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

Give me some hints. Thank you.

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    $\begingroup$ 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? $\endgroup$ – Kevin Arlin Jul 29 '13 at 6:29
  • $\begingroup$ you need a proof ? $\endgroup$ – Eli Elizirov Jul 29 '13 at 6:52
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    $\begingroup$ 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. $\endgroup$ – Andreas Blass Feb 8 '15 at 23:39
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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 $y$.

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  • $\begingroup$ Thank you for your answer. $\endgroup$ – Dutta Jul 29 '13 at 10:31
  • $\begingroup$ @Hagen von Eitzen what do you mean "by definition of product"? $\endgroup$ – Aryaman Jal Nov 1 '17 at 8:30
  • $\begingroup$ @HagenvonEitzen Could you entertain this follow-up question on your answer? Thank you. $\endgroup$ – Alex Vong Dec 19 '17 at 15:20
  • $\begingroup$ Just a quick question, im studying topology from Munkre's Book , and when he defines a path he defines a continuous map from an arbitary interval $[a,b]$ and not from $[0,1]$ , so from this defintion i can from each coordinate pick a path for each coordinate but i dont know if the paths will start in the same place, how do i deal with this problem? Is it due to the fact that the spaces are Homeomorphic so i can always assume theres is one going from $[0,1]$? $\endgroup$ – Lizard King Aug 9 '19 at 11:43
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    $\begingroup$ @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 '19 at 4:57
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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 $\langle x_1,y_1\rangle$.

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  • $\begingroup$ @Samprity: You’re welcome. $\endgroup$ – Brian M. Scott Jul 29 '13 at 10:33
  • $\begingroup$ +1 very clear answer $\endgroup$ – Lily Apr 8 '18 at 8:57

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