Skeleton of CW product $\Sigma^i X \times \Sigma^i Y$ I'm in trouble with an assertion in Hatcher's algebraic topology; let $X, Y$ be CW-complexes, and let's denote with $\Sigma$ the reduced suspension (it is the usual suspension $SX=X \times[0,1]/X \times\partial[0,1]$ with the "basepoint $\times I$" collapsed). The statement I'm in trouble with is the following: 
$\Sigma^i X \vee \Sigma^i Y$ is the $(2i-1)$-skeleton of $\Sigma^i X \times \Sigma^i Y$. 
I agree that this makes sense, but I don't know how to prove this (maybe it holds for any product $X \times Y$, without any suspension).
 A: This claim occurs in the proof of Proposition 4F.1 in the version from 2002 (section "Spectra and Homology Theories").
It is not true, because in general $\Sigma^i X \vee \Sigma^i Y$ has cells of dimension $ > 2i-1$.
However, Hatcher only wants to prove that
$(\ast)$ $\pi_{n+i}(\Sigma^i X \vee \Sigma^i Y) \approx \pi_{n+i}(\Sigma^i X \times \Sigma^i Y)$ for $n+i < 2i-1$.
But now Tyrone's comment applies.
By the way, the following statement is correct and suffices to prove $(\ast)$:
$\Sigma^i X \vee \Sigma^i Y$ and $\Sigma^i X \times \Sigma^i Y$ have the same $(2i-1)$-skeleton.
The cells of $ \Sigma^i X \times \Sigma^i Y$ have the form $e \times e'$ with cells $e$ of $\Sigma^i X$ and $e'$ of $\Sigma^i Y$. We have $\dim(e \times e') \le 2i-1$ only in case $e = \{ \ast \}$ and $\dim(e') \le 2i-1$ or $e' = \{ \ast \}$ and $\dim(e) \le 2i-1$ because $\Sigma^i Z$ has one $0$-cell $\{ \ast \}$ (the basepoint) and all other cells have dimension $\ge i$. 
$\Sigma^i X \vee \Sigma^i Y = \Sigma^i X \times \{ \ast \} \cup \{ \ast \} \times \Sigma^i Y \subset \Sigma^i X \times \Sigma^i Y$ consists of all cells $e \times \{ \ast \}, \{ \ast \} \times e'$ with cells $e$ of $\Sigma^i X$ and $e'$ of $\Sigma^i Y$. They have dimension $\le 2i-1$ if and only if their nontrivial factor $e$ resp. $e'$ has dimension $\le 2i-1$.
This proves the statement. Note that it can be generalized to the following:
If $X, Y$ are CW-complexes with one $0$-cell (being the base point) and no cells of dimension $1,\dots,i-1$, then $X \vee Y$ and $X \times Y$ have the same $(2i-1)$-skeleton.
