For CW complexes X, Y where $X$ is $m$-connected, $Y$ is $n$-connected, $\pi_r[X\vee Y] =\pi_r[X\times Y]$ for all $r\lt m+n+1$ via inclusion.

Question

It is known that if a CW complex $X$ has no cells in dim $\le k$, then it is $k-$ connected that is 1) X is path connected, 2) the homotopy groups $\pi_r(X)=0$ for all $r\le k$.

Suppose that the converse of the above statement is true and hence assume that $X$ has no cells in dim $\le m$ and that $Y$ has no cells in dim $\le n$. Then, the cellular structure of $X\times Y$ has no cells in dim $\le m+n$. But I'm not sure how to go from here.

Please help. Thanks.

Answer 1

It is impossible that a non-empty CW complex has no cells in dimension $\le k$; each non-empty CW complex must have a non-empty $0$-skeleton $X^0$.

Thus the first correction is to require that $X$ has no cells in dimension $r$ for $1 \le r \le k$. In particular, it does not have $1$-cells. Then, if $X$ has more than one $0$-cell, $X$ cannot be path connected. In fact, it is well-known that a CW complex is path-connected iff its $1$-skeleton is path-connected.

Therefore the correct statement is

If a CW complex $X$ has one $0$-cell (i.e. $X^0 = *$) and no cells in dimensions $r$ with $1 \le r \le k$ (i.e. $X^r = *$ for $r \le k$) then it is $k$- connected, that is the homotopy groups $\pi_r(X)=0$ for all $r\le k$.

Let us look at $X \vee Y$, where the wedge point is the single $0$-cell. $X \vee Y$ can be regarded as the subspace $X \times * \cup * \times Y$ of $X \times Y$. It has an obvious cellular structure; its $r$-cells are the products $e_X^r \times *$ and $* \times e_Y^r$, where the $e_X^r$ and $e_Y^r$ are the $r$-cells of $X$ and $Y$, respectively. Thus the $r$-skeleton of $X \vee Y$ is

$$(X \vee Y)^r = X^r \times * \cup * \times Y^r .$$

Also $X \times Y$ has an obvious cellular structure. Its $r$-cells are the products $e_X^i \times e_Y^j$ with $i + j = r$, where the $e_X^i$ and $e_Y^j$ are the $i$-cells of $X$ and $j$-cells of $Y$, respectively. Clearly all $e_X^r \times *$ and $* \times e_Y^r$ are $r$-cells of $X \times Y$. Let us call an $r$-cell $e_X^i \times e_Y^j$ (where $i + j = r$) non-trivial if $i > 0$ and $j > 0$. A necessary requirement for the existence of a non-trivial $r$-cell is that $i > m$ and $j > n$ because $X$ does not have $i$-cells for $1 \le i \le m$ and $Y$ does not have $j$-cells for $1 \le i \le n$. Hence there are no non-trivial $r$-cells for $1 \le r \le m + n +1$. This means that

$$(X \times Y)^{m+n+1} = X^{m+n+1} \times * \cup * \times Y^{m+n+1} = (X \vee Y)^{m+n+1}.$$

By cellular approximation we see that the inclusion-induced $$\pi_r(X \vee Y) \to \pi_rf(X \times Y)$$ is an isomorphism for all $r < m+n+1$.