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Let $q$ denote the quotient map and write $[X]$ for a point of Real projective space. For $i=1,\dots ,n+1$, take the open setIt is open in the quotient topology because its inverse image under $q$ is the open set $\{X:X_i\ne 0\}$. These sets cover, since a nonzero vector has at least one nonzero coordinate. Define the manifold chartThe hat means that the $i$th coordinate is omitted. Ratios are unchanged by multiplying $X$ by any nonzero real number, so the chart is well defined. Its inverse inserts $1$ into position $i$ and takes the resulting equivalence class. The ratios and this inverse are continuous, giving a homeomorphism $U_i\to {\mathbb{R}}^n$.

For a compact description of the smooth transition maps, write $a_i=1$ and let $a_k=X_k/X_i$ for $k\ne i$. On $U_i\cap U_j$ one has $a_j\ne 0$, andEach coordinate is a rational smooth function on the open domain $a_j\ne 0$; reversing $i,j$ gives a smooth inverse. Thus these $n+1$ charts form the standard affine atlas of real projective space.

For completeness, the underlying space is Hausdorff: normalizing $X$ identifies it with the antipodal quotient of the unit sphere, and the continuous injective map $[X]\mapsto X{X}^T/(X^{T}X)$ into the space of symmetric matrices separates any two distinct classes by disjoint open neighborhoods. It is also a second-countable space: pull back the countable bases of rational balls in these finitely many charts, and take their union. Therefore

Fix $z_0\in S^1$, and let ${\delta }_{z_0}\in C(S^1{)}^{\ast }$ be the point evaluation functional ${\delta }_{z_0}(f)=f(z_0)$. Define orbit averages in the dual space byEach ${\mu }_n$ has norm at most one and satisfies ${\mu }_n(1_{S^1})=1$. The space $C(S^1)$ is separable, so part (a) gives a subsequence converging pointwise to some $\mu \in C(S^1{)}^{\ast }$. In particular,For every $f\in C(S^1)$, the Cesaro average telescopes at the endpoints:whose absolute value is at most $2\Vert f{\Vert }_{\infty }/n$. Passing to the convergent subsequence provesThus $\mu$ is an invariant functional of a composition operator.

It need not be unique when $T$ is not the identity operator. For the nonidentity homeomorphism $\tau (z)=\overline{z}$, both $1$ and $-1$ are fixed. Hence the two distinct point evaluationshave value one on $1_{S^1}$ and are invariant under $T$. Therefore nonidentity does not imply uniqueness.

A closed set $C\subseteq X$ is compact because it is closed in the compact space $X$. Its image $f(C)$ is compact by the fact that a continuous image of a compact space is compact, and every compact subset of the Hausdorff space $Y$ is closed. Thus the continuous bijection $f$ is a closed map. Its inverse $f^{-1}:Y\to X$ is consequently continuous, so $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

Let $x,y\in H^n(S^n\times S^n;\mathbb{Z})$ be the pullbacks of the two orientation classes. The cohomology ring of a product of two spheres hasand $xy$ generates $H^{2n}\cong \mathbb{Z}$. WriteThe matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ lies in $G{L}_2(\mathbb{Z})$ because a homeomorphism induces a cohomology ring automorphism.

If $n$ is even, graded commutativity givesThus $ab=cd=0$. Invertibility forces $A$ to be diagonal or anti-diagonal, and its two nonzero entries must each be $\pm 1$. Hence there are exactly eight possible actions: the signed permutation matrices.

Evenness is necessary. For $n=1$, the product is the torus, and every $A\in G{L}_2(\mathbb{Z})$ is induced by an integral linear automorphism of the torus. For example, the infinitely many matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ give distinct actions on $H^1$.

Let $A\in G{L}_2(\mathbb{Z})$. If $r-r^{\prime }\in {\mathbb{Z}}^2$, then $A(r-r^{\prime })\in {\mathbb{Z}}^2$, sodefines a continuous map on $X={\mathbb{R}}^2/{\mathbb{Z}}^2$. Since $\det A=\pm 1$, the integral general linear group contains $A^{-1}$ as another integer matrix, and $f_{A^{-1}}$ is the inverse of $f_A$. Hence $f_A$ is a homeomorphism, and $f_A(x_0)=x_0$.

The map $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is the lift of $f_A$ fixing the origin. A loop whose lift ends at $m\in {\mathbb{Z}}^2$ is sent to a loop whose lift ends at $Am$. Under the isomorphism in part (a), the induced map is thereforeThis is precisely the integral linear automorphism of the torus associated with $A$.

Choose the intersection point $p=[({\theta }_0,{\theta }_1)]$ as the zero-cell. Let $a$ be the horizontal one-cell $S^1\times \{{\theta }_1\}$ and $b$ the vertical one-cell $\{{\theta }_0\}\times S^1$. The usual CW complex structure on the torus has one two-cell $c$ attached by the word $ab{a}^{-1}{b}^{-1}$, and the extra disk gives a two-cell $d$ attached along $b$ with degree one. Collapsing $a$ leaves one zero-cell $p$, one one-cell $b$, and the two two-cells $c,d$. The cellular boundary formula givesTherefore

The Excision theorem says that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq \mathrm{int}A$, then inclusion inducesIt lets us compute local homology in arbitrarily small neighborhoods.

Let $C\subseteq X$ be the image of $\{{\theta }_0\}\times S^1$. At a point of $X\setminus C$, a neighborhood is a disk, whose link is a circle. At a point of $C\setminus \{p\}$, three half-disks meet along their diameters, and the link is a Theta graph, with first homology ${\mathbb{Z}}^2$. At $p$, the two folds created by collapsing the horizontal circle give a dumbbell graph as link: two circles joined by an interval. Its first homology is again ${\mathbb{Z}}^2$. The local homology from a link therefore givesBecause local homology is invariant under a homeomorphism, every self-homeomorphism of $X$ preserves the rank-two locus $C$.

It must also preserve $p$. Indeed, $p$ is the unique point of $C$ whose sufficiently small local link is a dumbbell graph; every other point of $C$ has a theta-graph link. The connecting edge of a dumbbell graph is a bridge in a graph, whereas no edge of a theta graph is a bridge, so these local topological types are distinct.

The quotient maps define a continuous homomorphismIts kernel is ${\bigcap }_{U\in \mathcal{U}}U=\{1\}$, because $\mathcal{U}$ is a neighborhood basis and $G$ is Hausdorff. Thus $\Phi$ is injective. The standard compactness argument for inverse limits makes it surjective: a compatible family of cosets has the finite intersection property, and the corresponding closed cosets in compact $G$ have nonempty intersection. Finally, a continuous bijection from compact $G$ to the Hausdorff inverse limit is a homeomorphism. Henceas topological groups.

False. Excursion coding does not remember the speed of traversal. Let $f$ be any nonzero coding function and let $\varphi :[0,1]\to [0,1]$ be a nonidentity increasing homeomorphism. Set $g=f\circ \varphi$. ThenThus $[s]\mapsto [\varphi (s)]$ induces an isometry $T_g\to T_f$, although generally $g\ne f$.

Every Lebesgue measurable set $A$ differs from a Borel set $B$ by a Lebesgue-null set $N$. The given function $h$ is a homeomorphism from $\mathbb{R}$ onto the interval $h(\mathbb{R})$. Hence $h(B)$ is Borel relative to $h(\mathbb{R})$ and therefore Lebesgue measurable in $\mathbb{R}$. Moreover,and $h(N)$ is null by Lusin condition N. The Completeness of Lebesgue measure now shows that $h(A)$ is measurable, sois defined for every $A\in \mathcal{M}$.

Because $h$ is injective, it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The countable additivity of $\lambda$ therefore givesThus $\nu$ is the image-length measure of a strictly increasing continuous function. If $\lambda (A)=0$, the null-set hypothesis gives $\nu (A)=\lambda (h(A))=0$, so

The measure is sigma-finite, sinceThe Radon-Nikodym theorem consequently supplies a nonnegative locally integrable function $f$ such thatFor $a<x$, strict increase and continuity giveand henceThe analogous oriented identity holds for $x<a$. By differentiation of an indefinite Lebesgue integral,

Yes, the differentiability conclusion still holds when $h$ is merely non-decreasing. In fact, the Lebesgue theorem on differentiability of monotone functions says that every monotone function is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the Lebesgue-Stieltjes measure determined by

Because $f$ is a bijection and $e$ is a metric,is nonnegative, symmetric, obeys the triangle inequality, and vanishes exactly when $x=y$. Hence it is a metric. Its open sets are precisely the inverse images under $f$ of $e$-open sets in $N$. Since $f:(M,d)\to (N,e)$ is a homeomorphism, these are exactly the $d$-open sets. Thus $d^{\prime }$ and $d$ are equivalent metrics.

Leta homeomorphism $\mathbb{R}\to (0,1)$, and defineThis metric induces the standard topology on $\mathbb{R}$. However, the sequence $x_n=n$ is $d^{\prime }$-Cauchy because $h(n)\to 1$. If it converged to $x\in \mathbb{R}$, then continuity of $h$ would give $h(x)=1$, impossible. Therefore $(\mathbb{R},d^{\prime })$ is not complete.

Cut the octagon for $S_3$ along a diagonal joining the appropriate two vertex classes. The two resulting polygons can be reattached along that diagonal in the opposite order. Reading the new boundary word gives the square wordwhich is exactly the edge identification displayed for $S_4$. The cut-and-paste map is affine on the two pieces and respects every paired edge, so it descends to a homeomorphism $S_3\cong S_4$. Both are the Klein bottle.

After deleting an open disc, the punctured Klein bottle can be realized as a boundary connected sum of two embedded Möbius strips, and hence embeds in ${\mathbb{R}}^3$. The closed Klein bottle cannot embed: every connected closed surface embedded in ${\mathbb{R}}^3$ is two-sided and therefore orientable, by the Jordan-Brouwer separation theorem, whereas the Klein bottle is nonorientable. Thus

A compact space is a topological space in which every open cover has a finite subcover. A Hausdorff space is one in which every two distinct points have disjoint open neighbourhoods. A homeomorphism is a bijection that is continuous and whose inverse is continuous.

For an equivalence relation $R$ on $X$, let $X/R$ be the set of equivalence classes and letThe quotient topology declares $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. It follows directly from the definition that $q$ is continuous.

Suppose that the continuous map $f:X\to Y$ is constant on equivalence classes. The only possible factorisation iswhich is well-defined by the hypothesis and satisfies $F\circ q=f$. For every open $V\subseteq Y$,is open in $X$. The definition of the quotient topology therefore makes $F^{-1}(V)$ open, so $F$ is continuous. This proves the universal property of the quotient topology.

Let $h:A\to B$ be a homeomorphism, and let $f:B\to B$ be continuous. The conjugate mapis continuous. By hypothesis it has a fixed point $a\in A$. Thereforeand applying $h$ gives $f(h(a))=h(a)$. Hence every continuous self-map of $B$ has a fixed point. This is the homeomorphism invariance of the fixed-point property.

No. TakeIt contains more than one point and has the fixed-point property of a closed interval, but it is not homeomorphic to the closed unit disc. Indeed, deleting any point $(t,0)$ with $0<t<1$ disconnects $B$, so such a point is a cut point. Deleting one point from a two-dimensional closed disc leaves a path-connected space. Since being a cut point is preserved by homeomorphisms, the two spaces cannot be homeomorphic. This gives the required counterexample.

At a point of the open two-cell, the local link is a circle, so local homology in degree two has rank one. At a point of $Z$, a neighborhood consists of $n$ half-discs sharing their boundary interval; its link is the Theta graph ${\Theta }_n$. The local homology from a link calculation givesWhen $n>2$, this local rank characterizes the points of $Z$. Since local homology is invariant under a homeomorphism, every homeomorphism satisfies $f(Z)=Z$, and in particular $f(Z)\subset Z$.

For $n=2$, the two pages form an ordinary disc neighborhood, and $Y$ is ${\mathbb{RP}}^2$. The distinguished circle $Z$ is a projective line, but a projective linear homeomorphism can carry it to a different projective line. Thus the conclusion does not hold.

The unitary group acts on the Lagrangian Grassmannian by $A\cdot L=A(L)$. Every Lagrangian subspace has an orthonormal basis, and adjoining its $J_0$-image gives a unitary basis, so this action is transitive. The stabilizer of the standard real subspace ${\mathbb{R}}^n\subset {\mathbb{C}}^n$ consists exactly of real unitary matrices, namely $O(n)$. Henceis a continuous bijection from a compact space to a Hausdorff space and is therefore a homeomorphism.

For $n=1$, every line in ${\mathbb{R}}^2$ is Lagrangian. ThusConcretely, the line making angle $\theta$ with the real axis corresponds to $e^{2i\theta }$.

For a closed unital subalgebra $A\subseteq B$ containing $x$, invertibility in $A$ implies invertibility in $B$, so ${\sigma }_B(x)\subseteq {\sigma }_A(x)$. On a connected component of the resolvent set of $x$ in $B$, the set of $\lambda$ for which $(x-\lambda 1{)}^{-1}\in A$ is both open, by a local Neumann series, and closed, by closedness of $A$. It contains all sufficiently large $\lambda$, hence the entire unbounded component. Thus spectrum in a closed unital subalgebra says that ${\sigma }_A(x)$ is ${\sigma }_B(x)$ with some bounded complementary components filled in.

Now let $A$ be the Banach subalgebra generated by one element $x$ and put $K={\sigma }_A(x)$. If $D$ were a bounded component of $\mathbb{C}\setminus K$, choose $\lambda \in D$. Since $(x-\lambda 1{)}^{-1}\in A$, polynomials $p_n(x)$ converge to it. The polynomialssatisfy $q_n(\lambda )=0$ and $q_n(x)\to 1$. Applying the contractive Gelfand transform gives $q_n\to 1$ uniformly on $K$, hence on $\partial D$. But the maximum modulus principle applied to $1-q_n$ givesa contradiction. Therefore $\mathbb{C}\setminus K$ is connected.

The mapis continuous and surjective by part a. It is injective because characters agreeing on $x$ agree on every polynomial in $x$, hence by continuity on their norm closure $A$. The character space is compact by the Banach-Alaoglu theorem, while $K$ is Hausdorff, so this continuous bijection is a homeomorphism.

Under this identification, the Gelfand transform $\theta :A\to C(K)$ obeysby the calculation in part b. Since $f(x)\in A$, choose polynomials $p_n$ with $p_n(x)\to f(x)$. Contractivity of $\theta$ givesThus every function holomorphic near $K$ is uniformly approximable there by polynomials.

Use the finite-dimensional vector-space topology on $V^{\ast }$: relative to the dual basis, it is the ordinary Euclidean topology on ${\mathbb{R}}^{|I|}$. Each coordinate map $f\mapsto f(e_i)$ is continuous, sois closed, while $A_i$, $A_i^{-}$, and the finite intersection $C={\bigcap }_i{A}_i$ are open. Their closures are

Every ${\sigma }^{\ast }(w)$ is an invertible linear map with inverse ${\sigma }^{\ast }(w^{-1})$. Linear maps between finite-dimensional topological vector spaces are continuous, so both it and its inverse are continuous. Hence every ${\sigma }^{\ast }(w)$ is a homeomorphism.

For a compact Hausdorff space $K$, every character of the Banach algebra $C(K)$ is an evaluation characterat a unique $x\in K$. The map $x\mapsto {\delta }_x$ is a homeomorphism from $K$ onto ${\Phi }_{C(K)}$ with its Gelfand topology.

An orientation-preserving homeomorphism with locally $L^2$ weak derivatives and essentially bounded infinitesimal distortion. In holomorphic coordinates, its Beltrami coefficient satisfies $\Vert \mu {\Vert }_{\infty }<1$, and the maximal dilatation is $(1+\Vert \mu {\Vert }_{\infty })/(1-\Vert \mu {\Vert }_{\infty })$.