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.
Questions:
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?
Thanks!