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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.2011-12-08

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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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    @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.2013-06-17