# First fundamental group and connectedness of $X:=\mathbb{R^4}\setminus \{\pi_1 \cup \pi_2 \}$

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.

• 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) Commented Feb 14, 2019 at 9:47

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 $$\mathbb Z\times\mathbb Z$$.