Let $\{X_\alpha\}_{\alpha\in J}$ be a family of topological spaces and let $X=\prod_{\alpha\in J}X_\alpha$ be endowed with the product topology. Show that $X$ is path connected if and only if each $X_\alpha$ is path connected.

I know this is true if path coennected is changed by connection but I am very confused with this problem, I have tried to do the following:

$\Rightarrow $ let $x,y \in X_\alpha$ be then $(x_\alpha), (y_\alpha)\in \prod X_\alpha$ where $x_\alpha=x, y_\alpha=y$ and $x_\beta=0, y_\beta=0$ for any $\beta\neq\alpha$, and since P is path connected then there is continuous $f:[0,1]\rightarrow \prod X_\alpha$ such that $f(0)=(x_\alpha)$ and $f(1)=(y_\alpha)$, then if $\pi_\alpha:\prod X_\alpha \rightarrow X_\alpha$ is the projection, which is continuous, we have that $\pi_\alpha f:[0,1]\rightarrow X_\alpha$ is continuous and $\pi_\alpha (f(0))=x$ and $\pi_\alpha (f(1))=y$.

$\Leftarrow$ I do not know how to do this address, could someone help me please? Thank you very much.

  • 2
    $\begingroup$ The continuous image of a path-connected space is path connected takes care of the left to right implication. $\endgroup$ – Henno Brandsma Oct 22 '17 at 21:37

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 $Y_\alpha$.


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.

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