Suppose $X_1, X_2$ are to path connected spaces. I want to show that $X_1 \times X_2$ is path connected.
Let $(x_1,y_1)$ and $(x_2, y_2)$ be given.
By assumption i know there exist:
$\gamma_1: [a,b] \rightarrow X_1$ with $\gamma_1(a) = x_1$ and $\gamma_1(b) = x_2$,
$\gamma_2: [a',b'] \rightarrow X_2$ with $\gamma_2(a') = y_1$ and $\gamma_1(b') = y_2$.
Now I use the homeomorphism $\pi: [0,1] \rightarrow [a,b]$ given by $\pi(x) = (b-a)x +a$ to construct $y_1^{'}:[0,1] \rightarrow X_1$ given by $y_1^{'}(x) = (\gamma_1 \circ \pi) (x)$
I do the same for $\gamma_2: [a',b'] \rightarrow X_2$.
How do I then see that $\gamma: [0,1] \rightarrow X_1 \times X_2$ with $\gamma(t) = (\gamma_1'(t), \gamma_2'(t))$ is continuous which allow me to conclude that $X_1 \times X_2$ is path connected ? (I know $\gamma(0) = (x_1,y_1)$ and $\gamma(1) = (x_2,y_2)$, so I only need to determine that $\gamma$ is continuous ?)