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Take the normalizationThe first variation of geodesic energy says that the critical points on the fixed-endpoint path space are exactly the affinely parametrized geodesics. The kernel of the second variation of geodesic energy, or Riemannian index form, is the space of Jacobi fields vanishing at both endpoints. Hence a geodesic is a nondegenerate critical point exactly when that space is zero, or equivalently when its terminal endpoint is not a conjugate point to its initial endpoint along the given geodesic.

The Morse index theorem states that the index is the sum of the multiplicities of all conjugate points strictly inside the parameter interval:The multiplicity at the terminal endpoint gives nullity rather than an additional contribution to the index. These statements use the usual Sobolev $H^1$ completion of the path space; smooth paths have the same homotopy type.

First suppose $p\ne q$ as well as $p\ne -q$, and put $\theta =d(p,q)\in (0,\pi )$. There is a unique great circle through these two points. Every critical path runs along it with constant speed, possibly passing around it extra times. Ifthen a complete enumeration without repetitions isIts length is $L=|\ell |$. Equivalently the two positive length lists areThere are no other critical paths, because a nonconstant geodesic of the round sphere is a constant-speed great circle and its plane must contain $p,q$.

To compute conjugate points and indices of round-sphere geodesics, the round unit sphere has constant sectional curvature $1$. Along a geodesic of speed $L$, a normal Jacobi field in a parallel direction satisfies $y^{\prime \prime }+L^{2}y=0$. A field with $J(0)=0$ is therefore a constant multiple of $\sin (Lt)$ in each of the $n-1$ normal directions. Its tangential component satisfies $y^{\prime \prime }=0$ and contributes no endpoint-vanishing field. Consequently the conjugate points occur at $Lt=j\pi$, each with multiplicity $n-1$. Since neither length list contains an integer multiple of $\pi$, all the displayed critical paths are nondegenerate, and the Morse index theorem gives

The permitted case $p=q$ requires separate treatment. There is a constant geodesic, which has index and nullity zero. All other critical paths areFor each $r$, this is a family parametrized by $S^{n-1}$; both directions of traversal are included through $v$ and $-v$. The interior conjugate points number $2r-1$, and the endpoint is also conjugate. ThusThese are Morse-Bott critical manifolds for $n>1$, rather than nondegenerate critical points. For $n=1$ the same formulas give zero index and nullity, with the two isolated directions comprising $S^0$.

For the requested homology calculation, choose distinct nonantipodal endpoints. Concatenation with a fixed path back to the basepoint gives a homotopy equivalence between this fixed-endpoint path space and the based loop space $\Omega S^n$. We use the Morse cell-attachment theorem for geodesic energy: on a complete compact Riemannian manifold, the fixed-endpoint energy, when all its critical points are nondegenerate, gives a CW complex of the same homotopy type with one cell of dimension equal to the index of each critical point. One can obtain this theorem from finite-dimensional broken-geodesic approximations and ordinary Morse theory; critical energy values tend to infinity here.

The two index lists interleave to give exactly one cell in every dimension $j(n-1)$, $j\ge 0$. If $n>2$, these dimensions are separated by at least two. The cellular homology groups therefore have zero boundary maps, since no occupied cell dimension has an occupied dimension one lower. With integer coefficients,This determines the graded abelian groups. The argument does not need an identification of the multiplication on loop-space homology.

Finally apply the Freudenthal suspension theorem: for an $(n-1)$-connected based CW complex $X$, $n\ge 2$, the suspension homomorphismis an isomorphism for $k\le 2n-2$ and a surjection for $k=2n-1$. A sphere $S^n$ is $(n-1)$-connected, and its reduced suspension of a topological space is homeomorphic to $S^{n+1}$. HenceThe isomorphism is the suspension map. Its stable range is also reflected by the Morse theory cell structure of $\Omega S^{n+1}$: after its bottom $n$-cell, the next positive-dimensional cell has dimension $2n$. For the stated homology problem $n>2$, all the connectivity hypotheses apply.

Writing the suspension of a topological space as two cones and applying the Mayer–Vietoris theorem gives the reduced homology of a suspensionThus $H_0(\Sigma X)\cong \mathbb{Z}$ for nonempty $X$, $H_1(\Sigma X)\cong {\tilde{H}}_0(X)$, and the same shift holds in every higher degree. For a space of finite CW type, the reduced Euler characteristic changes sign, soSince $\chi ({\mathbb{CP}}^2)=3$,But the Euler characteristic of a product satisfies $\chi (A\times A)=\chi (A{)}^2$, a nonnegative perfect square. Neither $3$ nor $-1$ is such a square, so no ${\Sigma }^j{\mathbb{CP}}^2$ is homotopy equivalent to $A\times A$.

A homeomorphism $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is a proper map, so it extends over the one-point compactification to a homeomorphism $f^{+}:S^n\to S^n$ fixing infinity. Defineusing the degree of a continuous mapping. Since $f^{+}$ acts invertibly on $H_n(S^n;\mathbb{Z})\cong \mathbb{Z}$, its degree is $\pm 1$. Functoriality of induced homology maps givesFor $A\in \mathrm{GL}(n,\mathbb{R})$, path connectedness of each determinant-sign component reduces $A$ to the identity when $\det A>0$ and to one coordinate reflection when $\det A<0$. Hence

Suppose $h:Y\times Y\to {\mathbb{R}}^{2n+1}$ were a homeomorphism, and put $d=2n+1$. Then $Y^4\cong {\mathbb{R}}^d\times {\mathbb{R}}^d\cong {\mathbb{R}}^{2d}$. The square of the cyclic permutationinterchanges the two $Y^2$ factors. Under $h\times h$, it is conjugate to the linear factor swap $(u,v)\mapsto (v,u)$ on ${\mathbb{R}}^d\times {\mathbb{R}}^d$, whose determinant has sign $(-1{)}^{d^2}=-1$. Thus $\mathrm{deg}({\sigma }^2)=-1$. On the other hand, multiplicativity gives $\mathrm{deg}({\sigma }^2)=(\mathrm{deg}\sigma {)}^2=1$, a contradiction. Therefore no such $Y$ exists.

DefineThis is continuous and takes values on the unit sphere. It agrees at $(0,y)$ and $(1,y)$, is constant on the bottom edge, and is constant on the top edge. Thus it is constant on every $R$-class, so the universal property of the quotient topology gives a continuous map

For $0<y<1$, the third coordinate $\cos (\pi y)$ determines $y$, and the first two coordinates determine $x$ modulo one. Consequently the only equal values of $f$ in the open strip arise from $x=0$ and $x=1$. At $y=0$ the whole edge maps to the north pole, and at $y=1$ the whole edge maps to the south pole. These are exactly the identifications defining $R$, so $\overline{f}$ is injective. The spherical-coordinate formula also shows that it is surjective.

The square is compact, hence its quotient $X/R$ is compact by part (ii), while $S^2$ is Hausdorff as a subspace of ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now makes $\overline{f}$ a homeomorphism. Geometrically, identifying the vertical sides produces a cylinder and collapsing each boundary circle to a point produces the suspension of a topological space $S^1$, which is $S^2$. This proves the square quotient model of the two-sphere.