# Simply connected and direct products

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!

Hint: $$\pi_1(X \times Y, (x_0, y_0)) \cong \pi_1(X, x_0) \times \pi_1(Y, y_0)$$.
• I know that result... But I´m not sure that $\pi(X \times Y) = 0$ implies $\pi(X) = 0$. – user183002 Mar 21 at 20:17
• 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... – JHF Mar 21 at 20:24
• So....... It thas implies $\pi(X)$ only has one element? – user183002 Mar 21 at 20:26