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An $i$-dimensional singular simplex $\sigma :{\Delta }^i\to X$ is $\bar{p}$-allowable if ${\sigma }^{-1}(X_{N-k})$ is contained in the $(i-k+\bar{p}(k))$-skeleton of ${\Delta }^i$ for every $k\ge 2$. A negative skeleton is empty. Individual faces need not be allowable; this is why the definition of intersection chains separately imposes allowability on the total boundary.

For an $n$-sheeted covering space $f:X\to Y$, define the transfer of a singular simplex as the sum of its $n$ lifts. Restriction of these lifts to any face gives exactly the lifts of that face, so transfer is a chain map. Its induced map $\mathrm{tr}:H_i(Y;R)\to H_i(X;R)$ satisfies $f_{\ast }\mathrm{tr}=n\mathrm{id}$ for any commutative coefficient ring $R$. If $n$ is invertible in $R$, then $n^{-1}\mathrm{tr}$ is a right inverse of $f_{\ast }$. In particular a finite covering surjects onto the base's rational homology.

For a singular simplex $\sigma$ of dimension $n$ and a degree-$p$ cochain, use $\sigma ⌢\alpha =\alpha (\sigma |[v_0,\dots ,v_p])\sigma |[v_p,\dots ,v_n]$, setting the result to zero if $p>n$. A continuous map $f$ commutes with both face restrictions, and the pullback cochain evaluates by composing with $f$. Therefore $f_{\ast }(\sigma ⌢f^{\ast }\alpha )=(f\circ \sigma )⌢\alpha$ already on chains. Passing to homology and cohomology proves the displayed cap product identity.

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.

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.

For an $n$-dimensional singular simplex $\sigma$ and a degree-$p$ singular cochain $\alpha$, use the front-face/back-face conventionand set it to zero for $p>n$. Extend linearly over $R$. Passing to homology classes of cycles and cohomology classes of cocycles defines the cap product $H_n(X;R)\times H^p(X;R)\to H_{n-p}(X;R)$.

For a continuous map $f$, the cochain pullback satisfies $(f^{\ast }\alpha )(\tau )=\alpha (f\circ \tau )$, and the induced chain map sends $\sigma$ to $f\circ \sigma$. Since composition commutes with restrictions to faces,The identity holds on chains, so it descends directly to classes and proves the naturality of the cap product: $f_{\ast }(x⌢f^{\ast }\alpha )=f_{\ast }(x)⌢\alpha$. No additional well-definedness verification is required here.

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.