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).
Best Answer
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.