# Path connected product topology implies every topology is path connected

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.

• 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 $\beta\neq\alpha$ Jan 6, 2022 at 8:53

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.
• 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 $x\in X$. Jan 6, 2022 at 9:35
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.