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Suppose diffeomorphisms preserving capacities of solid ellipsoids converge uniformly on compact sets to a diffeomorphism $\varphi$. For $0<a<1$, uniform closeness and degree of a continuous mapping give ${\varphi }_j(aE)\subseteq \varphi (E)\subseteq {\varphi }_j(a^{-1}E)$. The capacity axioms squeeze $c(\varphi (E))$ between $a^{2}c(E)$ and $a^{-2}c(E)$. Letting $a$ tend to one proves equality. Rescaling a smooth $\varphi$ near a point and repeating the argument also proves capacity preservation by its derivative.

For connected oriented closed $d$-dimensional manifolds with $d\ge 2$, their connected sum removes an open $d$-dimensional ball from each and identifies the resulting boundary spheres by an orientation-reversing diffeomorphism. The remaining orientations fit together. Collapsing the separating sphere gives a pinch map to the wedge sum of the two closed manifolds, and projection to either summand has degree of a continuous mapping one. In intermediate positive degrees the cohomology is the direct sum of the summand groups. Products of classes from different summands vanish; products landing in top degree use the single common orientation class. These statements follow from excision, the Mayer–Vietoris sequence and the degree-one projections.

A Euclidean homeomorphism is a proper map and extends to its one-point compactification $S^n$. Its degree is the degree of a continuous mapping of this extension, and belongs to $\{1,-1\}$. A linear isomorphism has degree equal to the sign of its determinant.

A symplectic capacity assigns $c(X,\omega )\in [0,\infty ]$ to symplectic manifolds in a fixed dimension, is monotone under symplectic embeddings, satisfies $c(X,a\omega )=ac(X,\omega )$ for $a>0$, and is nontrivial in the sense thatThe second domain is the standard symplectic cylinder. A normalized symplectic capacity assigns $\pi$ to both unit domains, hence $\pi R^2$ to both radius-$R$ domains. The general axioms already give invariance under symplectomorphisms and quadratic scaling under coordinate dilation. For the rigidity argument, use the assumed existence of capacities in each dimension, including after adding a symplectic plane; normalization is not needed.

Let ${\varphi }_j$ be symplectomorphisms converging locally uniformly to a smooth diffeomorphism $\varphi$. Work in Darboux charts. For a sufficiently small solid ellipsoid $E$ centered at a chosen point and $0<a<1$, uniform convergence and degree of a continuous mapping give, for large $j$,For the second inclusion, ${\varphi }^{-1}{\varphi }_j$ is uniformly close to the identity on the larger ellipsoid. Its boundary stays outside $\overline{E}$ and is homotopic there to the identity, so it has degree one about every point of $E$ and must cover $E$. Monotonicity and scaling now give $a^{2}c(E)\le c(\varphi (E))\le a^{-2}c(E)$. Letting $a$ tend to one proves capacity preservation under uniform limits.

At the chosen point, write $T=D\varphi$. The rescaled maps $(\varphi (tz)-\varphi (0))/t$ preserve capacities of ellipsoids and converge uniformly on compact sets to $T$. Applying the same sandwich argument proves that $T$ preserves the capacities of all solid ellipsoids; so does $T^{-1}$.

We prove the required linear capacity rigidity explicitly. Write the standard symplectic form as $\Omega$. Suppose $\Omega (u,v)=1$ andComplete $u,v$ to a symplectic basis, and complete $T^{T}u/\lambda$, with the appropriately signed $T^{T}v/\lambda$, to another symplectic basis. If their basis matrices are $P,P^{\prime }$, then $A=(P^{\prime }{)}^{-1}{T}^{T}P$ has first two columns $\lambda e_1,\pm \lambda f_1$. Hence $A^T$ sends the unit ball into the symplectic cylinder of radius $\lambda$. This matrix differs from $T$ by symplectic linear maps, so it preserves capacities of ellipsoids. Its first coordinate pair is multiplied by $\lambda$, with a possible sign, and therefore $(A^T{)}^k$ sends the unit ball into the cylinder of radius ${\lambda }^k$. Monotonicity givesfor every $k$, contradicting finiteness of the right-hand unit-cylinder capacity as $k$ tends to infinity. For a normalized capacity, the first iterate already gives the contradiction.

A zero pairing is handled by a small perturbation of $u,v$. A pairing of absolute value greater than one gives the same contradiction for $T^{-1}$, after normalizing the transformed pair. Thusfor every pair. The squared bilinear forms agree, so their difference times their sum is the zero polynomial. The real polynomial ring is an integral domain; consequently one factor vanishes identically. Therefore $T$ is either symplectic or anti-symplectic.

To eliminate the negative sign, repeat the argument for ${\varphi }_j\times {\mathrm{id}}_{{\mathbb{R}}^2}$, which are symplectomorphisms for $\omega \oplus {\omega }_{{\mathbb{R}}^2}$. If ${\varphi }^{\ast }\omega =-\omega$ at the chosen point, the limit pulls this product form back to $-\omega \oplus {\omega }_{{\mathbb{R}}^2}$, which is neither the product form nor its negative. This contradicts the preceding derivative classification. Thus ${\varphi }^{\ast }\omega =\omega$ everywhere. We have proved the Eliashberg–Gromov rigidity theorem: the symplectomorphism group is closed in the $C^0$ topology on diffeomorphisms.

Finally put $E=W^{\perp }$, with perpendicularity taken for the standard Euclidean inner product. Its dimension is two, so being nonisotropic means that the restricted symplectic form is nondegenerate. For the standard compatible complex structure $J$, $W^{\omega }=JE$. If $Jv\in W\cap JE$, then $v\in E$ and $\omega (v,e)=0$ for every $e\in E$, forcing $v=0$. Thus $W$ is a symplectic subspace and admits a symplectic basis extended to the whole space.

In the resulting linear symplectic coordinates, $W$ is $\{q_1=p_1=0\}$. The projection of the bounded set $U$ to the $(q_1,p_1)$ plane is bounded, so $U+W$ lies in a symplectic cylinder of some finite radius $R$. As $U$ is nonempty and open, it contains a ball of some radius $r>0$, which is also contained in $U+W$. Monotonicity and scaling giveas in capacity of a bounded set thickened by a symplectic subspace. For a normalized symplectic capacity, the two bounds simplify to $\pi r^2$ and $\pi R^2$.

Suppose an odd continuous map $f:S^2\to S^1$ existed. Restrict to the equator, parametrized by $e^{i\theta }$. Choose a continuous real lift $F$ with $f(e^{i\theta })=e^{iF(\theta )}$. Oddness gives $F(\theta +\pi )-F(\theta )=\pi +2\pi k$ for a fixed integer $k$, since the integer-valued difference is continuous. Thus $F(\theta +2\pi )-F(\theta )=2\pi (1+2k)$, and the equatorial map has odd degree of a continuous mapping, hence nonzero degree. But it extends over a hemisphere, which is a disk, so the equatorial loop is null-homotopic and has degree zero. This contradiction proves the dimension-two Borsuk-Ulam theorem in the stated form.

If a continuous injection $h:S^2\to {\mathbb{R}}^2$ existed, then $h(x)-h(-x)$ would never vanish. Normalization would definea continuous odd map, already ruled out. In particular the sphere cannot be homeomorphic to any planar subset.

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.

Write $S^n$ as the union of slightly enlarged northern and southern hemispheres $U$ and $V$. Both are contractible, while $U\cap V$ deformation retracts onto $S^{n-1}$. The reduced Mayer–Vietoris theorem therefore givesStarting from ${\tilde{H}}_0(S^0;\mathbb{Z})\cong \mathbb{Z}$ proves the homology of a sphere:This uses no cellular homology. A reflection of $S^n$ reverses its orientation and has degree of a continuous mapping $-1$, so it induces the identity on $H_0$, multiplication by $-1$ on $H_n$, and the unique map between zero groups in every other degree.

For a CW complex with skeleta $X^k$, its cellular chain complex isThe differential is the connecting map to $H_{k-1}(X^{k-1})$ followed by passage to $H_{k-1}(X^{k-1},X^{k-2})$. Equivalently, the coefficient of a $(k-1)$-cell in the boundary of a $k$-cell is the degree obtained from its attaching map after collapsing the complement of that lower cell. This is the cellular boundary formula.

The quotient $D^k/(x\sim -x\text{ on }{S}^{k-1})$ builds ${\mathbb{RP}}^k$ from ${\mathbb{RP}}^{k-1}$ by one $k$-cell, so ${\mathbb{RP}}^n$ has one cell in each dimension $0,\dots ,n$. The two lifts of the attaching map contribute with relative sign $(-1{)}^k$, and the cellular homology of real projective space has differentialConsequentlyWith ${\mathbb{F}}_2=\mathbb{Z}/2$ coefficients every differential vanishes, and hencewith zero homology outside that range.

A pinch map collapses a separating subspace to a point so that the quotient splits into a wedge sum. For a connected sum of oriented manifolds, collapsing the separating boundary sphere gives a map $M\#N\to M\vee N$. The two wedge projections have degree of a continuous mapping one with compatible orientations. Consequently their pullbacks identify the two top orientation classes, while positive-degree classes from different summands have zero cup product.