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In the standard free basis of a group $a,b$ for the fundamental group of a once-punctured torus, its oriented boundary loop represents $ab{a}^{-1}{b}^{-1}$ up to orientation and conjugation. The square model proves this: radial deformation retraction takes the inner boundary once around the four sides, whose orientations read $a,b,a^{-1},b^{-1}$. Attaching a simply connected space along that boundary kills the normal closure of the commutator of group elements, whenever the hypotheses of the Seifert-van Kampen theorem hold.

On $\mathbb{C}\setminus [-1,1]$, take the principal square root with positive real part. The Möbius transformation $(z-1)/(z+1)$ has image $(\mathbb{C}\setminus (-\infty ,0])\setminus \{1\}$. The principal square root therefore maps to the right half-plane with $1$ omitted, and the final Möbius transformation gives exactly $0<|w|<1$. The inverse is $z=-(w+w^{-1})/2$. Adding $\infty$ to the source on the Riemann sphere fills the missing value $w=0$ and gives a conformal map onto the full disk. Without $\infty$, a loop surrounding the segment has nonzero winding number, so the source is not a simply connected space and cannot be conformally equivalent to a disk.

For a connective spectrum with ${\pi }_{0}E\cong \mathbb{Z}$, an $E$-homology isomorphism of simply connected spaces is a weak homotopy equivalence. Its simply connected homotopy mapping cone is $E$-acyclic. If it had a first nonzero integral homology group, the Hurewicz theorem and the bottom-degree generalized Hurewicz isomorphism would produce a nonzero $E$-homology group. Thus the cone is integrally acyclic and the homological Whitehead theorem applies.

For simply connected space $SU(5)$ broken to the connected Standard Model subgroup, the exact homotopy sequence gives ${\pi }_2(G/H)\cong {\pi }_1(H)\cong \mathbb{Z}$. Stable magnetic monopoles form. Their gauge and scalar cores have scales $(g\eta {)}^{-1}$ and $(\sqrt{\lambda }\eta {)}^{-1}$; their masses are $M_M\sim 4\pi \eta /g$ times a dimensionless profile factor. The physical subgroup is $(SU(3)\times SU(2)\times U(1))/{\mathbb{Z}}_6$, which retains an infinite-cyclic fundamental group.

// Target: cosmology.bigb

A path-connected region is multiply connected when some closed path cannot contract to a point within the region. An annulus and the complement of a solid torus are examples. The accessible region in the ideal magnetic Aharonov-Bohm effect has this obstruction. Such a region is not a simply connected space.

An exact differential $Mdx+Ndy$ is $d\Phi$ for a scalar potential function $\Phi$, so $M=\partial \Phi /\partial x$ and $N=\partial \Phi /\partial y$. Equality $\partial M/\partial y=\partial N/\partial x$ is necessary, and is sufficient on a simply connected space when these partial derivatives are continuous.

Here, after multiplying the ordinary differential equation by $dx$,On either half-plane $y>0$ or $y<0$, integrate $M$ with respect to $x$:Comparison of $\partial \Phi /\partial y$ with $N$ gives $h^{\prime }(y)=6y$, hence $h(y)=3y^2$ up to a constant. The solution is therefore the family of level setsLocally these define differentiable graphs $y(x)$ wherever $N\ne 0$, by the implicit function theorem. When $N=0$, $M\ne 0$ on this domain, so a finite derivative cannot satisfy $M+N{y}^{\prime }=0$ there; such points are vertical tangents of the implicit curves, rather than additional classical graph solutions.

Embed the planar vector field as $\mathbf{V}=(A(x,y),B(x,y),0)$, independent of $z$. Applying the Cartesian curl formula givesConsequently, for continuously differentiable components,This is the local closedness condition for the one-form $Adx+Bdy$. To infer global path independence from this condition alone, the domain also matters: it suffices that the planar domain is a simply connected space. The polynomial field in the next calculation is defined on the whole plane, so there is no missing-hole obstruction.

Set $A=10x^4-2x{y}^3$ and $B=-3x^2{y}^2$. Their mixed derivatives agree: $A_y=-6x{y}^2=B_x$. Both components are smooth on all of ${\mathbb{R}}^2$, a simply connected space, so the line integral is path independent.

Choose a horizontal segment from $(0,0)$ to $(2,0)$ followed by a vertical segment to $(2,1)$. On the first, $y=0$ and $dy=0$; on the second, $x=2$ and $dx=0$. ThusThe selected path and its orientation are shown below. No contribution comes from a differential whose coordinate is held constant on a segment.

The SU(5) grand unified theory has simply connected space $G=SU(5)$ and a connected Standard Model subgroup. More precisely that subgroup is $(SU(3)\times SU(2)\times U(1))/{\mathbb{Z}}_6$; the product notation suppresses a finite central quotient. The long exact sequence of homotopy groups of a fibration $H\to G\to G/H$ givesThus stable magnetic monopoles form, whereas this transition does not require walls or strings. The monopole core and mass scales in an SU(5) transition are set by the massive gauge and scalar fields:For order-one couplings, the core size is of order ${\eta }^{-1}$ and the mass is of order $\eta$ up to the often substantial factor $4\pi /g$. These heavy objects are nonrelativistic soon after formation and can survive as stable relics.

Choose basepoints and replace the spaces and map by CW complex models. If the hypothesis is stated for unreduced represented homology theory, the basepoint coefficient summands split naturally, so it also gives an isomorphism in reduced represented homology theory. Let $C$ be the homotopy mapping cone of $f$. Its reduced represented homology theory is zero in every degree, by the exact sequence of the corresponding cofiber sequence of spectra.

The mapping cone $C$ is a simply connected space: the mapping cylinder retracts onto $Z$, and coning off the path-connected space $Y$ adds no fundamental-group generators; the van Kampen calculation gives the quotient of ${\pi }_{1}Z$ by the image of ${\pi }_{1}Y$, which is trivial. Suppose some positive-degree integral homology group of $C$ were nonzero, and let $n$ be the least such degree. Then $n\ge 2$. Starting with ${\pi }_{1}C=0$, the Hurewicz theorem shows inductively that ${\pi }_{i}C=0$ for $i<n$: at each first possible degree $i$, its homotopy group is identified with the zero group $H_i(C;\mathbb{Z})$. It then identifiesApplying the result of part (a) to the $(n-1)$-connected based space $C$ gives ${\tilde{E}}_n(C)\cong {\pi }_{n}C\ne 0$, contradicting the vanishing of its represented homology theory.

Thus $C$ is integrally acyclic, and the ordinary homology exact sequence says that $f$ is an integral homology isomorphism. The allowed homological Whitehead theorem for simply connected spaces makes the map of CW complex models a homotopy equivalence. Returning to the original spaces therefore givesThe simply connected hypothesis is used both in the first-nonzero-degree Hurewicz theorem argument and in the final homological Whitehead theorem.

A simply connected space is path-connected and has trivial fundamental group, equivalently every based loop is null-homotopic. The Whitehead theorem says that a map of CW complexes inducing isomorphisms on all homotopy groups is a homotopy equivalence. Its homological Whitehead theorem form says that a map between simply connected CW complexes, in particular finite ones, is a homotopy equivalence if it induces isomorphisms on integral homology. The latter implication follows from the relative Hurewicz theorem and the ordinary Whitehead theorem; homotopy equivalences necessarily induce homology isomorphisms.

For the attached-cell space $X$, the cellular homology chain complex in positive degrees isDualizing gives the cellular cohomology complex $\mathbb{Z}\stackrel{0}{\to }\mathbb{Z}\stackrel{t\mapsto (2t,3t)}{\to }{\mathbb{Z}}^2$. The last map is injective and its image is primitive because $\gcd (2,3)=1$. Its cokernel is $\mathbb{Z}$, explicitly via $(a,b)\mapsto 3a-2b$. HenceThe van Kampen theorem gives ${\pi }_1(X)=⟨a\mid a^2=1,a^3=1⟩=1$. Moreover $H_2(X)=\ker (23)$ is generated by $(3,-2)$. Since $X$ is simply connected, the Hurewicz theorem gives ${\pi }_2(X)\cong H_2(X)$. Choose a map $S^2\to X$ representing that generator. It induces an isomorphism on $H_2$, and on $H_0$, while all other reduced homology groups vanish on both sides. The homological Whitehead theorem proves $\boxed{X\simeq S^2}$. This is a special case of coprime two-cell attachments to a circle.

It is not homeomorphic to the sphere. Indeed, if it were, its embedded one-skeleton would be a Jordan circle in $S^2$, with the two open cell interiors as its complementary regions. The Jordan curve theorem and its Schoenflies form identify the closures of those regions with discs. A characteristic map, homeomorphic on the interior of such a disc, has local degree $+1$ or $-1$ at an interior point; the boundary homomorphism in relative homology then makes its attaching map on the circle have degree $+1$ or $-1$. This contradicts degrees $2$ and $3$. This argument does not require choosing special covering representatives of the attaching maps.

Finally, two compact examples with the same additive cohomology areBoth have $H^0=\mathbb{Z}$, $H^1={\mathbb{Z}}^2$, $H^2=\mathbb{Z}$ and zero higher cohomology, by their cellular chain complexes. But their fundamental groups are respectively ${\mathbb{Z}}^2$ and the nonabelian free group on two generators, so they cannot be homotopy equivalent. This illustrates that cohomology groups do not determine homotopy type.

For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, the necessary and sufficient condition for a conservative vector field is $\nabla \times \mathbf{F}=0$. Necessity follows from equality of mixed partial derivatives of a potential; sufficiency uses the ${\mathbb{R}}^3$ being a simply connected space. On a general domain the topology cannot be omitted.

Here the relevant mixed partial derivatives areThus the curl vanishes. Integrating the first component in $x$ gives $\Phi =x^2\cos y-2x{z}^3+A(y,z)$. Matching the second component gives $A_y=3+2y{e}^z$, hence $A=3y+y^2{e}^z+B(z)$. Matching the last component forces $B^{\prime }=0$. A potential of a conservative vector field with the convention $\mathbf{F}=\nabla \Phi$ is consequentlyIf a physical potential is defined instead through $\mathbf{F}=-\nabla V$, it is $V=-\Phi$.

Let the free surface be at $r=R_1$ and take the Newtonian gravitational potential to vanish at infinity. Mass conservation fixes $R_1=(3M_1/(4\pi {\rho }_1){)}^{1/3}$. The enclosed mass is $m(r)=4\pi {\rho }_1{r}^3/3$, and hydrostatic equilibrium requires $d{p}_0/dr=-{\rho }_{1}Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, givingThe pressure is zero outside. A prescribed constant external pressure would simply add that constant to $p_0$.

The bulk Eulerian and Lagrangian fluid perturbations obey $\delta \rho =0$ because the mass density is uniform and the Lagrangian displacement has zero divergence. The linearized Euler momentum equation in the interior isTaking the curl gives ${\omega }^2\nabla \times \mathbf{\xi }=0$. Thus every nonzero-frequency normal mode has an irrotational flow displacement; the simply connected space formed by the interior admits $\mathbf{\xi }=\nabla U$ and incompressibility gives the Laplace equation ${\nabla }^{2}U=0$. For a regular center, one spherical harmonic component is $U=A_l{r}^l{Y}_l^m$, with $|m|\le l$.

It is essential to retain the surface gravity perturbation of a uniform-density sphere. Although the bulk mass density perturbation is zero, the displaced density discontinuity givesThe Poisson equation implies that $\delta \Phi$ is continuous and its outward radial derivative has the jumpUsing regularity at the center and decay at infinity, writeThe derivative jump is $-(2l+1)C_l{R}_1^{l-1}{Y}_l^m$, so $C_l=-4\pi G{\rho }_{1}l{A}_l/(2l+1)$.

Integrating the bulk Euler momentum equation gives $\delta p/{\rho }_1={\omega }^{2}U-\delta \Phi$ for $l\ge 1$. At the free surface the Lagrangian pressure perturbation vanishes:Combining these relations yields the Kelvin stellar mode frequencyThe $2l+1$ choices of $m$ are degenerate because of spherical symmetry. For $l=1$, $r{Y}_1^m$ is a linear combination of coordinates in a Cartesian coordinate system, so $\nabla U$ is a constant displacement: the zero frequency is rigid translation of the whole isolated body. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by incompressibility and a regular center.

The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential incompressible stellar surface modes. It is false for all neutral displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ has zero divergence, zero normal displacement at the surface, and $\delta p=\delta \Phi =0$, so it is a zero-frequency displacement, but $\nabla \times \mathbf{\xi }=2\mathbf{\Omega }\ne 0$. This does not change the Kelvin stellar mode spectrum just derived.

A domain is simply connected when every closed path in it can be continuously contracted to a point while remaining in the domain.

Such a domain is a simply connected space.