Show that $X=\prod X_\alpha$ is path connected if and only if each $X_\alpha$ is path connected

Solution 1:

Use the following theorem.

Let $X$ be a topological space and $(Y_\alpha)_{\alpha\in A}$ be a family of topological spaces. Then a function $f:X\to\prod_{\alpha\in A}Y_\alpha$ is continuous iff $\pi_\alpha\circ f$ is continuous for every $\alpha\in A$, where $\pi_\alpha$ denotes the canonical projection from $\prod_{\beta\in A}Y_\beta$ onto $Y_\alpha$.

Details:

Suppose $(X_\alpha)_{\alpha \in J}$ is a family of path connected topological spaces and fix $(x_\alpha),(y_\alpha)\in\prod X_\alpha$. For each $\alpha$ we may use the path connectedness of $X_\alpha$ to obtain a continuous function $f_\alpha:[0,1]\to X_\alpha$ such that $f_\alpha(0)=x_\alpha$ and $f_\alpha(1)=y_\alpha$. Define $f:[0,1]\to\prod X_\alpha$ by $f(t)=(f_\alpha(t))$ for each $t\in[0,1]$. Then $f(0)=(x_\alpha)$, $f(1)=(y_\alpha)$, and for each $\alpha\in J$ we have that $\pi_\alpha \circ f=f_\alpha$ is continuous. From the theorem we deduce that $f$ is continuous, which concludes the proof that $\prod X_\alpha$ is path connected.