Product Space is Path-connected iff Factor Spaces are Path-connected

Theorem
Let $\mathbb S = \left \langle {\left({S_i, \tau_i}\right)}\right\rangle_{i \in I}$ be a sequence of topological spaces for $i$ in some indexing set $I$ such that $\forall i \in I: S_i \ne \varnothing$.

Let $\displaystyle T = \left({S, \tau}\right) = \prod_{i \mathop \in I} \left({S_i, \tau_i}\right)$ be the product space of $\mathbb S$.

Then $T$ is a path-connected space each of $\left({S_i, \tau_i}\right)$ is a path-connected space.

Necessary Condition
Suppose $S_i$ is path-connected for each $i\in I$.

Let $x \ne y \in S$ be arbitrary.

Since each $S_i$ is path-connected, we have a continuous mapping:
 * $f_i: \left[{0 \,.\,.\, 1}\right] \to S_i$

such that $f_i \left({0}\right) = x_i$ and $f_i \left({1}\right) = y_i$ for all $i \in I$.

We then define:
 * $f : \left[{0 \,.\,.\, 1}\right] \to S$

by setting:
 * $\forall t \in \left[{0 \,.\,.\, 1}\right]: f \left({t}\right)_i = f_i \left({t}\right)$

We have that $f_i = \operatorname{pr}_i \circ f$ are all continuous.

So, by Function to Product Space is Continuous iff Composition with Projections are Continuous, it follows that $f$ is continuous.

Also:
 * $f \left({0}\right)_i = f_i \left({0}\right) = x_i$

and:
 * $f \left({1}\right)_i = f_i \left({1}\right) = y_i$

Therefore $f \left({0}\right) = x$ and $f \left({1}\right) = y$.

So there exists a continuous mapping:
 * $f: \left[{0 \,.\,.\, 1}\right] \to S$

such that $f \left({0}\right) = x$ and $f \left({1}\right) = y$.

Since $x$ and $y$ were arbitrary, we have shown that $S$ is path-connected.

Sufficient Condition
Conversely, suppose $T$ is a path-connected space.

Let $j$ be an arbitrary $j \in I$.

Let $x_j$ and $y_j$ be points in $S_j$ such that $x_j \ne y_j$.

We have that $\forall i \in I: S_i \ne \varnothing$.

Using the axiom of choice, a choice function can be set up to allow the choice of some $q_i \in S_i$ for each $i \ne j$.

We then define:
 * $\forall s \in T: \operatorname{pr}_i \left({s}\right) = q_i$

From Constant Mapping is Continuous, it follows that each of $\operatorname{pr}_i: T \to S_i$ where $i \ne j$ is continuous:

Thus there exist points in $x, y \in S$ such that:
 * $\operatorname{pr}_i \left({x}\right) = q_i$
 * $\operatorname{pr}_i \left({y}\right) = q_i$

for all $i \ne j$, and:
 * $\operatorname{pr}_j \left({x}\right) = x_j$
 * $\operatorname{pr}_j \left({y}\right) = y_j$

Since $T$ is path-connected, there exists a continuous mapping:
 * $f: \left[{0 \,.\,.\, 1}\right] \to S$

such that $f \left({0}\right) = x$ and $f \left({1}\right) = y$.

By Function to Product Space is Continuous iff Composition with Projections are Continuous:
 * $\operatorname{pr}_j \circ f: \left[{0 \,.\,.\, 1}\right] \to S_j$

is also continuous.

Furthermore:
 * $\left({\operatorname{pr}_j \circ f}\right) \left({0}\right) = \operatorname{pr}_j \left({x}\right) = x_j$

and:
 * $\left({\operatorname{pr}_j \circ f}\right) \left({1}\right) = \operatorname{pr}_j \left({y}\right) = y_j$

Thus since $j$ was arbitrary, we have shown that all $S_i$ are path-connected.

Also see

 * Finite Product Space is Connected iff Factors are Connected