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Suppose that $(X_i, \tau_{X_i})$ are path-connected topological spaces for all $i \in I$. I know that the product $\Pi_{i \in I}X_i$ with its product topology is path-connected. But is the converse true ? If $\Pi_{i \in I}X_i$ is path-connected, is every $(X_i, \tau_{X_i})$ path-connected ?

Show that $X=\prod X_\alpha$ is path connected if and only if each $X_\alpha$ is path connected The proof given here says that for $x_\alpha \in X_\alpha$, $x=(0,0,0,x_\alpha,0...,0) \in \Pi X_\alpha$. This is not true, the other topological spaces do not necessarily contain zeros. I think it might be possible to fix it though.

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  • $\begingroup$ You're right, the given vector need not exist in the product topology, but you can select (using the axiom of choice) a fixed point $x_\beta\in X_\beta$ for every $\beta$ and proceed as if $x_\beta=0$, for every $\beta\neq\alpha$ $\endgroup$
    – Alessandro
    Jan 6, 2022 at 8:53

2 Answers 2

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If $X$ is path-connected and $f: X \to Y$ is continuous and onto then $Y$ is path-connected. Just like for connected spaces. (Proof: let $y_1, y_2 \in Y$, find $x_1, x_2 \in X$ so that $f(x_1) = y_1, f(x_2)= y_2$, there is a continuous $p: [0,1] \to X$ with $p(0)=x_1, p(1)=x_2$ as $X$ is path-connected; then $f \circ p: [0,1]\to Y$ is also continuous and shows $Y$ is path-connected).

All $X_\alpha$ are continuous images of $\prod_{\alpha \in A} X$ under the projections $p_\alpha$. Though AC is still needed to show onto-ness of the projections.

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  • $\begingroup$ Well, you actually do choose points in your solution, don't you? When you say "$X_\alpha$ are continuous images of $\prod_{\alpha\in A}X_\alpha$...", you can prove that only by showing that given $x_\alpha\in X_\alpha$ we can choose $\{x_\beta\in X_\beta\}_{\beta\neq\alpha}$ to complete $x_\alpha$ to a vector $x\in X$. $\endgroup$
    – Alessandro
    Jan 6, 2022 at 9:35
  • $\begingroup$ @Alessandro Indeed we cannot avoid AC to show all projections are onto. But I think it's cleaner to separate that fact from the general preservation fact as I did. $\endgroup$ Jan 6, 2022 at 9:39
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As suggested by @Alessandro, axiom of choice gives us the possibility to take an element in every space. Then let $x,y \in X_\alpha$, we have that $X=(x_1,\cdots,x_\alpha,\cdots)$ and $Y=(y_1,\cdots, y_\alpha, \cdots)$ are in $\Pi X_i$ and are path connected via $f$. The projection along $p_{X_\alpha}$ gives us the conclusion.

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