Are Products of Arc Connected Spaces Arc Connected, or vice versa? Do Arc-Connected spaces behave well with respect to products? That is, does the following hold -

$\prod X_\alpha$ is arc connected iff $X_\alpha$ is arc connected $\forall \alpha$, where $X_\alpha$ is non-empty $\forall \alpha$

If not, can we construct a counterexample? I've tried looking for this in various textbooks, but I've not been able to find anything either supporting or contradicting the above claim.
 A: Let all $X_\alpha$ be arc-connected and consider $X=\prod_{\alpha \in A} X_\alpha$ and $x \neq y$ in $X$. Define $$A' = \{\alpha \in A: x_ \alpha \neq y_ \alpha\} \neq \emptyset$$
and for each $\alpha \in A'$ we pick a homeomorphism $h_ \alpha: [0,1] \to Z_ \alpha$ such that $h(0)= x_ \alpha$ and $h(1) = y_\alpha$ by arc-connectedness (and AC). Then $h_{A'} = \prod_{\alpha \in A'} h_ \alpha : [0,1]^{A'} \to \prod_{\alpha \in A'} X_\alpha$ is also a homeomorphism (a Cartesian product of homeomorphisms is easily seen to be a homeomomorphism too) and we clearly have a "homeomorphic path" $p:[0,1] \to [0,1]^{A’}$ defined by $p(t)_\alpha = t$ for all $\alpha \in A'$ that connects the $0$ to $1$ in that subproduct. (basically we are using the convexity of powers of $[0,1]$). Now define $h: [0,1] \to X$ by $\pi_\alpha(h(t)) = h_\alpha(t)$ when $\alpha \in A'$ and $\pi_\alpha(h(t)) = x_\alpha = y_\alpha$ for other $\alpha$. Then $h$ is a homeomorpic embedding of $[0,1]$ into $X$ sending $0$ to $x$ and $1$ to $y$.
The reverse is probably true but not so clear to me yet, a possible start:
Suppose that $X=\prod_{\alpha \in A} X_\alpha$ is arc-connected.
Pick a fixed point $q:=(q)_\alpha$ in $X$ (otherwise $X$ is empty and we cannot show anything interesting).
Fix $\alpha_0 \in A$, and let $x \neq y$ be points of $X_{\alpha_0}$. Define $\hat{x}$ and $\hat{y}$ in $X$ by extending them by $q$: e.g. for $\hat{x}$: $\hat{x}_\alpha = x$ if $\alpha = \alpha_0$ and $\hat{x}_\alpha = q_\alpha$ otherwise. Then there is a homeomorphism $h$ from $[0,1]$ into $X$ such that $h(0)=\hat{x}$ and $h(1)=\hat{y}$. I would like to conclude that $\pi_\alpha \circ h$ would be a homeomorphic embedding of $[0,1]$ from $x$ to $y$ but even a two-dimensional picture convinces me that this does not need to hold. So we'd have the "morph" $h$ into shape. I don't see as yet how, but maybe others do.
A: Sorry all, this is wrong. There is no arc from, for example, (0,0) to (1,0). What I thought was an arc is a 1-1 path, continuous but not a homeomorphism.
The other implication does not hold. Let $S$ be the Sierpinski space ($\{0,1\}$ with $\{0\}$ the only non-trivial open set) and let $Y$ be an arc-connected space with at least two points. $S \times Y$ is arc-connected but $S$ is not.
