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Let $X$ and $Y$ be two topological spaces. Is there any relation between Connectivity of $X\times Y$ and connectivity of $X$ and $Y$?

I think by Kunneth formula, it must be $Conn(X\times Y)\leq\min\{Conn(X), Conn(Y)$}$ and I am wondering, How can be far from each others?

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You have $\pi_n(X\times Y, (x_0, y_0)) \cong \pi_n(X, x_0) \times \pi_n(Y, y_0)$, so $\pi_n(X\times Y, (x_0, y_0))$ is trivial if and only if both $\pi_n(X, x_0)$ and $\pi_n(Y, y_0)$ are trivial. So indeed the connectivity of the product is always the minimum of the connectivities of the factors.


Since some people seem to not agree that this is enough to show the desired equality, let me elaborate.

For a pointed topological space $(X,x_0)$ we say $X$ is $n$-connected when $\pi_k(X,x_0)=0$ for all $k\le n$. We define the connectivity of $X$ as the largest $n$ such that $X$ is $n$-connected.

Now the connectivity of $X\times Y$ is the largest $n$ such that $\pi_k(X\times Y, (x_0,y_0))=0$ for all $k\le n$. Now, as pointed out above, homotopy groups behave well under products, i.e. $\pi_n(X\times Y, (x_0, y_0)) \cong \pi_n(X, x_0) \times \pi_n(Y, y_0)$, so the connectivity of $X\times Y$ is the largest $n$ such that both $X$ and $Y$ are $n$-connectecd, which is in fact the minimum of the connectivities.

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    $\begingroup$ @DanielValenzuela Why? I added some elaboration. $\endgroup$ – Christoph Oct 20 '15 at 15:54
  • $\begingroup$ I thought the question was about being connected (not n-connected). $\endgroup$ – Daniel Valenzuela Oct 20 '15 at 19:54

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