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