Will product of path connected topological spaces be necessarily path connected? Why or why not?

Give me some hints. Thank you.

  • 2
    $\begingroup$ What's the natural way to define a path between $(x,y)$ and $(x',y')$ in $X\times Y$ where $X$ and $Y$ are path-connected spaces? More generally, between $(x_i)_{i\in I}$ and $(x'_i)$ in $\prod X_i, X_i$ all path-connected? Is this natural definition always going to be continuous? $\endgroup$ Jul 29, 2013 at 6:29
  • $\begingroup$ you need a proof ? $\endgroup$ Jul 29, 2013 at 6:52
  • 5
    $\begingroup$ It's interesting that Hagen von Eitzen and Brian M. Scott gave two different constructions, which seem to be good for different purposes. Brian's seems (to me) easier to visualize, while Hagen's proves a stronger result, namely that arbitrary products, even over infinite index sets, preserve path connectedness. $\endgroup$ Feb 8, 2015 at 23:39

2 Answers 2


This is very straightforward from applying the definitions involved: Assume $X=\prod_{i\in I}X_i$ with $X_i$ path-connected. Let $x=(x_i)_{i\in I}$, $y=(y_i)_{i\in I}$ be two points in $X$. By assumption, thee exist continuous paths $\gamma_i\colon[0,1]\to X_i$ with $\gamma_i(0)=x_i$ and $\gamma_i(1)=y_i$. By definition of product, there exists a unique continuous $\gamma\colon[0,1]\to X$ such that $\pi_i\circ\gamma=\gamma_i$ for all $i\in I$ where $\pi_i$ is the projection $X\to X_i$. That makes $\gamma$ a path from $x$ to $y$.

  • $\begingroup$ Thank you for your answer. $\endgroup$
    – Supriyo
    Jul 29, 2013 at 10:31
  • $\begingroup$ @Hagen von Eitzen what do you mean "by definition of product"? $\endgroup$ Nov 1, 2017 at 8:30
  • $\begingroup$ @HagenvonEitzen Could you entertain this follow-up question on your answer? Thank you. $\endgroup$
    – Alex Vong
    Dec 19, 2017 at 15:20
  • 2
    $\begingroup$ @Pedro yes. Since, there exist a homeomorphism b/n $[0,1]$ and $[a, b] $. Any map $f: [a, b] \rightarrow X$ can be extended to a cont. map b/n $f: [0,1] \rightarrow X$ through $f$. $\endgroup$
    – MUH
    Sep 23, 2019 at 4:57
  • 1
    $\begingroup$ Does the same result hold if we were working in box topology $\endgroup$
    – James C
    Jun 14, 2020 at 18:43

HINT: To get from $\langle x_0,y_0\rangle\in X\times Y$ to $\langle x_1,y_1\rangle\in X\times Y$, you can go from $\langle x_0,y_0\rangle$ to $\langle x_1,y_0\rangle$ to $\langle x_1,y_1\rangle$.

  • $\begingroup$ +1 very clear answer $\endgroup$
    – jasmine
    Apr 8, 2018 at 8:57

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