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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 $d^{\prime }f-(-1{)}^{j}fd$.

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.

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.

For chain maps $f,g:C_{\ast }\to D_{\ast }$, a chain homotopy from $g$ to $f$ is a family of group homomorphisms $s_q:C_q\to D_{q+1}$ satisfyingIf $z$ is a chain cycle, then $(f-g)(z)=d^{D}s(z)$ is a chain boundary. Consequently $[f(z)]=[g(z)]$, so $\boxed{f_{\ast }=g_{\ast }}$ on every homology group.

Order the vertices of the simplex as $v_0,\dots ,v_n$. Its simplicial chain complex hasand zero groups otherwise. An arbitrary ordering represents the sign of the permutation times the increasing ordering. Its boundary operator isIn $d^2$ each face obtained by deleting the vertices in positions $r<t$ occurs twice. Deleting $r$ first gives sign $(-1{)}^r(-1{)}^{t-1}$, whereas deleting $t$ first gives $(-1{)}^t(-1{)}^r$. These signs are opposite, so all terms cancel. The case $d_0{d}_1=0$ follows directly from $d_0=0$. Thus $\boxed{d^2=0}$.

For the homology calculation, add the augmentation $\epsilon :S_0\to \mathbb{Z}$ sending every vertex to $1$. Regard $\mathbb{Z}$ as the group in degree $-1$ of an augmented chain complex. Define the simplicial cone chain contraction byIf the simplex does not contain $v_0$, expansion of its coned boundary gives $ds(\sigma )=\sigma -sd(\sigma )$. If it contains $v_0$, the only face on which $s$ is nonzero is the face omitting $v_0$, and $sd(\sigma )=\sigma$. In degree zero this uses the augmentation, and in degree $-1$ it says $\epsilon s_{-1}=1$. HenceEvery positive-degree chain cycle is therefore a chain boundary. In degree zero, the same identity says $z-\epsilon (z)[v_0]=d{s}_0(z)$, so the augmentation induces an isomorphism $H_0\to \mathbb{Z}$, with inverse $1\mapsto [v_0]$. Thus, including $n=0$,This is a direct chain homotopy calculation and uses no result about cellular homology.

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.