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For a locally compact space, replace finite singular chains by chains for which every compact subset meets the images of only finitely many simplices with nonzero coefficients. The boundary operator preserves this local finiteness, giving the Borel-Moore chain complex and its homology. On a compact space it reduces to ordinary singular homology. A noncompact manifold has a locally finite fundamental class, with its orientation local system when necessary. The chain groups and homology are zero in negative degrees.

For a continuous map $f:X\to Y$, precomposition with the induced map on singular chains sends a singular cochain on $Y$ to one on $X$. It commutes with the coboundary and induces the induced map on cohomology.

Suppose $X={\bigcup }_a{X}_a$ is directed by inclusion and every compact subset of $X$ lies in some $X_a$. Every finite singular chain then lies in one $X_a$, so $C_{\ast }(X)={\underrightarrow{\lim }}_a{C}_{\ast }(X_a)$. Exactness of filtered colimits gives

The group $I{S}_i^{\bar{p}}(X)$ consists of finite singular chains whose nonzero simplices are allowable singular simplices and whose boundary, after cancellation, also consists of allowable simplices. Since ${\partial }^2=0$, these groups form a chain complex. Its homology is intersection homology.

In the Deligne-sheaf convention, $I{H}_{\bar{p}}^i(X)={\mathbb{H}}^i(X,I{C}_{\bar{p}})$, with $I{C}_{\bar{p}}$ normalized in degree zero on the regular stratum. For a compact oriented $N$-dimensional pseudomanifold over a field, $I{H}_{\bar{p}}^i(X)\cong I{H}_{N-i}^{\bar{p}}(X)$ and it is dual to $I{H}_i^{\bar{q}}(X)$, where $\bar{p}(k)+\bar{q}(k)=k-2$. Thus the dual of lower-middle singular chains uses the upper-middle Deligne sheaf. The two middle conventions agree on Witt spaces.

Use integral singular chains. The inclusion $Y\subseteq X$ identifies $C_k(Y)$ with the subgroup of $C_k(X)$ generated by the singular simplices whose images lie in $Y$. It is a chain subcomplex. DefineThis boundary is well defined because $\partial C_k(Y)\subseteq C_{k-1}(Y)$, and its square is zero because ${\partial }^2=0$. Its homology is, by definition, $H_k(X,Y)$. A relative cycle has a representative $c$ with $\partial c\in C_{k-1}(Y)$; representatives of the same relative homology class differ by $\partial b+y$, where $b\in C_{k+1}(X)$ and $y\in C_k(Y)$.

Inclusion and quotient induce homomorphisms $i_{\ast }:H_k(Y)\to H_k(X)$ and $j_{\ast }:H_k(X)\to H_k(X,Y)$. Define the relative homology connecting homomorphism byHere $\partial c$ is a cycle in $Y$ since ${\partial }^{2}c=0$. If $c$ is changed to $c+\partial b+y$, its boundary changes by $\partial y$, a boundary in $Y$, proving independence of representatives. The formula is additive, so $\delta$ is a homomorphism.

The long exact sequence in relative homology isand ends inWe prove all three types of exactness directly, so no unproved abstract lemma about chain complexes is required.

At $H_k(X)$, $j_{\ast }{i}_{\ast }=0$ because a chain in $Y$ is zero in the relative chain group. Conversely, let an absolute cycle $c$ have $j_{\ast }[c]=0$. The definition of a relative boundary gives $c=\partial b+y$ with $y\in C_k(Y)$. Since $c$ is a cycle, $\partial y=0$. Thus $[c]=i_{\ast }[y]$, proving $\ker j_{\ast }=\mathrm{im}{i}_{\ast }$.

At $H_k(X,Y)$, $\delta j_{\ast }=0$ because an absolute cycle has zero boundary. Conversely, suppose $\delta [c]=0$. Then $\partial c=\partial y$ for some $y\in C_k(Y)$. Therefore $c-y$ is an absolute cycle and $j_{\ast }[c-y]=[c]$. Hence $\ker \delta =\mathrm{im}{j}_{\ast }$.

At $H_{k-1}(Y)$, $i_{\ast }\delta =0$ because $\partial c$ is a boundary in $X$. Conversely, if a cycle $y\in C_{k-1}(Y)$ becomes zero in $H_{k-1}(X)$, then $y=\partial c$ for some $c\in C_k(X)$. That $c$ is a relative cycle, and $\delta [c]=[y]$. Thus $\ker i_{\ast }=\mathrm{im}\delta$.

These arguments also prove the assertions at shifted indices. Finally, every zero-dimensional relative cycle lifts to a chain in $C_0(X)$, and every zero-chain is an absolute cycle; this proves surjectivity onto $H_0(X,Y)$. All maps, representative changes, compositions and exactness assertions have consequently been justified from the chain definitions.

Suppose a nontrivial finite group acts freely on ${\mathbb{R}}^n$ by homeomorphisms. Choose a nonidentity element of finite order $m$, a prime $p$ dividing $m$, and the power of that element with exponent $m/p$. It has order $p$, so a subgroup $C_p$ also acts freely.

Let $M={\mathbb{R}}^n/C_p$. A finite free action is properly discontinuous: around any point choose mutually disjoint neighborhoods of the finitely many orbit points, then intersect their inverse translates to obtain an evenly covered neighborhood. Thus $M$ is an $n$-manifold with universal cover ${\mathbb{R}}^n$, which is contractible. The group-cohomology argument of Question 2 applies even if no cell structure has been chosen: singular simplices in $M$ lift to the cover, their lifts form free deck-group orbits, and the augmented singular chains of the contractible cover form an exact free resolution. Therefore

We calculate the right-hand side explicitly. Write $R=\mathbb{Z}[C_p]$, let $t$ generate $C_p$, and set $N=1+t+\cdots +t^{p-1}$. The periodic resolution of a finite cyclic group isIts exactness can be seen in coefficients. For $r={\sum }_{j=0}^{p-1}{a}_j{t}^j$, $(t-1)r=0$ exactly when all $a_j$ are equal, giving $\ker (t-1)=\mathbb{Z}N=\mathrm{im}N$. Also $Nr=\epsilon (r)N$, so $\ker N=\ker \epsilon$. If $\epsilon (r)=0$, thenHence $\ker N=\ker \epsilon =(t-1)R$, proving exactness everywhere.

Apply ${\mathrm{Hom}}_R(-,\mathbb{Z})$ with the trivial action. Multiplication by $t-1$ induces zero, and multiplication by $N$ induces multiplication by $p$. The cochain complex is thereforeIts positive even-degree cohomology iswhile its positive odd-degree cohomology is zero.

This contradicts the dimension of the quotient manifold. The noncompact form of Poincaré duality for noncompact manifolds, with the orientation local system when needed, isHere Borel-Moore homology is the homology of locally finite chains. Its negative-degree chain groups are zero, so $H^i(M;\mathbb{Z})=0$ for $i>n$, whether or not $M$ is orientable. Taking any $2j>n$ contradicts the displayed cyclic group calculation. ThusThis proves finite groups cannot act freely on Euclidean space for general homeomorphisms; it does not assume that the action consists of affine maps or isometries.

Use rational coefficients and finite singular chains. For an $N$-dimensional stratified pseudomanifold with filtration $X_j$, the lower middle perversity isA singular simplex $\sigma :{\Delta }^i\to X$ is an allowable singular simplex when, for every $k\ge 2$, ${\sigma }^{-1}(X_{N-k})$ lies in the $(i-k+\bar{m}(k))$-skeleton of ${\Delta }^i$; negative-dimensional skeletons are empty. Let $A_i$ be the span of these simplices in $S_i(X;\mathbb{Q})$. The intersection chains areThe condition on $\partial c$ is imposed after combining equal simplices and cancelling coefficients: the faces of an allowable simplex need not themselves be allowable. If $c\in I{S}_i$, then $\partial c\in A_{i-1}$ and $\partial (\partial c)=0$, so $\partial c\in I{S}_{i-1}$. Thus $I{S}_{\ast }$ is a chain subcomplex of $S_{\ast }$, and $I{H}_i(X)=H_i(I{S}_{\ast })$ defines intersection homology.

Write $v$ for the vertex of $cY$. Its codimension is $2n$, so the vertex condition for an allowable $i$-simplex isIn particular all simplices of degree at most $n$ avoid $v$. For $i<n$, both the degree-$i$ chains and the degree-$(i+1)$ chains that can bound them lie in $cY\setminus \{v\}=Y\times (0,1)$. The two chain complexes therefore have the same cycles and boundaries in degree $i$, proving the low-degree isomorphism.

For $i\ge n$, let $z$ be an allowable $i$-cycle. Cone every simplex of $z$ radially to $v$, using the cone on ${\Delta }^i$ as ${\Delta }^{i+1}$, and write the resulting chain as $cz$. At the vertex, coning an old inverse image in the $(i-n-1)$-skeleton gives a subset of the $(i-n)$-skeleton, and the new cone vertex also lies in this skeleton because $i-n\ge 0$. For every other singular stratum, coning the inverse image increases its permitted skeleton dimension by one, exactly as increasing the simplex dimension does. Thus $cz$ is allowable. The usual cone identity is $\partial (cz)=z-c(\partial z)=z$; since $z$ is allowable, the boundary condition for intersection chains holds as well. Every cycle is consequently a boundary. This proves the intersection homology cone formula here:The assumption $\dim Y=2n-1$ entails $n\ge 1$, so no augmented degree-zero exception arises in this coning argument.

For the suspension of a topological space, the two endpoints are collapsed separately, giving two distinct vertices. Use the following standard computational facts for intersection homology: a product with an interval has the same intersection homology as the original stratified space, via the projection and its stratum-preserving homotopy inverse; and an open cover has a Mayer–Vietoris sequence whose overlap map is the signed pair of inclusion maps. Cover the suspension by its two open cones. Their overlap is $Y\times (-\epsilon ,\epsilon )$. Below degree $n$, the overlap map is, under the interval identifications, $a\mapsto (a,-a)$ in $I{H}_i(Y)\oplus I{H}_i(Y)$, hence is injective with cokernel $I{H}_i(Y)$. In degree $n$ both cone groups vanish and the next overlap map, in degree $n-1$, is still injective. Above degree $n$ both adjacent cone groups vanish. Exactness therefore givesThis differs from ordinary reduced suspension homology: below the cutoff the cycles cannot use either vertex, and in the middle degree coning kills them. If $Y$ is a compact oriented Witt space, its Poincare duality pairs $I{H}_i(Y)$ with $I{H}_{2n-1-i}(Y)$. The displayed groups of the suspension are correspondingly paired in complementary degrees $i$ and $2n-i$, while its middle group is zero. The new vertices have even codimension $2n$ and thus impose no additional Witt space condition; the suspension is again Witt. Its middle intersection form has zero rank and, when its dimension is divisible by four, zero signature.

Use the short exact sequence of coefficient groupsThe groups of singular chains are free, so applying the cochain construction gives a short exact sequence of cochain complexes. Its long exact sequence in cohomology has the required terms and reduction map $r_{\ast }$. Define the Bockstein homomorphism to be its connecting homomorphism.

Explicitly, represent $x$ by a mod-$p$ cocycle $a$ and lift it to an integral singular cochain $A$. Since $a$ is a cocycle, $\delta A$ is divisible by $p$. ThenThis is exactly the lift-and-coboundary construction for the coefficient sequence above: the mod-$p^2$ coboundary of $A$ lies in the image of $j$. The quotient cochain is a cocycle, since ${\delta }^{2}A=0$ and integral cochains have no $p$-torsion. Replacing $A$ by another lift adds $pC$ and changes the quotient by $\delta C$ modulo $p$; replacing the representative by a coboundary likewise leaves its cohomology class unchanged. Thus the definition is independent of choices.

A degree-zero cocycle is a function on points constant on each path component. Choose an integer representative for its value on each component. This produces an integral zero-cocycle lift, with zero coboundary. Hence the Bockstein vanishes on $H^0(X;{\mathbb{F}}_p)$ for every space.

For a nonzero example in every positive degree $i$, take the Moore space $X=S^i{\cup }_p{e}^{i+1}$, attaching the cell by a map of mapping degree $p$. Its positive-degree integral cellular cochain complex has the differential $\mathbb{Z}\stackrel{p}{\to }\mathbb{Z}$ from degree $i$ to $i+1$; when $i=1$, the preceding differential from degree zero is zero. Modulo $p$ the displayed differential vanishes, and $H^i(X;{\mathbb{F}}_p)=H^{i+1}(X;{\mathbb{F}}_p)={\mathbb{F}}_p$. Lift the degree-$i$ generator to the cellular cochain with value one. Its coboundary has value $p$, so dividing by $p$ gives the generator in degree $i+1$. Naturality of cellular and singular cohomology with coefficient sequences identifies this with the Bockstein on a cyclic Moore space. Therefore $\beta$ is nonzero, indeed an isomorphism, in the requested degree for every prime $p$.

Finally take integral lifts $A,B$ of cocycles representing $x\in H^i(X;{\mathbb{F}}_p)$ and $y\in H^j(X;{\mathbb{F}}_p)$. Write $\delta A=p{A}_1$ and $\delta B=p{B}_1$. The cochain cup product obeys the graded Leibniz identity, soDivide by $p$, reduce modulo $p$ and pass to cohomology. This proves the Bockstein derivation rule

For a singular simplex $\sigma :{\Delta }^i\to Y$, there are exactly $n$ lifts to $X$: choose any of the $n$ points above one vertex, then use the lifting property, since the simplex is simply connected. Defineand extend $R$-linearly to singular chains. The restrictions of these lifts to any face are exactly the $n$ lifts of that face, each appearing once. Indeed a lift is determined uniquely by its value at any point of that face, and every possible value in the corresponding fiber extends to a lift of the whole simplex. Therefore $\partial T_i=T_{i-1}\partial$, so $T$ is a chain map and induces the homological transfer for a finite covering, denoted $f^{\ast }$ in the question.

On chains, every lifted simplex projects back to the original simplex. Thus $f_{\#}{T}_i=n\mathrm{id}$, and on homology this givesOver $\mathbb{Q}$, the map $n^{-1}{f}^{\ast }$ is a right inverse of $f_{\ast }$, proving surjectivity in every degree. Consequently each rational Betti number of the covering space is at least that of its base. This transfer goes in the opposite direction to the usual homological pushforward and is not the ordinary cohomological pullback.

For equality, use the connected double covering space $S^1\to S^1$, $z\mapsto z^2$. Both spaces have $b_0=b_1=1$ and all other Betti numbers zero. For strict increase, take a connected double covering space of the genus-two closed orientable surface, whose existence was proved in 1(b). The genus of a finite cover of a closed orientable surface gives covering genus $1+2(2-1)=3$. Its first Betti number is six, whereas that of the genus-two base is four; their zeroth and second Betti numbers are both one. Thus a connected double cover can preserve every Betti number or strictly increase some of them.

Unless a coefficient group is displayed, use integral singular cohomology. The standard CW complex structure on infinite-dimensional real projective space has one cell in each nonnegative dimension. Its cellular chain complex has boundary $d_j=2$ for positive even $j$ and $d_j=0$ for odd $j$. The cellular cohomology differential ${\delta }^j$ is therefore zero for even $j$ and multiplication by two for odd $j$. ConsequentlyMore generally, for an abelian group $A$, the positive odd groups are $A[2]=\{a:2a=0\}$ and the positive even groups are $A/2A$. In particular $H^j({\mathbb{RP}}^{\infty };{\mathbb{Z}}_2)={\mathbb{Z}}_2$ in every nonnegative degree.

For the required Bockstein homomorphism, use the short exact sequenceHere $m\ge 2$; for $m=1$ all three coefficient groups are zero. Since singular chains are free abelian groups, applying cochains gives a short exact sequence of cochain complexes. Its connecting homomorphism defines $\beta$ and its long exact sequence is precisely the required one, with the other maps induced by $\iota$ and $q$.

Explicitly, represent a class by a cocycle $u\in C^n(X;{\mathbb{Z}}_m)$ and choose a lift $\tilde{u}\in C^n(X;{\mathbb{Z}}_{m^2})$. Since $q(\delta \tilde{u})=0$, there is a unique cochain $v$ with $\iota (v)=\delta \tilde{u}$. Injectivity of $\iota$ and ${\delta }^2=0$ show $\delta v=0$. DefineChanging the lift by $\iota (w)$ changes $v$ by the coboundary $\delta w$. Changing the representative $u$ by a coboundary can be lifted by a coboundary as well and leaves the resulting class unchanged. Thus this is a well-defined group homomorphism, and the standard cochain lifting argument gives exactness.

Compute the Bockstein homomorphism on infinite-dimensional real projective space using its cellular cohomology complex. A generator with coefficients ${\mathbb{Z}}_2$ is represented by $1$ in degree $n$, lifted to $1\in {\mathbb{Z}}_4$. Its coboundary is $0$ for even $n$ and $2$ for odd $n$. Dividing via $\iota (1)=2$ givesThus the odd-degree maps are isomorphisms. The comparison between cellular cohomology and singular cohomology is natural with respect to coefficient maps, so this computes the same connecting homomorphism constructed above.

For the sphere homology calculation use singular homology, homotopy invariance of homology, and the reduced Mayer–Vietoris sequence, not cellular homology. For a point, its chain group for singular homology is $\mathbb{Z}$ in every nonnegative degree, and its boundary operator is multiplication by ${\sum }_{i=0}^q(-1{)}^i$, hence alternately zero and one in positive degrees. Its homology is therefore $\mathbb{Z}$ in degree zero and zero above. Singular chains of a disjoint union split into the summand chain complexes. Start with $S^0$, the disjoint union of two points: $H_0(S^0;\mathbb{Z})={\mathbb{Z}}^2$, all positive groups vanish, and ${\tilde{H}}_0(S^0;\mathbb{Z})=\mathbb{Z}$. For $n\ge 1$, remove the north and south poles to obtain an open cover $U,V$ of $S^n$. Each member is contractible, while $U\cap V$ has a deformation retraction to $S^{n-1}$, including the disconnected intersection when $n=1$. The reduced exact sequence therefore suppliesAlso $S^n$ is connected, so its reduced degree-zero group vanishes. Induction givesThe separate $S^0$ calculation avoids conflating its two degree-zero generators.

Choose an orientation of $S^n$, with $n\ge 1$, and let $[S^n]$ be its fundamental class. The mapping degree is the integer determined by $f_{\ast }[S^n]=\mathrm{deg}(f)[S^n]$. To define the local degree at $x_0$, set $y_0=f(x_0)$ and require $x_0$ to be isolated in $f^{-1}(y_0)$. Choose an open neighborhood $B$ containing no other point of that fiber. Then $f$ defines a map of pairsThe degree-$n$ relative homology groups on both sides are $\mathbb{Z}$, by a manifold chart and excision. The local orientation of a manifold chooses their generators. The induced map multiplies these generators by an integer, denoted ${\mathrm{deg}}_{x_0}f$. Excision and naturality show independence of the choice of $B$. Differentiability is unnecessary; if $f$ is smooth and $D{f}_{x_0}$ is invertible, the inverse function theorem gives ${\mathrm{deg}}_{x_0}f=\mathrm{sign}\det D{f}_{x_0}$ relative to the orientations. Without isolation in the fiber this pointwise definition need not apply.

For the degree as a sum of local degrees, assume $y\in S^n$ has finite fiber $F=\{x_1,\dots ,x_s\}$. Disjoint coordinate neighborhoods and excision identifyThe absolute fundamental class maps to the tuple of local orientation generators. Under the relative map induced by $f$, the $j$th generator goes to ${\mathrm{deg}}_{x_j}f$ times the generator at $y$, and the resulting map from the direct sum adds these contributions. On the other hand, naturality says that first applying the absolute map $f_{\ast }$ and then passing to the local group at $y$ gives $\mathrm{deg}(f)$ times that generator. ThusIf the fiber is empty the sum is zero: the map factors through the contractible punctured sphere and has degree zero. If every point of a fiber is isolated, compactness makes that closed fiber finite, so the theorem applies. In particular, a regular value of a smooth map gives the signed count of its inverse images. For $n=0$, define a mapping degree on the reduced generator $[p_{+}]-[p_{-}]$: the identity has mapping degree one, the swap of the two points minus one, and either constant map zero. With local orientation signs positive at $p_{+}$ and negative at $p_{-}$, the same sum formula holds. The connected-sphere convention above is the standard positive-dimensional one.

Every singular simplex has compact image, and a singular chain is a finite sum of simplices. The image of any chain is therefore compact and lies in some $X_a$. Directedness puts any finite collection of chains into one common $X_b$, soBecause filtered colimits of abelian groups are exact, kernels and images commute with this colimit. Taking homology gives the homology of a directed union:

For an open $U\subseteq {\mathbb{R}}^N$, use the directed family of finite unions of closed rational cubes contained in $U$. Every compact subset of $U$ lies in one such finite polyhedron, and each polyhedron has finitely generated cellular homology. There are only countably many of them, so their direct limit is countable. Thus every $H_i(U;\mathbb{Z})$ is countable.

Cohomology behaves differently because $\mathrm{Hom}$ turns a direct sum into a direct product. The connected open sethas one independent loop around each puncture, so $H_1(U;\mathbb{Z})\cong {⨁}_{n\ge 1}\mathbb{Z}$. Since $H_0(U;\mathbb{Z})=\mathbb{Z}$, the universal coefficient theorem for cohomology giveswhich is uncountable. This is the first cohomology of the countably punctured plane.

A relative homology class is represented by a singular chain whose boundary lies in $A$. Two representatives give the same class when their difference is a boundary operator image plus a chain in $A$. Thus a path with endpoints in $A$ can represent a class even when its endpoints are different.

A relative cycle is represented by a singular chain $b$ whose boundary lies in $A$. Define $\delta [\bar{b}]=[\partial b]$. If a representative changes by $\partial c+a$ with $a\in C_n(A)$, its boundary changes by $\partial a$, an $A$-boundary. Thus the map is well-defined and additive. Its image is exactly the kernel of $H_{n-1}(A)\to H_{n-1}(B)$: a cycle in $A$ dies in $B$ precisely when it is the boundary of a relative cycle lifted to $B$. This proves the corresponding exactness in the long exact sequence in relative homology directly.

Triangulate ${\Delta }^q\times I$ by the simplices with vertices $(v_0,0),\dots ,(v_i,0),(v_i,1),\dots ,(v_q,1)$ and coefficients $(-1{)}^i$. Composing with a homotopy between f and g gives a degree-one map P on singular chains. Shared faces cancel; the remaining faces give the displayed chain homotopy identity. Hence homotopic maps induce the same map on singular homology.