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A topological space $X$ is simply connected if and only if $X$ is path-connected and the fundamental group of $X$ at each point is trivial, that is, $\pi(X,x_0) = 0$ for any $x_0 \in X$.

Now, we know that if $X,Y$ are simply connected, then $X \times Y$ is simply connected. My question: Is it also true the converse?

Since $X \times Y$ is path-connected we get $X,Y$ are path-connected. There exist a way to assert $\pi(X) = \pi(Y) = 0$?

Thanks!

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Hint: $\pi_1(X \times Y, (x_0, y_0)) \cong \pi_1(X, x_0) \times \pi_1(Y, y_0)$.

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  • $\begingroup$ I know that result... But I´m not sure that $\pi(X \times Y) = 0$ implies $\pi(X) = 0$. $\endgroup$ – user183002 Mar 21 at 20:17
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    $\begingroup$ Try counting the size of the groups on either side of the isomorphism. We know that $\pi_1(X \times Y)$ has one element, so... $\endgroup$ – JHF Mar 21 at 20:24
  • $\begingroup$ So....... It thas implies $\pi(X)$ only has one element? $\endgroup$ – user183002 Mar 21 at 20:26
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    $\begingroup$ Yes. The product of two groups is trivial iff the two factors are trivial. $\endgroup$ – JHF Mar 21 at 20:28

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