Path connected product topology implies every topology is path connected

Question

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.

Answer 1

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.

Answer 2

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.