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A cellular map $f:X\to Y$ between CW complexes satisfies $f(X^q)\subseteq Y^q$ for every skeleton. It induces a chain map $f_{\#}:C_{\ast }^{\mathrm{cell}}(X)\to C_{\ast }^{\mathrm{cell}}(Y)$.

A graded Hom functor construction consists of maps shifting degree by $j$. One homological convention is $d_{M}f=f{d}_C+(-1{)}^{j-1}{d}_{C^{\prime }}f$. Its degree-zero cycles are chain maps, and its degree-zero boundaries are null-homotopic chain maps, so its zeroth homology is the module of chain homotopy classes. This convention differs by degree-dependent signs from the alternative convention $d^{\prime }f-(-1{)}^{j}fd$.

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 chain map $f:C\to D$, use $M_i=C_{i-1}\oplus D_i$ and $d(x,y)=(d_{C}x,(-1{)}^{i}f(x)+d_{D}y)$. The chain map identity makes $d^2=0$. The short exact sequence of chain complexes $0\to D\to M\to C_{\ast -1}\to 0$ has connecting homomorphism $(-1{)}^i{f}_{\ast }$ in degree $i$. Its long exact sequence in homology shows that the homology of $M$ vanishes precisely when every $f_{\ast }$ is an isomorphism.

The topological mapping cone of $f:X\to Y$ is $Y{\cup }_{f}CX$, where $CX$ is the cone on $X$ and its base is identified with $Y$ through $f$. For nonempty $X$, the cone vertex supplies a distinguished basepoint. This construction is related to, but distinct from, the mapping cone of a chain map.

A chain complex consists of abelian groups $C_q$ and homomorphisms ${\partial }_q:C_q\to C_{q-1}$ with ${\partial }_{q-1}{\partial }_q=0$. Its groups of chain cycles and chain boundaries are $Z_q=\ker {\partial }_q$ and $B_q=\mathrm{im}{\partial }_{q+1}$; the identity ensures $B_q\subseteq Z_q$. Its homology groups areFor a nonnegative complex, take $C_q=0$ for $q<0$.

The singular chain group $C_q(X)$ is the free abelian group on continuous maps $\sigma :{\Delta }^q\to X$, called singular simplices. With ${\delta }_i$ the inclusion of the face opposite vertex $i$, defineDeleting any two vertices in the two possible orders gives the same face with opposite signs, proving ${\partial }^2=0$. This defines the singular chain complex and its singular homology. A continuous map $f$ induces the chain map $f_{\#}(\sigma )=f\circ \sigma$; it commutes with the boundary because composition commutes with restriction to faces.

To prove homotopy invariance of homology, let $F:X\times I\to Y$ have endpoints $f,g$. Triangulate ${\Delta }^q\times I$ by the $q+1$ simplices with ordered verticesLet $p_i:{\Delta }^{q+1}\to {\Delta }^q\times I$ be their affine parameterizations. The singular prism operator isOn taking boundaries, shared internal faces of consecutive prism simplices cancel. The top and bottom faces give $g_{\#}\sigma -f_{\#}\sigma$, and the remaining faces are $-P_{q-1}(\partial \sigma )$ with their induced signs. ThusFor a chain cycle $z$, $g_{\#}z-f_{\#}z=\partial Pz$ is a chain boundary, so $f_{\ast }=g_{\ast }$ on homology. If $f$ is a homotopy equivalence, its homotopy inverse therefore induces an inverse to $f_{\ast }$, proving that singular homology depends only on homotopy type.

Use the standard CW complex for an orientable genus-three surface: one vertex, six edges $a_1,b_1,a_2,b_2,a_3,b_3$, and a two-cell whose attaching word is ${\prod }_{j=1}^3[a_j,b_j]$. The cellular homology theorem identifies this cellular calculation with singular homology. The boundary of every edge is zero, and each edge's total exponent in the attaching word is zero, so the two-cell boundary is zero. HenceChoose $C_1$ to be the nonseparating curve represented by $a_1$, and choose $C_2$ as a small circle bounding a disc in a surface patch. Then $[C_1]$ is a nonzero primitive homology class, whereas the inclusion of $C_2$ induces zero on $H_1$. If a homotopy equivalence carried $C_1$ onto $C_2$, its restriction would factor through $C_2$ and kill $[C_1]$, contradicting injectivity on first homology. This is the homology obstruction to carrying surface curves onto one another.

There is nevertheless such a continuous map. Map the edge $a_1$ homeomorphically onto $S^1$ and all other edges to its base point. The two-cell attaching word then has winding number zero in $S^1$, hence is null-homotopic and the map extends to $r:{\Sigma }_3\to S^1$. Compose $r$ with a homeomorphism $S^1\to C_2$ and the inclusion of $C_2$ in the surface. This gives $g(C_1)=C_2$. Thus an arbitrary map can carry the first curve onto the second, but a homotopy equivalence cannot.

For an unambiguous drawing, realize ${\Sigma }_3$ as the double of a manifold formed from a disc with three holes. Its two copies are glued along all four boundary circles. The red curve below follows an arc from the outer boundary to the first hole on the front copy and returns along the matching arc on the back copy; it is a nonseparating curve. The blue circle is entirely in a disc patch on the front copy and is contractible.

Let $\pi :E\to B$ be an oriented real vector bundle of rank $r$ over a paracompact base. A Thom class is a class $U\in H^r(E,E\setminus B;\mathbb{Z})$ restricting on every oriented fiber pair $({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ to its positive generator. The Thom isomorphism theorem states that this class exists uniquely with those fiber restrictions and thatHere $B$ is identified with the zero section. Equivalently use the disk/sphere bundle pair. The compact-support version is $H_c^q(B;\mathbb{Z})\cong H_c^{q+r}(E;\mathbb{Z})$, with fiberwise compact support in the Thom construction. If the bundle is not oriented, the integral assertion requires the corresponding orientation local system.

For a closed co-oriented embedded submanifold $Y$ of codimension $r$, co-orientation means an orientation of its normal bundle. A tubular neighborhood identifies a neighborhood of $Y$ with that bundle. Its Thom class, transferred by excision, is a class in $H^r(M,M\setminus Y;\mathbb{Z})$. For compact $Y$ this maps into the direct limit defining compactly supported cohomology, giving $\boxed{{\epsilon }_Y\in H_c^r(M;\mathbb{Z})}$. The construction is independent of the tubular neighborhood and uses the chosen co-orientation. In an oriented ambient manifold it is the Poincare dual of the oriented submanifold. Here “closed submanifold” has the usual compact-without-boundary meaning; if it meant merely a closed subset and $Y$ were noncompact, the supported relative class would still exist but need not have compact support.

We next prove the fixed-point implication without imposing an unprinted orientability assumption on $M$. SetUse a finite triangulation of the closed smooth manifold. The elementary Hopf trace identity says that the alternating trace of a chain endomorphism equals its alternating trace on homology. To verify it, split each finite-dimensional chain space as boundaries, a complement representing homology, and a complement mapping isomorphically onto the preceding boundaries. The traces on these two boundary summands cancel in adjacent degrees, leaving exactly the homology traces. The cohomology traces equal the homology traces because the induced maps are dual over $\mathbb{Q}$.

If $f$ has no fixed points, compactness gives $\delta ={\min }_{x}d(x,f(x))>0$. Choose a triangulation $K$ of mesh less than $\delta /4$ and a sufficiently fine subdivision $K^{\prime }$ admitting a simplicial approximation $F:K^{\prime }\to K$ uniformly within $\delta /4$ of $f$. The subdivision chain map $s:C_{\ast }(K)\to C_{\ast }(K^{\prime })$ sends each simplex to simplices inside it. For a vertex $v$ of such a small simplex, $d(v,F(v))>3\delta /4$, so $F(v)$ cannot lie in the original coarse simplex. Consequently $F_{\#}s$ has zero diagonal coefficient on every coarse simplex. Its trace vanishes in every degree, while it induces $f_{\ast }$ on homology. The trace identity yields $L(f)=0$. Taking the contrapositive provesThis is the Lefschetz fixed-point theorem, established here rather than invoked. For oriented $M$ the same identity is the graph-diagonal formula for the Lefschetz number: $L(f)=⟨(1,f{)}^{\ast }{\epsilon }_{\Delta },[M]⟩$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same cohomology maps, so their Lefschetz numbers agree.

For a based space $(X,x_0)$ and $k\ge 1$, the homotopy group ${\pi }_k(X,x_0)$ is the set of based homotopy classes of maps $(I^k,\partial I^k)\to (X,x_0)$, equivalently based maps $S^k\to X$. Concatenating along one coordinate defines its group operation; reversing that coordinate gives the inverse. The group is abelian for $k\ge 2$. The symbol ${\pi }_0(X)$ records path components and is generally only a pointed set. An Eilenberg–MacLane space $K(G,n)$ is a connected CW space with ${\pi }_n=G$ and all other positive homotopy groups zero; for $n>1$ the group $G$ must be abelian.

Projection onto coordinates givesa map to a product is exactly a pair of maps, and a based homotopy to the product is exactly a pair of based homotopies. These identifications respect concatenation. Thus $K(G,n)\times K(H,n)$ has exactly the homotopy groups required of $K(G\times H,n)$. Taking CW models and using the allowed uniqueness of Eilenberg–MacLane spaces gives

Let $p$ be prime. Take $S^{\infty }$ in complex coordinates, with $C_p$ acting by scalar multiplication by a primitive $p$th root of unity. This is a free action. Every map from a finite-dimensional sphere lands in a finite stage and becomes null-homotopic in a sufficiently higher sphere, so $S^{\infty }$ is weakly contractible; the Whitehead theorem for CW complexes makes it contractible. Therefore the infinite lens space $S^{\infty }/C_p$ has contractible universal cover, fundamental group $C_p$, and no higher homotopy: it is a model of $K(C_p,1)$.

Here is the group cohomology computation, including the ring structure. For the cyclic group generated by $g$, the standard free resolution alternates the maps $g-1$ and $N=1+g+\cdots +g^{p-1}$. Applying ${\mathrm{Hom}}_{\mathbb{Z}[C_p]}(-,{\mathbb{F}}_p)$ makes both maps zero, so each cohomology degree is one-dimensional. The period-two identity chain map represents a class $b$ in degree two, and cup multiplication by this periodicity class is an isomorphism from each degree to the degree two higher. For odd $p$, a nonzero degree-one class $a$ has $a^2=0$ by graded commutativity, while $b^j$ and $a{b}^j$ are nonzero by periodicity. For $p=2$ the model is homotopy equivalent to ${\mathbb{RP}}^{\infty }$, whose ring is the infinite version of that computed in question 2. Consequently the mod-p cohomology ring of a cyclic group isThe standard results used are the contractible-cover description of a $K(G,1)$, its identification of singular cohomology with group cohomology, and the cyclic periodic resolution with cup-periodicity just stated.

The product equivalence and the assumed Künneth theorem over the field giveThe tensor product is graded: for odd $p$, $a_1{a}_2=-a_2{a}_1$ and each $a_i^2=0$. The dimensions in degree $d$ are $d+1$: for $p=2$ count $t_1^i{t}_2^{d-i}$; for odd $p$ the generating series is $(1+t{)}^2/(1-t^2{)}^2=(1-t{)}^{-2}$.

If an $n$-dimensional CW complex has exactly one cell in dimension $n+1$, then $C^{n+1}(X;{\mathbb{F}}_p)={\mathbb{F}}_p$. Cellular cohomology is the quotient of the kernel of its outgoing coboundary by the image of the incoming one, a quotient of a subspace of this one-dimensional vector space. ThusThis one-cell bound on next-degree cohomology is valid even with infinitely many higher cells; the printed finite-per-dimension hypothesis is more than it needs.

Now a free finite-group action makes $S^n\to M=S^n/G$ a covering. For $n>1$ the sphere is simply connected, so ${\pi }_1(M)=G$, ${\pi }_i(M)=0$ for $1<i<n$, and ${\pi }_n(M)=\mathbb{Z}$. The covering also identifies every higher homotopy group with that of $S^n$. Use the homotopy-group effect of attaching higher cells: attaching $(i+1)$-cells, $i\ge 2$, preserves homotopy groups below $i$ and quotients ${\pi }_i$ by the subgroup generated by the attaching classes and their fundamental-group translates. Attach a single $(n+1)$-cell along a generator of ${\pi }_n(M)$; its translates do not require additional cells since the original generator already generates the underlying group. This kills ${\pi }_n$ without changing ${\pi }_1$ or the lower groups.

Successively in each higher degree attach cells representing generators of the remaining homotopy group. In the union every higher homotopy group vanishes, since a map and its relevant homotopy have compact domain and are handled at a finite stage. The result is a $K(G,1)$ with exactly one $(n+1)$-cell. No finiteness assumption on the number of later cells is needed for the preceding cochain bound. If desired, one can retain finitely many per degree using the Serre finiteness theorem for homotopy groups: a simply connected CW complex of finite type has finitely generated homotopy groups. At each finite stage the universal cover is simply connected and finite type, since $G$ is finite; choosing finitely many generators then preserves finite type inductively. The first stage needs just the stated ${\pi }_n(S^n)=\mathbb{Z}$.

If $G=C_p\times C_p$, the resulting $K(G,1)$ would have $\dim H^{n+1}\le 1$ by its single cell, but its already computed cohomology has dimension $n+2>1$ in that degree. This contradiction provesThis is the free sphere actions exclude elementary abelian subgroups of rank two obstruction.

Let $Z={\mathbb{CP}}^5/{\mathbb{CP}}^2$. Its quotient cell structure has one cell in each of dimensions $0,6,8,10$. In particular $Z$ is five-connected. The relative cohomology sequence, or its cellular cochain complex, givesThe subscripts denote degrees. Under the quotient map back to ${\mathbb{CP}}^5$, the three positive classes pull back to $x^3,x^4,x^5$. All positive products vanish, since their degree is at least twelve. Thus the cohomology ring of a collapsed projective subspace in this case is the square-zero algebra $R$ on these three classes.

Now compute its Sullivan minimal model $(\Lambda V,d)$, using cohomological grading and the convention that even generators generate a polynomial ring and odd generators an exterior algebra. Five-connectivity gives $V^j=0$ for $j\le 5$. Since a minimal differential is decomposable, it cannot have a nonzero value below degree twelve. Matching cohomology through degree ten therefore forceswith no other generators in degrees at most ten. The first products must then be killed by generatorsThese follow by killing respectively the unwanted degree-twelve, degree-fourteen, and two degree-sixteen classes. Subsequent generators kill $bc,c^2$ and the resulting syzygies. For example $av-bu$ is closed of degree nineteen, so it requires a further generator of degree eighteen. Thus the full minimal model is not just the algebra on the first seven generators.

There is nevertheless a simple map from the full model to $R$: send $a,b,c$ to the three displayed classes and every other generator to zero. It is a chain map, since every generator differential is a sum of products of positive-degree elements, and all positive products in $R$ vanish. It induces an isomorphism in cohomology: the entire cohomology of $\Lambda V$ is concentrated in degrees $0,6,8,10$, and its indicated basis maps to the indicated basis of $R$. This proves formality of a topological space and identifies the full minimal model as a minimal resolution of $R$, rather than only matching the initial part of its cohomology.

An explicit description of that full resolution is useful. Take the free graded Lie algebraIts Chevalley–Eilenberg cochain algebra $C^{\ast }(L)$ has one generator of degree $|\ell |+1$ for every homogeneous Lie basis word $\ell$, and its quadratic differential is dual to the bracket, with the usual graded signs. Rescaling the bracket-word duals gives exactly the generators and differentials listed above. Positive Lie degrees make this algebra Sullivan minimal. Its cohomology is $R$: the universal enveloping algebra is $T(p,q,r)$, and the augmentation module has the exact free resolutionExactness is immediate because every nonempty tensor word has a unique last letter. Applying ${\mathrm{Hom}}_{T(p,q,r)}(-,\mathbb{Q})$ gives just the unit and the three dual classes in total degrees $6,8,10$; products of positive classes land in an absent Ext degree two and are zero. Chevalley–Eilenberg cochains of a graded Lie algebra compute this Ext, so this also verifies the full model's cohomology.

The wedge sum $W=S^6\vee S^8\vee S^{10}$ has the same connectivity, cohomology ring, and initial closed minimal generators. The same full-model-to-$R$ argument applies to $W$, so its minimal model is also $C^{\ast }(L)$, by uniqueness of Sullivan minimal models of quasi-isomorphic simply connected rational cochain algebras. The Sullivan minimal-model classification identifies isomorphic minimal models with the same rational homotopy type. ConsequentlyThe explicit infinite minimal resolution above is shared by both spaces.

It is enough to assume that $h_k:{\pi }_k(X)\otimes \mathbb{Q}\to H_k(X;\mathbb{Q})$ is surjective for every $k\ge 2$; an integrally surjective Hurewicz homomorphism certainly has this property. First give the minimal-model argument suggested by the hint. Let $A=(\Lambda V,d)$ be a simply connected Sullivan minimal model, with $V$ in degrees at least two. Put $A^{+}={⨁}_{k>0}{A}^k$. Its indecomposable quotient of an augmented algebra isMinimality makes the induced differential on $V$ zero, so projection is a chain map $q:A^{+}\to (V,0)$. The Sullivan model identification says that the induced mapis dual to rational Hurewicz homomorphism. Under the usual finite-type duality convention, the assumed surjectivity makes $q_{\ast }$ injective.

A product of two positive cohomology classes has a decomposable cocycle representative, so its image under $q_{\ast }$ is zero. Injectivity therefore givesThis deduction uses the hypothesis; zero cup products by themselves would not yet prove formality of a topological space.

Choose a homogeneous cohomology basis $h_{\lambda }$ and cocycle representatives $z_{\lambda }\in A$. Their indecomposable parts $q(z_{\lambda })$ are linearly independent. Extend them to a graded basis of $V$ and replace the corresponding generators by the actual cocycles $z_{\lambda }$. This is a valid change of free algebra generators: each $z_{\lambda }$ differs from its indecomposable part by products of generators of strictly smaller degree, so degree induction constructs the inverse substitution. Call these closed generators $C$, and the remaining generators $N$.

Define a map of graded algebrasIts target has zero differential. Every differential in $A$ is decomposable, and all products of positive target classes vanish, so $\varphi$ is a chain map. If $z$ is any positive cocycle, its indecomposable part is in the span of the $q(z_{\lambda })$, because these form the image of $q_{\ast }$. Write $q(z)={\sum }_{\lambda }{c}_{\lambda }q(z_{\lambda })$. Then $\varphi (z)={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$, while injectivity of $q_{\ast }$ gives $[z]={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$. Thus $\varphi$ induces the identity on cohomology and is a quasi-isomorphism. This proves that $X$ is a formal space.

The rational wedge sum assertion has a direct topological proof, which also avoids finite-dimensional duality restrictions in the preceding model argument. Choose a homogeneous basis $b_{\lambda }$ of ${\tilde{H}}_{\ast }(X;\mathbb{Q})$. By surjectivity, each basis class is the image under the Hurewicz homomorphism of an element of ${\pi }_{n_{\lambda }}(X)\otimes \mathbb{Q}$. In $X_{\mathbb{Q}}$ this element is represented by an actual sphere map $S^{n_{\lambda }}\to X_{\mathbb{Q}}$. Simple connectivity and $H_1=0$ ensure $n_{\lambda }\ge 2$. The wedge sum of these maps givesand induces an isomorphism of rational homology by the choice of basis. The rational Whitehead theorem now makes it a rational homotopy equivalence, givingA wedge sum of spheres is a formal space. For each individual sphere, its augmented minimal model maps by a quasi-isomorphism to its sphere cochain algebra, and also to its cohomology algebra, by killing the extra odd generator in the even-sphere case. Take based simplicial models of the spheres. Their wedge sum is a simplicial pushout over the common basepoint, so the algebra of rational polynomial differential forms is the corresponding fibre product of augmented sphere algebras. Taking fibre products of these model maps gives a quasi-isomorphism zigzag to the square-zero cohomology algebra. This also works for infinitely many sphere summands: the positive parts are products of the individual complexes, and products of vector-space exact sequences over a field are exact. In the finite case its minimal resolution is the minimal model of a wedge of simply connected spheres. Thus the topological proof establishes both conclusions without imposing finite rank on the sphere basis. For the zero reduced-homology case the wedge sum is a point. Both the minimal-model construction and the explicit sphere map explain why the surjectivity hypothesis is decisive.

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.

Here the usual homological algebra convention is that the right-exact functor $F$ is additive. Module categories have enough projective modules, because any module is a quotient of a free module. Choose a projective resolutionApply $F$ to the unaugmented chain complex and define the left derived functors byAdditivity ensures that $F$ preserves zero maps, sums of maps and homotopy identities, so the transformed differentials still compose to zero.

For a module homomorphism $u:M\to N$, lift it to a chain map between chosen projective resolutions. In degree zero, lift through $Q_0\twoheadrightarrow N$ using projectivity of $P_0$. In each higher degree, the already constructed map lands in the next kernel; exactness of $Q_{\bullet }$ and projectivity of $P_i$ supply the next lift. Any two lifts are chain homotopic: their degree-zero difference lands in $\ker (Q_0\to N)=\mathrm{im}{d}_1^Q$, so it lifts to $h_0:P_0\to Q_1$. Inductively lift the remaining difference after subtracting the preceding homotopy term. This givesApplying $F$ preserves this identity, so the lifts induce the same homology map. Identity lifts and composable lifts show that $L_{i}F$ is a functor. Comparison lifts between two resolutions over the identity of $M$ are inverse up to chain homotopy; hence the construction is independent of the chosen resolution up to its canonical natural isomorphism.

Right exactness identifies $L_{0}F(M)=\mathrm{coker}(F(P_1)\to F(P_0))$ naturally with $F(M)$. For a projective module $P$, the length-zero resolution gives $L_{i}F(P)=0$ for $i>0$. For a short exact sequence $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$, the horseshoe lemma builds a degreewise split short exact sequence of projective resolutions. Concretely, lift the projective generators covering $M^{\prime \prime }$ into $M$ and combine them with generators covering $M^{\prime }$; repeat on the kernels. The middle projective term in each degree is the direct sum of the two outer ones. An additive functor preserves these split sequences, and lifting cycles, taking their differentials and returning to the subcomplex produces the connecting maps inThis is the natural long exact derived-functor sequence. In particular, if $0\to K\to P\to M\to 0$ has $P$ projective, then $L_{i}F(M)\cong L_{i-1}F(K)$ for $i\ge 2$, and $L_{1}F(M)=\ker (F(K)\to F(P))$. These are dimension-shifting formulas. Positive degrees can be killed by the surjection $P\to M$. If $F$ is exact, it preserves every resolution's exactness and all its positive left derived functors vanish. Conversely, if $L_{1}F$ vanishes on all modules, the displayed long exact sequence makes $F$ exact. A natural transformation between additive right-exact functors induces transformations on the left derived functors by applying it termwise.

For extension of scalars $F(M)=B{\otimes }_{A}M$, these derived functors are the Tor functors ${\mathrm{Tor}}_i^A(B,M)$. When $B=A/(a)$ and $a$ is a non-zero-divisor, there is the length-one free resolutionAlthough the construction above resolves $M$, one may instead use this resolution of $B$. To justify that calculation, tensor a free resolution of $M$ with the two-term resolution of $B$ to form a double complex. Taking homology first along the resolution of $B$ leaves $B{\otimes }_A{P}_{\bullet }$; taking it first along the resolution of $M$ leaves $M\stackrel{a}{\to }M$. The free terms preserve exactness under tensoring. The double complex is first-quadrant with finitely many terms on each total diagonal, so both filtrations compute the same total homology. This proves the required balancing directly.

The kernel and cokernel of multiplication by $a$ now give Tor for a principal non-zero-divisor quotient:These are naturally $B$-modules. For $M=A$, the degree-one group is zero; for $M=B$, multiplication by $a$ is zero and both degrees zero and one are $B$. Thus, when $B\ne 0$, degree one explicitly detects the failure of this base-change functor to be exact.

The chain map identity $d_{D}f=f{d}_C$ gives, for $(x,y)\in M_i$,Thus $M$ is a chain complex; it is a sign convention for the algebraic mapping cone.

Put $E_i=C_{i-1}$ with differential $d_E=d_C$. Inclusion into the second summand and projection onto the first give a short exact sequence of chain complexesIts connecting homomorphism is particularly explicit. If $x\in C_{i-1}$ is a chain cycle, lift it to $(x,0)\in M_i$ and take its boundary. The second component is $(-1{)}^{i}f(x)$, so the long exact sequence in homology containsIf every $H_i(M)$ vanishes, the exact sequence makes every $f_{\ast }$ both injective and surjective. Conversely, if every $f_{\ast }$ is an isomorphism, the map $H_i(D)\to H_i(M)$ has zero image, while $H_i(M)\to H_{i-1}(C)$ has zero image and zero kernel; hence $H_i(M)=0$.

One can also see the converse directly on chain cycles. If $(x,y)\in M_i$ is a chain cycle, then $d_{C}x=0$ and $d_{D}y=-(-1{)}^{i}f(x)$. Injectivity of the induced map on homology $f_{\ast }$ gives $x=d_{C}z$. Subtracting the chain boundary $d_f(z,0)$ leaves $(0,t)$ with $d_{D}t=0$. Surjectivity of the induced map on homology $f_{\ast }$ lets us choose a chain cycle $w\in C_i$ and $v\in D_{i+1}$ such thatThen $(0,t)=d_f(w,v)$. Thus every chain cycle in $M$ is a chain boundary, proving the mapping cone acyclicity criterion:

For each prime number $p$, reduction of the mapping cone is the mapping cone of the reduced chain map. Part (a), whose proof works over any coefficient ring, therefore givesEvery $M_i$ is a finitely generated free abelian group. Multiplication by $p$ consequently gives a short exact sequence of chain complexesThe neighboring terms $H_{i+1}(M\otimes {\mathbb{F}}_p)$ and $H_i(M\otimes {\mathbb{F}}_p)$ in its long exact sequence in homology both vanish. Thus multiplication by $p$ on $H_i(M)$ is an isomorphism for every prime number $p$.

The group $H_i(M)$ is a finitely generated abelian group, since it is a quotient of a subgroup of $M_i$. Write it, using the Fundamental theorem of finitely generated abelian groups, aswith $T_i$ finite. If $b_i>0$, multiplication by any prime number is not surjective. If $T_i\ne 0$, choose a prime number dividing its order; multiplication by that prime is not injective. Both contradict the preceding isomorphism. Hence $H_i(M)=0$ for every $i$.

Applying part (a) integrally now proves the stronger natural conclusionThis is prime coefficient detection of quasi-isomorphisms. The finite-generation hypothesis is used precisely when excluding nonzero groups on which every prime acts invertibly.

First check the signs directly. For $(x,y)\in M_i$, applying the proposed differential twice givesbecause $f$ is a chain map. Thus $M$ is a chain complex.

Let $C^{\prime }$ be the shifted chain complex with $C_i^{\prime }=C_{i-1}$ and differential $d_C$; here there is no minus sign in that differential. Inclusion in the second summand and projection onto the first give a short exact sequence of chain complexesTo compute the connecting map in its long exact sequence in homology, lift a cycle $x\in C_{i-1}$ to $(x,0)\in M_i$. Its boundary is $(0,(-1{)}^{i}f(x))$. Hence the connecting map $H_i(C^{\prime })\to H_{i-1}(D)$ is $(-1{)}^i{f}_{\ast }$, and the relevant exact portion isIf every $f_{\ast }$ is an isomorphism, exactness makes $H_i(M)=0$. Conversely, if every $H_i(M)$ is zero, the adjacent exact portions make every $f_{\ast }$ both injective and surjective. ThereforeThis is the acyclicity criterion for a mapping cone, with the degree-dependent signs adjusted to the given convention.

For the assertion about spaces, use the cellular approximation theorem to replace $f$ by a cellular map, and take the induced map of the finite free cellular chain complexes. Tensoring these complexes and their cone with ${\mathbb{F}}_p$ gives the corresponding mod-$p$ complexes and cone. The same exact-sequence argument works over ${\mathbb{F}}_p$, so the assumed homology isomorphisms implyThe universal coefficient theorem for homology givesIn particular $H_i(M)\otimes {\mathbb{F}}_p=0$ for every prime. Each $H_i(M)$ is a finitely generated abelian group. A nonzero free summand would survive tensoring with every ${\mathbb{F}}_p$, while a nonzero finite cyclic summand would survive for a prime dividing its order. Thus $H_i(M)=0$ in every degree, by detection of integral acyclicity modulo primes. Applying the cone criterion once more proves the integral homology map is an isomorphism in every degree. Finite generation is what makes detection by all prime fields sufficient.

The tensor differential. Use homological grading, so each differential lowers degree by one. The tensor product of chain complexes hasfor $x\in C_p$. This Koszul sign rule givesThus the graded tensor product is a chain complex.

The Hom differential. Write $M_j={\prod }_p{\mathrm{Hom}}_R(C_p,C_{p+j}^{\prime })$. For a degree-$j$ element of this graded Hom complex of chain complexes, use the prescribed differentialThe next application uses degree $j-1$, soIn degree zero, $d_{M}f=f{d}_C-d_{C^{\prime }}f$, so its kernel consists exactly of chain maps. A degree-one element $h$ has $d_{M}h=h{d}_C+d_{C^{\prime }}h$, which is exactly the change between two chain maps related by a chain homotopy. ConsequentlyThis is natural: precomposition and postcomposition by chain maps preserve both degree-zero cycles and degree-zero boundaries.

The dual complex and the sign adjustment. Define the reversed dual chain complexHere $d_X\varphi$ is a functional on $C_{1-i}$, so it belongs to $X_{i-1}$, and $d_X^2=0$. The absence of an additional sign in $d_X$ is intentional.

Interpret finite generation of the free chain complexes as finiteness of their total graded free modules. Then only finitely many degrees occur, and the finite-free evaluation isomorphisms assemble into a graded isomorphismwhere $\varphi \in X_i$, $y\in C_j^{\prime }$, and this component is zero outside $C_{-i}$. The dual module construction and evaluation make $\psi$ natural. Finite rank is needed for evaluation to be an isomorphism; finite total support also makes the sums on the tensor side agree with the products on the Hom functor side.

Under $\psi$, the Hom functor differential has the formThe usual tensor product of chain complexes instead has second coefficient $(-1{)}^i$. SetThis is integer-valued for every $j\in \mathbb{Z}$, including negative $j$, and satisfies $\rho (j)-\rho (j-1)=j-1$. Conjugating the usual tensor differential by $A$ leaves the first term unchanged and changes its second coefficient to $(-1{)}^{i+j-1}$. Hence $d_M=\psi AD{A}^{-1}{\psi }^{-1}$, giving the sign conjugation for the tensor-Hom identificationThis is an isomorphism of chain complexes, not merely of their homology. If one instead assumes only degreewise finite rank with unbounded grading, the ordinary tensor need not identify with the product defining $M$; the finite-total convention is essential to this last conclusion.

A simplicial approximation to $f:|K|\to |L|$ is a simplicial map $g:K\to L$ satisfying the open-star conditionfor every vertex $v$. Equivalently, whenever $f(x)$ lies in the interior of a simplex of $L$, $|g|(x)$ lies in that simplex. In particular $f$ and $|g|$ are joined by the simplexwise straight-line homotopy.

The Simplicial approximation theorem says that, for finite $K$, every continuous $f$ has such an approximation after sufficiently many barycentric subdivisions of $K$. For general complexes one uses a subdivision adapted to the open-star cover; a single uniform number of barycentric subdivisions is the compact-domain version.

Given triangulations $X\cong |K|$, $Y\cong |L|$, choose a simplicial approximation $g$ on a suitable subdivision $K^{\prime }$. Its chain map sends an oriented simplex $[v_0,\dots ,v_r]$ to $[g(v_0),\dots ,g(v_r)]$, with repeated-vertex simplices sent to zero. It commutes with the boundary operator and induces $g_{\ast }$. Compose this with the subdivision isomorphism $H_r(K)\to H_r(K^{\prime })$ and the triangulation identifications to define $f_{\ast }:H_r(X)\to H_r(Y)$.

For $z\mapsto z^n$, a positively oriented fundamental loop winds $n$ times around the target circle. With integer coefficients,All higher homology groups vanish. The $H_1$ result can also be read from a subdivision into $n$ copies of a target fundamental cycle.