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The punctured plane has fundamental group $Z$ , while punctured $R^{n}$ is simply connected for $n \geq 3$

Statement

For $n \geq 2$ , put $P_{n} = R^{n} ∖ {0}$ and let $e_{0} = (1, 0, \dots, 0)$ .

The punctured plane satisfies $π_{1} (P_{2}, e_{0}) ≅ Z$ .
For every $n \geq 3$ , the space $P_{n}$ is path-connected and $π_{1} (P_{n}, x)$ is trivial for every $x \in P_{n}$ ; hence $P_{n}$ is simply connected.

Facts & Assumptions

Given: A natural number $n \geq 2$ , the punctured Euclidean space $P_{n}$ , its unit sphere $S^{n - 1}$ , and the standard point $e_{0} \in S^{n - 1}$ .

[L1]

For $n \geq 1$ , radial normalization $r (x) = x / ∥ x ∥_{2}$ is a retraction $P_{n} \to S^{n - 1}$ , and $H (x, t) = ((1 - t) + t / ∥ x ∥_{2}) x$ is a deformation retraction of $P_{n}$ onto $S^{n - 1}$ (For $n \geq 1$ , radial normalisation is a deformation retraction of $R^{n} ∖ {0}$ onto $S^{n - 1}$ ).

[L2]

If $A$ is a deformation retract of $X$ , the inclusion and retraction induce mutually inverse fundamental-group isomorphisms at every basepoint of $A$ (A retract induces an injection on fundamental groups, and a deformation retract induces an isomorphism).

[L3]

The geometric unit circle based at $e_{0} = (1, 0)$ has fundamental group isomorphic to $Z$ (The trigonometric loops give $π_{1} ({(x, y) : x^{2} + y^{2} = 1}, (1, 0)) ≅ Z$ ).

[L4]

For every $m \geq 2$ , the sphere $S^{m}$ is simply connected ( $S^{n}$ is simply connected for every $n \geq 2$ ).

[L5]

Loop concatenation makes each fundamental group a group, with constant-loop identity and path reversal representing inverses (Loop classes form the group $π_{1} (X, x_{0})$ under concatenation).

[F1]

For $n \geq 1$ , the first standard unit vector $e_{0} \in R^{n}$ has first coordinate $1$ and all other coordinates $0$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

Proof

technique · direct

1.1

For $n = 2$ , [L1] and [L2] identify $π_{1} (P_{2}, e_{0})$ with $π_{1} (S^{1}, e_{0})$ , and [L3] identifies the latter with $Z$ .

L1L2L3F1

1.2

Let $n \geq 3$ and $y \in S^{n - 1}$ . Since $n - 1 \geq 2$ , [L4] says that $S^{n - 1}$ is simply connected, so $π_{1} (S^{n - 1}, y)$ is trivial; [L1] and [L2] therefore make $π_{1} (P_{n}, y)$ trivial.

givenL1L2L4

2.1

For an arbitrary $x \in P_{n}$ , the path $γ_{x} (t) = ((1 - t) + t / ∥ x ∥_{2}) x$ runs in $P_{n}$ from $x$ to $r (x)$ . Concatenating an endpoint-fixed homotopy with the fixed paths ${\overline{γ}}_{x}$ and $γ_{x}$ preserves it, so $Φ_{x} ([α]) = [{\overline{γ}}_{x} * α * γ_{x}]$ is well defined. The piecewise formula $K (s, t) = γ_{x} (2 s (1 - t))$ for $s \leq 1 / 2$ and $K (s, t) = γ_{x} (2 (1 - s) (1 - t))$ for $s \geq 1 / 2$ contracts $γ_{x} * {\overline{γ}}_{x}$ to the constant path at $x$ ; applying the same formula to ${\overline{γ}}_{x}$ contracts ${\overline{γ}}_{x} * γ_{x}$ at $r (x)$ . Hence the product of $Φ_{x} ([α])$ and $Φ_{x} ([β])$ cancels its middle $γ_{x} * {\overline{γ}}_{x}$ and equals $Φ_{x} ([α * β])$ , while $[δ] \mapsto [γ_{x} * δ * {\overline{γ}}_{x}]$ is a two-sided inverse. Thus $Φ_{x} : π_{1} (P_{n}, x) \to π_{1} (P_{n}, r (x))$ is an isomorphism, and step 1.2 makes $π_{1} (P_{n}, x)$ trivial.

step 1.2L1L5construct

3.1

Given $x, z \in P_{n}$ , follow $γ_{x}$ to $r (x)$ , a sphere path from $r (x)$ to $r (z)$ supplied by the path-connectedness in [L4], and the reverse of $γ_{z}$ . This gives a path from $x$ to $z$ , so $P_{n}$ is path-connected. Together with step 2.1, this proves simple connectedness and completes both clauses.

step 2.1L1L4construct∎

Depends on

Used by

ℝ² is not homeomorphic to ℝⁿ for n≠2 Theorem

Dependency tree · two levels

44 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen Hatcher, Algebraic Topology, Proposition 1.14 and Corollary 1.16 (standard reference, not scraped)

The punctured plane has fundamental group ZZ, while punctured RnRn is simply connected for n≥3n≥3