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

Theorem
Let $\SS = \family {\struct{S_i, \tau_i}}_{i \in I}$ be an indexed family of topological spaces for $i$ in some indexing set $I$ such that $\forall i \in I: S_i \ne \varnothing$.

Let $\displaystyle T = \struct{S, \tau} = \prod_{i \mathop \in I} \struct{S_i, \tau_i}$ be the product space of $\SS$.

Then $T$ is a path-connected space each of $\struct{S_i, \tau_i}$ 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:
 * $f_i: \closedint {0} {1} \to S_i$

such that $\map {f_i} 0 = x_i$ and $\map {f_i} 1 = y_i$ for all $i \in I$.

We then define:
 * $f : \closedint {0} {1} \to S$

by setting:
 * $\forall t \in \closedint {0} {1} : \map f t_i = \map {f_i} t$

We have that $f_i = \pr_i \circ f$ are all continuous.

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

Also:
 * $\map f 0_i = \map {f_i} 0 = x_i$

and:
 * $\map f 1_i = \map {f_i} 1 = y_i$

Therefore $\map f 0 = x$ and $\map f 1 = y$.

So there exists a continuous mapping:
 * $f: \closedint {0} {1} \to S$

such that $\map f 0 = x$ and $\map f 1 = y$.

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 $\forall i \in I: S_i \ne \O$.

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:
 * $\forall s \in T: \map {\pr_i} s = q_i$

From Constant Mapping is Continuous, it follows that each of $\pr_i: T \to S_i$ where $i \ne j$ is continuous:

Thus there exist points in $x, y \in S$ such that:
 * $\map {\pr_i} x = q_i$
 * $\map {\pr_i} y = q_i$

for all $i \ne j$, and:
 * $\map {\pr_j} x = x_j$
 * $\map {\pr_j} y = y_j$

Since $T$ is path-connected, there exists a continuous mapping:
 * $f: \closedint {0} {1} \to S$

such that $\map f 0 = x$ and $\map f 1 = y$.

By Function to Product Space is Continuous iff Composition with Projections are Continuous:
 * $\pr_j \circ f: \closedint {0} {1} \to S_j$

is also continuous.

Furthermore:
 * $\map {\pr_j \circ f} 0 = \map {\pr_j} x = x_j$

and:
 * $\map {\pr_j \circ f} 1 = \map {\pr_j} y = y_j$

Thus since $x_j$ and $y_j$ were arbitrary, we have shown that $S_j$ is path-connected. 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