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After thinking about it for a while, I believe that the following space is not simply connected, but don't know how to prove it (either way).

Let $Q = [0, 1]^2$ be the unit square. The space is

$$X = \{ (z_1, z_2) \in Q^2 ~:~ z_1 \neq z_2 \},$$

which is like a $4$-dimensional cube minus a $2$-dimensional "diagonal".

Question: How do you prove that $X$ is or isn't simply connected?

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  • $\begingroup$ The fundamental group is $\mathbb{Z}$. It has a cicle as a strong deformation retract. $\endgroup$ Aug 1, 2018 at 19:04

1 Answer 1

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There are some details to be filled in here, but here goes.

You can rotate your $X$ so that your "plane $z_1=z_2$" becomes a coordinate $2$-plane, say $P=\{(0,0,x_3,x_4):x_3,x_4\in\Bbb R\}$. Then $X$ is mapped to a homeomorphic $X'$ with $X'\subseteq \Bbb R^4-P$. But $\Bbb R^4-P$ is homotopy equivalent to a $\Bbb R^2-O$ and so to the circle $S^1$ which has fundamental group $\Bbb Z$.

Idea: take a loop in $\Bbb R^4-P$ which does not contract to a point in that set, and manoeuvre it so that it lies inside $X'$. That will show that $\pi_1(X)\cong \pi_1(X')\ne0$.

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  • $\begingroup$ This is very good. I kept trying to visualize where the knots would get entangled in $\mathbb{R}^4$ when all I needed was some linear algebra. Thank you. $\endgroup$
    – Lee
    Aug 1, 2018 at 18:59

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