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A continuous map between CW complexes is homotopic to a cellular map. If the map is already cellular on a subcomplex, the homotopy can be taken relative to that subcomplex. In particular, a map from a $d$-dimensional CW complex can be deformed into the $d$-skeleton of the target. This permits finite cellular chain complexes to compute induced homology maps, and turns maps to Complex projective space into maps to an appropriate finite-dimensional projective skeleton.

For a cellular map between nonempty CW complexes, the reduced cellular chain complex of its topological mapping cone has differentialHere $C_{-1}(X)=0$ and the degree-zero group identifies a vertex $y$ with the reduced chain $y-v$, where $v$ is the cone vertex. Thus this is the mapping cone of $f_{\#}$, in the displayed ordering of the summands.

The cellular approximation theorem says that a continuous map between CW complexes is homotopic to a cellular map, meaning that it sends each $q$-skeleton into the target's $q$-skeleton. Give $S^n$, for $n>0$, its CW complex structure with one zero-cell and one $n$-cell. Its $n$-skeleton is the whole sphere, so the theorem applies and gives the requested result. For $n=0$ use the two zero-cells of $S^0$. In particular the conclusion isThere is no local-finiteness assumption on $X$.

For completeness, here is a direct proof of the needed case of cellular approximation, including the point-avoidance step. We use three standard facts, stated explicitly. First, a compact subset of a CW complex lies in a finite subcomplex. Second, a continuous map from a smooth manifold into Euclidean space has arbitrarily close smooth approximations, and smooth cutoffs exist around compact subsets of open sets. Third, the Sard theorem says the critical values of a smooth map have measure zero; when the domain dimension is smaller than the target dimension, its entire image has measure zero.

The first fact has a useful short proof. If a compact set met infinitely many open cells, choose one point in each of countably many distinct cells. Closure-finiteness makes every closed cell meet the chosen set, and every subset of it, in finitely many points. The weak topology of a CW complex then makes all those subsets closed. The chosen set is an infinite closed discrete subset of a compact space, a contradiction. Taking the closures of the finitely many cells met proves the claim.

Now $f(S^n)$ lies in a finite subcomplex $Y$. Take a top-dimensional open cell $e^d$ of $Y$ with $d>n$, and identify its interior with the open unit ball in ${\mathbb{R}}^d$ using its characteristic map. On the open set $U=f^{-1}(e^d)$, write $F:U\to B^d$ for the coordinate expression. Choose $0<r^{\prime }<r^{\prime \prime }<1$ and a smooth cutoff $\chi$ supported in $F^{-1}(B_{r^{\prime \prime }})$ and equal to one near the compact set $F^{-1}({\overline{B}}_{r^{\prime }})$. By smooth approximation of maps into Euclidean space, choose a smooth $h:U\to {\mathbb{R}}^d$ with $|h-F|<\delta$, where $\delta <\min (r^{\prime }/2,1-r^{\prime \prime })$. Replace $F$ byThe straight-line interpolation is a homotopy supported away from the boundary of the cell and stays inside that cell. Where $\chi \ne 1$, we have $|F|>r^{\prime }$ and hence $|G|>r^{\prime }/2$. By the Sard theorem, choose $z\in B_{r^{\prime }/4}$ outside $h(U)$. Where $\chi =1$ we have $G=h$, and elsewhere $|G|>r^{\prime }/2$, so the modified map avoids $z$ everywhere.

Radial retraction of $D^d\setminus \{z\}$ onto its boundary fixes that boundary. It therefore descends through the characteristic map, including its possibly noninjective boundary identifications, and fixes all other cells of $Y$. Composing gives a homotopy whose endpoint misses all of $e^d$. Remove every cell of dimension greater than $n$ in descending order. There are finitely many such cells, so their homotopies concatenate to yield $g:S^n\to Y^n\subseteq X_n$. This proves dimension reduction of sphere maps into CW skeleta. The approximation step is essential: an arbitrary continuous image of $S^n$ can fill a higher-dimensional region, so dimension alone would not supply an omitted point.

Put $m=2n$ and let $\alpha$ be the identity class of $S^m$. With the product-cell orientation used in part (i), form the folding map $\nabla :S^m\vee S^m\to S^m$, which is identity on each summand. The top cell of $S^m\times S^m$ is attached by the product of the two inclusions, so folding gives the attaching map $[\alpha ,\alpha {]}_W$. Thus there is a cellular mapwhich is the fold on the lower skeleton and degree one on the top relative cell.

Let $a,b\in H^m(S^m\times S^m;\mathbb{Z})$ be the classes from the two factors, and let $x\in H^m(C;\mathbb{Z})$ restrict to the generator of the bottom sphere. Orient the top class $z\in H^{2m}(C;\mathbb{Z})$ so that $F^{\ast }z=ab$. Then $F^{\ast }x=a+b$. Since $a^2=b^2=0$ and $m$ is even, graded commutative algebra givesThe map $F^{\ast }$ is an isomorphism in degree $2m$, because of its degree-one top-cell map. Therefore $x^2=2z$. By the definition of the Hopf invariant this provesThe shifted Lie convention in part (i) multiplies the geometric bracket by $(-1{)}^m=1$, so it gives the same square and the same invariant. Reversing the orientation chosen for the attached top cell would reverse both its generator and the numerical sign convention; with the standard orientation the value is the stated positive two.

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.

Use the standard complex orientations and the product orientation. Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ satisfy $a^2=b^2=0$ and $⟨ab,[S^2\times S^2]⟩=1$, and let $x\in H^2({\mathbb{CP}}^2;\mathbb{Z})$ satisfy $⟨x^2,[{\mathbb{CP}}^2]⟩=1$. These are the cohomology rings of the product of two spheres and the Complex projective plane.

For a map from the Complex projective plane, write $f^{\ast }a=rx$ and $f^{\ast }b=sx$. Naturality of the cup product gives $r^2{x}^2=f^{\ast }(a^2)=0$ and $s^2{x}^2=f^{\ast }(b^2)=0$. Since $x^2$ has infinite order, $r=s=0$. It follows that $f^{\ast }(ab)=0$, so every such map has degree $0$.

In the reverse direction, write $g^{\ast }x=ra+sb$. ThenEvery even mapping degree really occurs. To construct it, identify $S^2$ with ${\mathbb{CP}}^1$ and use the Segre embeddingThe generator $x_3\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ pulls back to $a+b$: restricting to either factor gives a projective line and hence coefficient $1$. By the cellular approximation theorem, $\sigma$ is homotopic to a cellular map. Its four-dimensional domain then maps into the four-skeleton ${\mathbb{CP}}^2$ of ${\mathbb{CP}}^3$. Denote that map by $g_0$; restriction of $x_3$ to ${\mathbb{CP}}^2$ is $x$, so $g_0^{\ast }x=a+b$ and $\mathrm{deg}{g}_0=2$.

For any integer $m$, choose a map $d_m:S^2\to S^2$ of mapping degree $m$. For $m>0$ one may use $z\mapsto z^m$ on the Riemann sphere, for $m<0$ use $z\mapsto {\overline{z}}^{|m|}$, and for $m=0$ use a constant map. Then $g_m=g_0\circ (d_m\times \mathrm{id})$ satisfies $g_m^{\ast }x=ma+b$, so the possible degrees are exactly all even integers, with $g_m$ realizing $2m$. The cellular approximation here deforms this particular map into the skeleton; it does not require a retraction of ${\mathbb{CP}}^3$ onto ${\mathbb{CP}}^2$.

Finally, take the connected sum of oriented manifolds with both summands carrying their standard complex orientations. Its degree-two cohomology has generators $u,v$ withThis follows by choosing the generators supported away from the two balls used to form the connected sum, so mixed cup products vanish while each square gives the common orientation class. WriteIf $\mathrm{deg}h=n$, the three cup product relations sayTaking determinants gives $-(\det P{)}^2=n^2$, which forces $n=0$. Therefore the only possible degree to the connected sum is $0$, realized by a constant map. This is a degree constraint from intersection forms: the indefinite intersection form of the sphere product cannot pull back a definite form with a nonzero degree multiplier.

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.

The $q$-skeleton of a CW complex is the subcomplex consisting of its topological cells of dimension at most $q$. A cellular map preserves these skeleta. The cellular approximation theorem lets continuous maps respect this filtration up to homotopy.