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A compact subset of a CW complex is contained in a finite CW complex that is a subcomplex of the original complex. Otherwise choose one point in each of infinitely many cells met by the subset. The closure-finiteness axiom makes each closed cell meet the chosen set in finitely many points. The weak topology axiom then makes the chosen set, and each of its subsets, closed. It is consequently an infinite closed discrete subspace of the compact subset, a contradiction. Only finitely many cells are met; their closures form a finite subcomplex containing the subset.

For a connected finite CW complex $X$, choose a point $p_e$ in each open cell $e$. An Euler chain is a singular chain $\xi$ of degree one withIt exists exactly when the Euler characteristic of $X$ is zero. Necessity follows because the total coefficient sum of every one-chain boundary is zero, whereas the displayed zero-chain has coefficient sum $\chi (X)$.

For sufficiency, choose a base point $b$ and paths ${\gamma }_e$ from $b$ to $p_e$. Then $\xi ={\sum }_e(-1{)}^{\dim e}{\gamma }_e$ has boundary equal to the required sum minus $\chi (X)b$, so it works when $\chi (X)=0$. Connected CW complexes are path connected, which justifies choosing these paths.

With the cell points fixed, two Euler chains differ by a cycle and define the same Euler structure when that cycle is a boundary. Adding homology classes makes Euler structures a torsor for $H_1(X;\mathbb{Z})$. This extra choice removes an ambiguity in Turaev torsion.^[ref]

For a connected finite CW complex $X$ with Euler characteristic zero, choose a point $p_e$ in each cell. An Euler chain is a singular chain $\xi$ of degree one satisfyingWith these points fixed, two Euler chains define the same Euler structure when their difference is a chain boundary. Adding cycles makes their classes a set with a simply transitive group action of $H_1(X;\mathbb{Z})$. This choice fixes the cellular-lift ambiguity in Turaev torsion; a homology orientation fixes its sign.^[ref]

Let $X$ be simply connected, with finitely generated integral homology groups, and suppose $H_i(X;\mathbb{Z})=0$ for $i>n$, where $n\ge 2$. Then $X$ admits a weak homotopy equivalence $C\to X$ from a finite CW complex of dimension at most $n+1$. If $X$ already has CW complex homotopy type, this map is a homotopy equivalence.

Start with a finite $n$-dimensional CW complex $C_n$ and an $n$-connected map $f:C_n\to X$, constructed by the finite type CW approximation. This means that $f$ induces isomorphisms on homotopy groups below $n$ and a surjection in degree $n$; it does not mean that $C_n$ itself is $n$-connected.

The kernel of $H_n(C_n;\mathbb{Z})\to H_n(X;\mathbb{Z})$ is a finitely generated free abelian group, because $H_n(C_n;\mathbb{Z})$ is a subgroup of its finite free top cellular chain group. The Relative Hurewicz theorem represents a basis of this kernel by attaching $(n+1)$-cells whose maps extend into $X$. Their cellular boundary is injective, killing exactly that kernel without adding top homology. All integral homology groups now agree, so the homological Whitehead theorem proves the conclusion. The acyclic simply connected case is already weakly equivalent to a point.

For a finite CW complex, every element in the kernel of the rank map in topological K-theory is a nilpotent element. One proof adjoins one positive-dimensional cell at a time: the restriction kernel is square-zero by the relative product in topological K-theory. For connected spaces the rank-zero kernel is Reduced topological K-theory; for disconnected spaces the kernel of restriction to just one basepoint can contain nonzero idempotents.

A finite CW complex can be defined without assuming separation properties in advance. Start with a finite discrete set $X^0$. Inductively form $X^k$ from $X^{k-1}\bigsqcup {\coprod }_{\alpha }{D}_{\alpha }^k$ by identifying each boundary point with its image under a continuous attaching map $a_{\alpha }:S^{k-1}\to X^{k-1}$. Only finitely many disks are attached in total, and $X=X^N$ for some finite $N$. The interiors of the disks are its cells. Its weak topology of a CW complex is the final topology of the characteristic maps: a subset is closed exactly when its inverse image in every characteristic disk is closed. In this finite construction it is the topology obtained by the successive quotients. Closure-finiteness is automatic because there are only finitely many cells.

Here is a direct proof of both requested topological properties, without presupposing that arbitrary attaching maps give a triangulable space. Suppose the compact previous stage $Y$ embeds by $e:Y\hookrightarrow {\mathbb{R}}^m$, and attach one $k$-cell, $k\ge 1$. Extend the boundary coordinate map radially byIt is continuous at zero because the image of the sphere is bounded. Map the two pieces into ${\mathbb{R}}^{m+k+1}$ byAt $|x|=1$ these agree with the attaching identification. They therefore induce a continuous map on the quotient. Its restriction to the old stage is injective. An interior point has positive last coordinate $t=1-|x|$, and the preceding coordinate block recovers $x$ by division by $t$. Hence it cannot coincide with an old point or another interior point. The induced map is injective.

The quotient is compact, being the image of a compact disjoint union. A continuous injection into Euclidean space is Hausdorff, by inverse images of disjoint separating neighborhoods; it is also a homeomorphism onto its image because closed subsets of the compact source have compact, hence closed, images in the Hausdorff target. Finite discrete $X^0$ embeds in $\mathbb{R}$, and induction over the finite list of cells proves $X$ is Hausdorff and its weak topology is exactly an induced Euclidean subspace topology. This is Euclidean embedding by finite cell attachment; no optimal ambient dimension is needed.

For Real projective space, filter by ${\mathbb{RP}}^0\subset \cdots \subset {\mathbb{RP}}^n$. The complement of ${\mathbb{RP}}^{k-1}$ in ${\mathbb{RP}}^k$ is an affine $k$-cell. A characteristic disk is a hemisphere of $S^k$, whose boundary is attached by the antipodal quotient $S^{k-1}\to {\mathbb{RP}}^{k-1}$. Thus there is one cell in every dimension $0,\dots ,n$.

The incidence number is the sum of the two sheet contributions. Fixing the orientation of one sheet gives degree one; the other differs by the antipodal map on $S^{k-1}$, whose degree is $(-1{)}^k$. Consequently the cellular homology of real projective space hasTaking kernels modulo images gives, for $n\ge 1$,Over $\mathbb{R}$, multiplication by two is invertible, soOver ${\mathbb{F}}_2=\mathbb{Z}/2$, every differential vanishes, giving $H_q({\mathbb{RP}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2$ for $0\le q\le n$, and zero otherwise. For $n=0$ the space is a point and only degree-zero homology is present, with the chosen coefficient group.

Use countability of negative curves. A smooth projective surface over $\mathbb{C}$ is a compact smooth manifold homotopy equivalent to a finite CW complex, so its integral cohomology group $H^2(X,\mathbb{Z})$ is finitely generated and therefore countable. Distinct irreducible curves $C,D$ with $C^2<0$ cannot have the same cohomology class: that would give $C\cdot D=C^2<0$, while distinct effective curves have nonnegative intersection multiplicities. Assigning a negative curve its cohomology class is thus injective, proving there are at most countably many negative curves.

For zero self-intersection, uncountable families do occur. On the product of two projective lines ${\mathbb{P}}^1\times {\mathbb{P}}^1$, every fiber ${\mathbb{P}}^1\times \{p\}$ has self-intersection zero, since a different fiber is disjoint and has the same numerical equivalence of divisors class. The parameter $p\in {\mathbb{P}}^1(\mathbb{C})$ varies over an uncountable set. Thus yes, there can be uncountably many such curves.

Write $v$ for the cone vertex. The quotient is the topological mapping cone $C_f=Y{\cup }_{f}CX$. First assume $X\ne \varnothing$, as is implicit in having this vertex; a map then also forces $Y\ne \varnothing$.

Take the open cover consisting of the image of $X\times [0,2/3)$ and the image of $Y\bigsqcup (X\times (1/3,1])$. Denote these sets by $U,V$. The first contracts to $v$, the second has a deformation retraction onto $Y$, and $U\cap V\cong X\times (1/3,2/3)$ retracts onto $X$. Under these identifications the two inclusion maps induce the zero map into positive-degree homology of $U$ and $f_{\ast }$ into the homology of $V$. The Mayer–Vietoris theorem consequently gives, in positive degrees,The same assertion holds in degree zero, but needs care: the ordinary Mayer-Vietoris map isSince ${\epsilon }_Y{f}_{\ast }a={\epsilon }_{X}a$, its kernel equals $\ker f_{\ast }$. On quotienting $H_0(Z;G)$ by the direct summand $G[v]$, its cokernel becomes $\mathrm{coker}{f}_{\ast }$. That quotient is canonically identified with reduced homology by subtracting the augmentation times $[v]$. Thus the last nonzero part isIn particular the map from $H_0(Y;G)$ sends a component class $[y]$ to $[y]-[v]$. This proves the mapping cone exact sequence; finite generation of $G$ is not needed for this construction.

There is a literal empty-space exception in the printed hypotheses. If $X=\varnothing$ and $Y$ is a point, the given quotient is just $Y$, and the asserted degree-zero exact sequence would be $0\to G\to 0$. Thus the displayed reduced sequence presupposes nonempty $X$. It holds also for empty $X$ if one defines its cone to include a new isolated vertex. The later homology-isomorphism claim itself remains true for empty finite complexes, with the cases involving one empty complex dealt with directly in degree zero.

Now take nonempty finite CW complexes $X,Y$. Put $A_q={\tilde{H}}_q(Z;\mathbb{Z})$. The exact sequence giveswhere $H_{-1}=0$. All $A_q$ are finitely generated abelian groups, since the groups for $X,Y$ are finitely generated. Moreover, by exactness,The reduced universal coefficient theorem for homology givesIf all $A_q$ vanish, so does the middle term for every prime, and the exact sequence with coefficients proves that $f_{\ast }$ is an isomorphism modulo every prime. Conversely, those isomorphisms imply that the middle terms vanish, hence $A_q\otimes {\mathbb{F}}_p=0$ for every prime $p$. By the Fundamental theorem of finitely generated abelian groups, a nonzero free summand would survive modulo every prime, and a cyclic torsion summand would survive modulo a prime dividing its order. Thus $A_q=0$ for every $q$, provingThis argument uses the exact sequence to obtain finite generation; it does not assume in advance that the cone of an arbitrary continuous map is a finite CW complex.

Finally suppose $f$ is a cellular map. Keep the cells of $Y$, add the vertex $v$, and add a $(q+1)$-cell for the cone on every $q$-cell of $X$. Its top face attaches to $Y$ through $f$, and its other faces attach to cones on lower-dimensional cells. The cellular condition puts these attachments in the appropriate skeleton. This constructs a finite CW complex structure on $Z$.

Let $A_{\ast }=C_{\ast }^{\mathrm{cell}}(X)$ and $B_{\ast }=C_{\ast }^{\mathrm{cell}}(Y)$ be the ordinary cellular chain complexes, with $A_{-1}=0$. Orient each coned cell so that its boundary has the form $f_{\#}(e)-c(d_{A}e)$. Using $[y]-[v]$ as the reduced generator for each vertex of $Y$ givesIn particular ${\tilde{C}}_0^{\mathrm{cell}}(Z)=B_0$, and the boundary of a cone edge is its image vertex minus $v$. The chain map identity $d_B{f}_{\#}=f_{\#}{d}_A$ givesThis is the cellular chain complex of a mapping cone, equivalently the algebraic mapping cone with the displayed summand and sign conventions.

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 a finite covering map $p:Y\to X$, the K-theory transfer of a finite covering obeysOn each fibre, this is the natural linear isomorphism distributing a tensor product over the direct sum of the fibers of $b$. It gives the formula for vector bundles and hence for their virtual differences.

Now suppose $X$ is a finite CW complex and the covering map has constant degree $n\ge 1$. The class $p_{!}(1)$ is the permutation vector bundle of rank $n$, so $\eta =p_{!}(1)-n$ has rank zero. By nilpotence of rank-zero K-theory classes, some power of $\eta$ is zero. After localization of a ring at $n$, the class $p_{!}(1)=n+\eta$ is a unit, with inverse given by a finite geometric sum in $-\eta /n$. If $p^{\ast }a=0$, the projection formula gives $a{p}_{!}(1)=0$, hence $a=0$ after this localization. Thus pullback is injective after inverting $n$. The finite-CW hypothesis supplies the nilpotence step.

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.

For CW pairs $(X,A)$ and $(Y,B)$, write their product pair asUse the compactly generated product, equivalently the weak CW complex topology on products of cells. For arbitrary infinite CW complexes this can be finer than the ordinary product topology. The ordinary product topology already agrees when one factor is locally compact, or when both factors have countably many cells; in particular there is no distinction for finite CW complexes.^[ref]

The quotient of this pair is $(X/A)\wedge (Y/B)$. Setwise, the collapsed subset is exactly the set of pairs with at least one coordinate in its collapsed subspace; the compatible quotient constructions give the homeomorphism in the compactly generated space convention. In a multiplicative generalized cohomology theory, external products of relative classes of degrees $k,m$ lie in degree $k+m$ for this pair. Pullback along a diagonal map gives the relative cup product.

Let $E_0,E_1$ be complex vector bundles of rank $d\ge 1$ over a finite CW complex $X$ with $\dim X\le 2d$. Iffor some $r$, then $E_0\cong E_1$: in this dimension range, stable isomorphism already determines the actual vector bundle.

The stabilization $BU(d)\to BU$ of the classifying space for rank-$d$ complex vector bundles is an n-connected map with $n=2d+1$. One sees the range from the fibration with fiber $U(d+1)/U(d)\cong S^{2d+1}$ and then iterates stabilization. The vanishing of the fiber's homotopy groups through degree $2d$ allows a stable homotopy between classifying maps to lift over $X\times [0,1]$, relative to its ends. Thus the original classifying maps are homotopic and the bundles are isomorphic.

The rank bound is useful when invariants such as the Chern character first identify stable classes: it supplies the final cancellation step in the stated rank and dimension range.