homology of smash product of Eilenberg-Maclane spaces Let $K_n=K(\mathbb{Z},n)$ be the Eilenberg-Maclane space. Prove:
(1) $K_m\wedge K_n$ is $(m+n-1)$-connected.
(2) $H_{m+n}(K_m\wedge K_n;\mathbb{Z})= H_{m+n}(K_m\times K_n;\mathbb{Z})$.
How to prove these two?
 A: $K_n = K(\Bbb Z, n)$ has a cell structure with no cells in dimensions $0 < d < n$ (and only one in dimension $n$), given by attaching progressively higher-dimensional cells to $S^n$ to kill off the higher homotopy groups. Then $K_m \wedge K_n = (K_m \times K_n)/(K_m \vee K_n)$ has no cells in dimensions $0 < d < m+n$, since $K_m \times K_n$ has one cell in dimension $0$, one in each of dimensions $m$ and $n$ corresponding to $K_m \vee K_n$, and more in dimension at least $m+n$. So by cellular approximation all maps $S^p \rightarrow K_m \wedge K_n$ are null-homotopic for $p<m+n$, i.e., $K_m \wedge K_n$ is $(m+n-1)$-connected.
For the second question, the cellular chain complex for $H_*(K_m \wedge K_n; \Bbb Z)$ and $H_*(K_m \times K_n; \Bbb Z)$ is identical in degree at least $m+n$; and because $K_m \wedge K_n$ has no cells of dimension $m+n-1$, the map $C_{m+n}(K_m \wedge K_n; \Bbb Z) \rightarrow C_{m+n-1}(K_m \wedge K_n; \Bbb Z)$ is the zero map. So it suffices to check that $C_{m+n}(K_m \times K_n; \Bbb Z) \rightarrow C_{m+n-1}(K_m \times K_n; \Bbb Z)$ is the zero map as well, even though the latter group needn't be zero. One can do this explicitly because if $C_{m+n-1}(K_m \times K_n; \Bbb Z) \neq 0$ one of $m,n = 1$; suppose it's $n$, and by restricting to the $m+n$-skeleton the question is simply to determine the degree of the attaching map $\partial (D^m \times D^1) \to S^m \vee S^1$, i.e., that it's zero.
An alternate method would be to note that because the $(m+n)$-skeleton of $K_m \wedge K_n$ is $S^{m+n}$, the first nonzero homotopy group is $\pi_{m+n}(K_m \wedge K_n) = \Bbb Z$, so by Hurewicz $H_{m+n}(K_m \wedge K_n; \Bbb Z) \cong \Bbb Z$. Then one computes $H_{m+n}(K_m \times K_n; \Bbb Z) \cong \Bbb Z$ using the Künneth theorem.
