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Let $\lbrace X_\alpha\rbrace_{\alpha\in\Lambda}$ be a set of simply connected spaces.

Is it true that $\pi_1(\prod\limits_{\alpha\in\Lambda} X_\alpha)=0$?

cf) I know that $\pi_1(X\times Y)=\pi_1(X)\times\pi_1(Y)$.

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This wikipage topospaces.subwiki.org/wiki/Simply_connected_space claims that it is true, but it gives neither reference, nor proof. –  Martin Sleziak Jul 5 '11 at 7:22

1 Answer 1

up vote 5 down vote accepted

If $f:S^1\rightarrow \prod\limits_{\alpha\in\Lambda} X_\alpha$. then we can define $f_\alpha:S^1 \rightarrow X_\alpha$. But using $\pi_1(X_\alpha)=0$, we get a homotopy $h_\alpha:S^1\times I \rightarrow X_\alpha$ from $f_\alpha$ to a constant.

But then, by the universal property of the product, that means we have a $h:S^1 \times I \rightarrow \prod\limits_{\alpha\in\Lambda} X_\alpha$ which projects to the $h_\alpha$. Also, $h(s,0)=f(s)$ and $h(s,1)$ is constant. So $h$ is a homotopy from $f$ to a constant, so the product is simply connected.

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