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 ?)

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    $\begingroup$ Read math.stackexchange.com/questions/454627/… $\endgroup$ – Rubi Shnol Mar 10 '15 at 11:57
  • $\begingroup$ But how do I conclude that $\gamma$ is continuous ? $\endgroup$ – Shuzheng Mar 10 '15 at 11:58
  • $\begingroup$ If you have two points $(x_1, x_2), (x_1', x_2')$ in $X_1 \times X_2$, have you considered using the path $\gamma = (\gamma_1, \gamma_2)$ where $\gamma_1$ and $\gamma_2$ are two paths between $x_1, x_2$ in $X_1$ and $x_1', x_2'$ in $X_2$, respectively? $\endgroup$ – Balarka Sen Mar 10 '15 at 11:59
  • $\begingroup$ A pair of function is continuous if each component is: just use the definition checking the preimage of an open. $\endgroup$ – User3773 Mar 10 '15 at 11:59
  • $\begingroup$ I dont know this result ? Is it easy to prove ? $\endgroup$ – Shuzheng Mar 10 '15 at 12:02

You can simplify things by selecting paths $\gamma_1: [0,1]\to X_1$ with $\gamma_1$ from $x_1$ to $x_2$ and $\gamma_2: [0,1] \to X_2$ from $y_1$ to $y_2$. Then set $$\gamma: [0, 2] \to X_1\times X_2,\quad t\mapsto \begin{cases}(\gamma_1(t),y_1) & t\le 1 \\ (x_2, \gamma_2(t-1)) & t > 1\end{cases}$$ Continuity now is trivial from the component-wise continuity.

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