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A topological space is connected if it is not the union of two disjoint nonempty open sets, or equivalently has no nontrivial subset that is both an open set and a closed set.

Let $K_x$ be the union of all connected subspaces containing $x$. A union of connected subsets with a common point is connected: if a separation of a topological space existed, every member containing $x$ would lie wholly in the same side. Hence $K_x$ is connected and contains every connected subspace through $x$, so it is the connected component of $x$.

If $K_x\cap K_y\ne \varnothing$, their union is connected and maximality gives $K_x=K_y$. Otherwise they are disjoint. Since every $x$ belongs to $K_x$, the connected components partition $X$.

Apply Serre duality for compact complex manifolds to $\mathcal{L}\otimes K_M$, where $K_M$ is the canonical bundle:We can show the group on the right vanishes directly from the embedding. Since ${\mathcal{L}}^{-1}=i^{\ast }\mathcal{O}(-1)$, the tautological bundle inclusion gives an inclusion of holomorphic vector bundlesA global holomorphic section of ${\mathcal{L}}^{-1}$ consequently determines $r+1$ global holomorphic functions on $M$. Compactness and the maximum modulus principle make these functions constant on every connected component. On such a component the section is therefore a constant vector $v$ lying in the tautological line $i(x)$ for every $x$.

If $v\ne 0$, all the points $i(x)$ equal $[v]$. This contradicts the fact that $i$ is an embedding of a component of positive dimension $n\ge 1$. Hence $v=0$ on every component, and $H^0(M,{\mathcal{L}}^{-1})=0$. Duality proves

For degree zero, compactness and the maximum modulus principle make holomorphic functions constant on each connected component. For degree two, a holomorphic differential form is $\bar{\partial }$-closed and its $\partial$ derivative has type $(3,0)$, which vanishes on a complex surface. Degrees above two have no nonzero holomorphic forms. It remains to treat a holomorphic differential form $\alpha$ of degree one.

Set $\beta =d\alpha =\partial \alpha$. Its coefficients are holomorphic, so $\beta$ is a holomorphic differential form of degree two, hence closed by the degree-two observation. Its conjugate is closed too. The Stokes theorem givesThe integrand is nonnegative in the complex orientation. In local coordinates, if $\beta =b(z)d{z}_1\wedge d{z}_2$ with $z_j=x_j+i{y}_j$, thenIt is strictly positive wherever $\beta$ is nonzero. Its zero integral therefore forces $\beta =0$ everywhere. Thus holomorphic forms on a compact complex surface are closed, even when the surface is not Kähler:

Let ${\mathcal{Z}}_M^1$ be the sheaf of closed holomorphic one-forms, namely $\ker (\partial :{\Omega }_M^1\to {\Omega }_M^2)$. The local exactness proved in part (b) gives the short exact sequence of sheavesIts long exact sequence in sheaf cohomology beginsThe first arrow is an isomorphism: on a compact complex manifold, a global holomorphic function is constant on each connected component, just as a global section of the constant sheaf is. Thus $\delta$ is injective. By part (c), every global holomorphic differential form of degree one is closed, so $H^0(M,{\mathcal{Z}}_M^1)=H^0(M,{\Omega }_M^1)$.

Concretely, for a holomorphic differential form of degree one choose local holomorphic primitives $f_i$. The connecting class $\delta (\alpha )$ is represented by the locally constant Čech cocycle $f_j-f_i$ on overlaps. The following arrow is induced by the inclusion of the constant sheaf into the sheaf of holomorphic functions. Exactness is inherited from the long exact sequence, giving

All Topological K-theory groups below are complex and are indexed modulo two using Bott periodicity. For $n\ge 1$, Complex K-theory of a sphere givesFor an even-dimensional sphere, choose a generator $u$ of Reduced topological K-theory. Its square vanishes: the Chern character sends $u$ to a top-degree cohomology class, whose square is zero, and the Chern character is injective on this torsion-free abelian group. Thus the ring answers areThe exceptional zero-dimensional sphere consists of two points: $K^0(S^0)=\mathbb{Z}\times \mathbb{Z}$ with coordinatewise multiplication and $K^{-1}(S^0)=0$. In particular its reduced generator is idempotent rather than square-zero.

For the Euler characteristic identity, induct over the cells of a finite CW complex. The starting CW complex consisting of $v$ vertices has $K^0\cong {\mathbb{Z}}^v$, $K^{-1}=0$, and $\chi =v$. If $X$ is obtained from $A$ by attaching one $d$-dimensional cell, the quotient is $S^d$, and the K-theory six-term exact sequence isThese abelian groups are finitely generated by induction. Tensoring the exact sequence with $\mathbb{Q}$ preserves exactness. The alternating sum of dimensions in a cyclic six-term exact sequence is zero, soTherefore

There is a genuine convention issue in the nilpotence assertion. With the usual definition of Reduced topological K-theory as the kernel of restriction to one basepoint, the assertion needs $X$ to be connected. For $X=S^0$ based at its first point, $(0,1)$ lies in ${\tilde{K}}^0(X)$ and satisfies $(0,1{)}^m=(0,1)$ for every $m\ge 1$. Thus it is not a nilpotent element. For an arbitrary finite CW complex, the correct assertion uses the rank map in topological K-theory on every connected component:For a connected space, $I(X)={\tilde{K}}^0(X)$.

Here is an induction proving the corrected assertion. On the vertices, $I(X)=0$. For a positive-dimensional cell attachment $A\subset X$, the ideal $J=\ker (K^0(X)\to K^0(A))$ is square-zero. Indeed, lift two elements of $J$ to relative topological K-theory $K^0(X,A)$. Their product is induced by the reduced diagonalThis map is null-homotopic, since $d<2d$ and $S^{2d}$ is $(2d-1)$-connected. The relative product in topological K-theory is therefore zero, giving $J^2=0$. If $a\in I(X)$, its restriction belongs to $I(A)$, so induction gives $(a{|}_A{)}^m=0$ for some $m$. Then $a^m\in J$ and $a^{2m}=0$. We have proved every Topological K-theory class of rank zero on every connected component is a nilpotent element, and hence the requested result for connected $X$.

For the K-theory transfer of a finite covering, let $V\to Y$ be a vector bundle and defineOn an evenly covered neighborhood the $n$ sheets identify this with a direct sum of $n$ local vector bundles. On overlaps the identifications permute the sheets and apply the transition maps of $V$, so these local descriptions glue to a vector bundle on $X$. This operation respects direct sums, hence extends to a homomorphism of Grothendieck groupsIt obeys the projection formula for the K-theory transfer:For actual vector bundles, this follows fiberwise by distributing the tensor product of vector bundles over the direct sum; it then extends to their Grothendieck groups.

Let $P=p_{!}(1_Y)$ be the permutation vector bundle of the covering. It has rank $n$ on every connected component, but need not be a trivial vector bundle; in particular one cannot replace $p_{!}{p}^{\ast }$ by multiplication by $n$ integrally. Instead writeThe componentwise nilpotence result from the preceding solution gives ${\eta }^N=0$ for some $N$. In the localization of a ring $K^0(X)\otimes \mathbb{Z}[1/n]$, the finite geometric seriesis an inverse. The projection formula for the K-theory transfer givesThus $p^{\ast }(a)=0$ implies $a=0$ after inverting $n$. More explicitly, $(n+\eta {)}^{-1}{p}_{!}$ is a left inverse, proving

A compact real tree is a compact metric space $(T,d)$ such that any $x,y\in T$ are joined by a unique arc, and that arc is isometric to $[0,d(x,y)]$. The multiplicity of a point in a real tree $a\in T$ is the number of connected components of $T\setminus \{a\}$.

False. Join, at one common endpoint $o$, a line segment of length $1/n$ for every positive integer $n$, and use the intrinsic path metric. This is a real tree. It is totally bounded, because outside the first finitely many arms every point lies arbitrarily close to $o$, and it is complete; hence it is compact. Removing $o$ leaves one connected component for every arm, so $o$ has countably infinite multiplicity of a point in a real tree.

The implicit function theorem states that if $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a differentiable bijection with differentiable inverse $g$, the chain rule applied to $g\circ f$ and $f\circ g$ givesThus $D{f}_x$ is an isomorphism with inverse $D{g}_{f(x)}$.

If a continuously differentiable map $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has invertible derivative everywhere, the inverse function theorem makes it a local diffeomorphism. In particular it is an open map, so its image is open. Its image need not be closed: $x\mapsto \arctan x$ has nonzero derivative everywhere and image $(-\pi /2,\pi /2)$.

For the given map of elementary symmetric polynomials,and direct evaluation of the determinant givesHence the critical set isIts complement is the Zariski-open set on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A continuous path cannot change an ordering without crossing $C$. Therefore ${\mathbb{R}}^3\setminus C$ has exactlyconnected components.

As a smooth manifold, $G\cong {\mathbb{R}}^{\ast }\times \mathbb{R}$. It has two connected components, distinguished by the sign of $a$: each of $a>0$ and $a<0$ is connected, and a continuous path in $G$ cannot cross $a=0$.

The rank of a vector bundle is locally constant on the base. Taking the difference of the ranks extends to the Grothendieck group of vector bundles, giving the rank map in Topological K-theory. For a finite CW complex, its kernel consists of classes of rank zero on every connected component.

A compact Lie group has finitely many connected components, each an oriented compact manifold of dimension $n$. Componentwise integration identifies its top-degree de Rham cohomology with one real coordinate per component. For a connected group, if top-degree classes are represented by invariant forms, the invariant volume form on a Lie group spans the cohomology: invariant top forms are one-dimensional, and its positive integral proves that its class is nonzero by the Generalized Stokes theorem. A disconnected group cannot have an invariant representative of every top-degree class, since left translations permute its components. For example, $S^1\times \mathbb{Z}/2$ has two independent degree-one classes but only one independent invariant top form.