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On $\mathbb{R^4}$ consider $\pi_1 := \{x_1=x_2=0\}$ and $\pi_2 :=\{x_3=x_4=0\}$. Let $X:=\mathbb{R^4}\setminus \{\pi_1 \cup \pi_2 \}$ .

Show that $X$ is arc-connected and find $\pi_1 \left(X\right)$

I am almost sure that $X$ could be written as $\left(\mathbb{R^2} \setminus \{\left(0,0\right)\} \right) \times \left(\mathbb{R^2} \setminus \{\left(0,0\right)\} \right)$, which is arc-connected. But I can't show that properly. And what about the fundamental group? Thank you.

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  • $\begingroup$ I changed the word in the title from "connection" to "connectedness". You seem to be asking about the latter, while the former is a certain object in differential geometry: en.wikipedia.org/wiki/Connection_(mathematics) $\endgroup$
    – lisyarus
    Commented Feb 14, 2019 at 9:47

1 Answer 1

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You are right. Note that $(x_1,x_2,x_3,x_4)\in X$ iff $(x_1,x_2,x_3,x_4)\notin \pi_1$ and $(x_1,x_2,x_3,x_4)\notin \pi_2$ iff $(x_1,x_2)\neq (0,0)$ and $(x_3,x_4)\neq(0,0)$, which is exactly equivalent to $(x_1,x_2,x_3,x_4)\in (\mathbb R^2\smallsetminus\{(0,0)\})\times(\mathbb R^2\smallsetminus\{(0,0)\})$. Now $X$ is path-connected as a product of path-connected spaces, and fundamental group of a product is the product of fundamental groups, so in this case it is $\mathbb Z\times\mathbb Z$.

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