $B$ are contractible. Furthermore, the intersection $$ A \cap B \;=\; \{x_0\} \ast \{y_0\} $$ is just a line segment. It follows that $A \cup B$ is a contractible subcomplex of $X\ast Y$, and therefore $(X\ast Y)/(A\cup B)$ is homotopy equivalent to $X\ast Y$.

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