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Does smashing of a pointed CW complex $X$ with an arbitrary pointed CW complex $Y$ increase the connectivity?

The connectivity of a pointed space $X$ is the maximal number $\operatorname{con}(X)$ such that $\pi_i(X)=0$ for all $0\leq i\leq\operatorname{con}(X)$.

More precisely, the question is: $\operatorname{con}(X\wedge Y)\geq\operatorname{con}(X)$?

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Whoops, ignore my answer - I was not aware of the "smash product," so I thought it was a loose term for the pointed join. – Thomas Andrews Dec 8 '11 at 18:09
up vote 6 down vote accepted

Yes. In fact, $\operatorname{conn}(X\wedge Y)\ge\operatorname{conn}(X)+\operatorname{conn}(Y)+1$ (if both $X$ and $Y$ are connected).

(Indeed, if $X$ is $n$-connected, it's homotopy equivalent to a CW-complex $X'$ with one $0$-cell and no cells in dimensions $1\le s\le n$. Now note that $S^k\wedge S^l=S^{k+l}$, so $X'\wedge Y'$ is homotopy equivalent to $X\wedge Y$ and doesn't have cells in dimensions $1\le s\le n+m+1$.)

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Cool, thanks. How do you prove, that if $X$ is $n$-connected, it has a cell structure with only one $0$-cell and no $\leq n$-cells else? – Daniel Dreiberg Dec 8 '11 at 18:19
@DanielDreiberg It's a well-known consequence of cellular approximation (for example, it's Example 4.14 after Proposition 4.13 in Hatcher's book). The idea is: induction by n + "$X/A\cong X$ when $A$ is contractible" lemma. – Grigory M Dec 8 '11 at 18:31
Ah, yes, right. Thanks. – Daniel Dreiberg Dec 8 '11 at 19:04
By the way, you have to assume connectedness since $\mathrm{conn}(S^0)=\mathrm{conn}(S^0\wedge S^0)=0\neq 1$. – Daniel Dreiberg Dec 9 '11 at 10:04
@DanielDreiberg: The usual convention is that all nonempty but disconnected spaces, like $S^0$, are $(-1)$-connected (because $\pi_0$ is not trivial) which makes the formula work out. – Tyler Lawson Jun 17 '13 at 1:49

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