Wedge of two circles

Definition

Abstract definition

The wedge of two circles (sometimes also called the figure of eight) is defined as the wedge sum of two circles with respect to any basepoint in both. Note that since a circle is a homogeneous space, the choice of basepoint does not affect the homeomorphism type of the wedge.

The wedge of two circles is denoted $S^1 \vee S^1$.

Definition as a subset of Euclidean space

The 'wedge of two circles can be viewed concretely as the following subset of the Euclidean plane:

$\! \{ (x,y) \mid [(x - 1)^2 + y^2 - 1][(x + 1)^2 + y^2 - 1] = 0 \}$

Here, the first factor corresponds to the circle with center $(1,0)$ and radius $1$, and the second factor corresponds to the circle with center $(-1,0)$ and radius $1$. The two circles are tangent to each other at $(0,0)$, which is the point where the circles are wedged together.

Equivalent spaces

Space How strongly is it equivalent to the wedge of two circles?
The space obtained by removing two points from the Euclidean plane. Homotopy-equivalent
The space obtained by removing three points from the 2-sphere Homotopy-equivalent
quotient space of disjoint union of two squares by the identification of one side of each. Homotopy-equivalent

Algebraic topology

Homology groups

The homology groups with coefficients in $\mathbb{Z}$ are as follows: $H_0(S^1 \vee S^1) \cong \mathbb{Z}$,$H_1(S^1 \vee S^1) \cong \mathbb{Z} \oplus \mathbb{Z}$, and all higher homology groups are zero.

More generally, the homology groups with coefficients in any module $M$ over a commutative unital ring $R$ are as follows: $H_0(S^1 \vee S^1;M) \cong M$, $H_1(S^1 \vee S^1;M) \cong M \oplus M$, and all higher homology groups are zero.

These results follows from the homology computation for spheres (specifically, the circle case) and reduced homology of wedge sum relative to basepoints with neighborhoods that deformation retract to them is direct sum of reduced homologies (which in turn follows from the Mayer-Vietoris homology sequence).

Cohomology groups

The cohomology groups with coefficients in $\mathbb{Z}$ are as follows: $H^0(S^1 \vee S^1) \cong \mathbb{Z}$,$H^1(S^1 \vee S^1) \cong \mathbb{Z} \oplus \mathbb{Z}$, and all higher cohomology groups are zero.

More generally, the cohomology groups with coefficients in a commutative unital ring $R$ are as follows: $H_0(S^1 \vee S^1;R) \cong R$, $H^(S^1 \vee S^1;R) \cong R \oplus R$, and all higher cohomology groups are zero. The cohomology group is $R[x,y]/(x^2,y^2,xy)$ where $x$ and $y$ are a basis for the free module $R \oplus R$ over $R$.

Homotopy groups

The set of path components $\pi_0(S^1 \vee S^1)$ is the one-point set. The fundamental group $\pi_1(S^1 \vee S^1)$ is isomorphic to free group:F2 -- this follows from the fact that fundamental group of wedge sum relative to basepoints with neighborhoods that deformation retract to them is free product of fundamental groups (which in turn follows from the Seifert-van Kampen theorem).