Simply connected topological space

Not to be confused with: a contractible topological space

Definition

Let $(X,\mathcal{ J })$ be a topological space, we say $X$ is simply connected if:

• $X$ is a path-connected topological space, and
• $\pi_1(X)$ is the trivial group

See next

• $(X,\mathcal{ J })$ is simply connected $\iff$ $\forall p,q\in C([0,1],X)\big[\big(p(0)\eq q(0)\wedge p(1)\eq q(1)\big)\implies p\simeq q\ (\text{rel }\{0,1\}))\big]$ - note $C([0,1],X)$ is the set of all paths into $X$
• If a topological space is simply connected then any retraction of that space is simply connected
  • and its contrapositive: If a retraction of a topological space is not simply connected then the space itself is not simply connected

Examples of simply connected spaces

• All convex subsets of $\mathbb{R}^n$ are simply connected topological subspaces
  • As $\mathbb{R}^n$ is itself convex, we see as a corollary: $\mathbb{R}^n$ is a simply connected topological space
• $\mathbb{S}^n$ is simply connected for $n\ge 2$
• $\mathbb{R}^n-\{0\}$ is a simply connected topological subspace for $n\ge 3$
  • Follows from $\mathbb{R}^n-\{0\}$ strongly deformation retracts to $\mathbb{S}^{n-1}$, which means $\mathbb{R}^n-\{0\}$ and $\mathbb{S}^{n-1}$ are homotopy equivalent topological spaces and then homotopy invariance of the fundamental group tells us $\pi_1(\mathbb{R}^n-\{0\})\cong\pi_1(\mathbb{S}^{n-1})$ and we know $\mathbb{S}^n$ is simply connected for $n\ge 2$ from above.
• $\bar{\mathbb{B} }^n-\{0\}$ is a simply connected topological subspace for $n\ge 3$
  • Follows from $\bar{\mathbb{B} }^n-\{0\}$ strongly deformation retracts to $\mathbb{S}^{n-1}$, which means $\bar{\mathbb{B} }^n-\{0\}$ and $\mathbb{S}^{n-1}$ are homotopy equivalent topological spaces and then homotopy invariance of the fundamental group tells us $\pi_1(\bar{\mathbb{B} }^n-\{0\})\cong\pi_1(\mathbb{S}^{n-1})$ and we know $\mathbb{S}^n$ is simply connected for $n\ge 2$ from above.