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The usual compact closed-manifold hypothesis is needed for the first assertion, or a proper bounded-below Morse exhaustion in an appropriate noncompact version. As literally printed without any such condition, $f(x)=x$ on $\mathbb{R}$ is a Morse function with no critical points, while $b_0=1$. We prove the intended closed-manifold statement, which applies to the surface product later in the question.

By the Morse handle-attachment theorem, crossing an index-$i$ critical point attaches an $i$-handle, and no topological change occurs between critical levels. Retraction to the core disks of handles gives a CW complex model with $c_i$ cells in dimension $i$. Its cellular chain complex has $C_i\cong {\mathbb{Z}}^{c_i}$. Tensor with $\mathbb{Q}$, which computes the ranks $b_i$ of integral homology. Since homology is a quotient of a vector subspace of $C_i\otimes \mathbb{Q}$, $b_i\le c_i$. This argument has not used the Morse inequalities.

The strong Morse inequalities and top-dimensional equality areTheir weak form is $c_i\ge b_i$. To see the algebra behind the strong form, put $r_i={\mathrm{rank}}_{\mathbb{Q}}{\partial }_i$ in the Morse chain complex, with $r_0=r_{m+1}=0$. The kernel-image dimensions give $c_i=b_i+r_i+r_{i+1}$, so the alternating difference at $j$ is exactly $r_{j+1}\ge 0$. Conversely, given the strong inequalities, denote those alternating differences by $S_j$, with $S_{-1}=0$. Then $c_j-b_j=S_j+S_{j-1}\ge 0$, proving that they imply every weak inequality.

They are not equivalent as numerical restrictions. For the Betti numbers $b=(1,0,0,1)$ of $S^3$, the formal counts $c=(2,0,0,2)$ satisfy all weak inequalities and the Euler equality, but at $j=1$ give $c_1-c_0=-2<-1=b_1-b_0$. They therefore cannot be the counts of a Morse function on $S^3$; the strong inequalities detect that additional obstruction.

If all the strong inequalities are equalities, all $r_i$ vanish and $c_i=b_i$. An integer boundary matrix with rational rank zero is the zero matrix. Thus the integral complex itself has zero differential, not merely its rationalization. Consequentlyand higher homology vanishes. In particular there is no homology torsion. This is an integral consequence of a Perfect Morse function, rather than just an assertion about ranks.

For ${\Sigma }_g$, the integral homology is free with Betti numbers $(1,2g,1)$. The Künneth theorem therefore has no torsion terms, and the product's Poincaré polynomial isSumming the weak inequalities yields the lower boundChoose a surface Morse function $h$ with one minimum, $2g$ saddles and one maximum. It can be built from one closed disk, the standard $2g$ one-handles, and a closing closed disk, using the local forms $u^2+v^2$, $-u^2+v^2$, and $-u^2-v^2$ at the handle centres. Regular collars interpolate between the handles without new critical points. Now set $F(x,y)=h(x)+h(y)$. Its critical points are the ordered pairs of those of $h$, and its Hessian matrix is a block direct sum; the indices add. This product of Morse functions has exactly $(2g+2{)}^2$ critical points and realizes every displayed Betti count. Small bumps constant near the critical pairs can separate their values without altering the Hessians or creating new critical points.

The same minimum does not apply to arbitrary smooth functions. Here is an explicit finite-critical-point counterexample, avoiding any assumption that degeneracy leaves the Morse lower bound intact. On $T^2=(\mathbb{R}/\pi \mathbb{Z}{)}^2$, takeIt is well defined because shifting either coordinate by $\pi$ reverses two of its factors. Its derivatives are $\sin v\sin (2u-v)$ and $\sin u\sin (u-2v)$. If either $\sin u$ or $\sin v$ vanishes, both derivatives vanish only at $(0,0)$. Otherwise $2u-v$ and $u-2v$ must be multiples of $\pi$, giving exactly $(\pi /3,2\pi /3)$ and $(2\pi /3,\pi /3)$ in addition to the origin. At the first point the Hessian is $\sqrt{3}\left(\begin{array}{cc}1& -1/2\\ -1/2& 1\end{array}\right)$, positive definite; at the second it is its negative. They are a minimum and maximum. The origin is an isolated degenerate monkey saddle, with leading term $uv(u-v)$. Thus this three-critical-point function on a torus has exactly three critical points.

For $g>1$, perform connected-sum gluing of functions between this torus and a perfect Morse surface of genus $g-1$: delete a small neighbourhood of the torus maximum and of the other surface's minimum, and join their regular boundary circles by a monotone cylinder. Collar coordinates ensure nonzero differential throughout the neck, so precisely those two critical points are lost and none are added. The result on ${\Sigma }_g$ has $3+[2(g-1)+2]-2=2g+1$ critical points, retaining the degenerate point. For $g=1$, simply use $h_0$. Calling this smooth function $\tilde{h}$ and the earlier perfect Morse function $h$, the smooth product $\tilde{h}(x)+h(y)$ hascritical points. This explicitly proves the negative answer to the unrestricted smooth-minimum question; it does not assert that this smaller illustrative count is sharp.

Use the standard CW complex structure on the real projective plane, with one cell $e_r$ for each $r=0,1,2$. The attaching map of the two-cell winds twice around the one-cell, so the cellular chain complex has $d{e}_2=2e_1$ and $d{e}_1=0$.

The product cells $e_{ij}=e_i\times e_j$, $0\le i,j\le 2$, give nine cells, in dimensions $0,1,2,3,4$ with counts $1,2,3,2,1$. The cellular chains of a product of finite CW complexes have boundary $d(e_i\times e_j)=d{e}_i\times e_j+(-1{)}^i{e}_i\times d{e}_j$. In the ordered basesthe nonzero matrices areIn particular $d_2{d}_3=d_3{d}_4=0$. The degree-one homology is ${\mathbb{Z}}^2/2{\mathbb{Z}}^2$. The degree-two cellular cycles are $\mathbb{Z}{e}_{11}$, and the cellular boundaries there are $2\mathbb{Z}{e}_{11}$. The degree-three cycles are $\mathbb{Z}(e_{21}+e_{12})$, whose double is the boundary of $e_{22}$. Finally $d_4$ is injective. This computes the integral homology of two real projective planes:The degree-three torsion is the Tor functor contribution in the Künneth theorem; it would be missed by retaining only tensor products of the two factors' homology groups.

Use integral cellular homology. All three vertices become a single vertex, and all three oriented edges become one oriented circle, with generator $a$. The face goes round that circle three times in the same direction. Thus the quotient has a CW complex structure with one cell in each of dimensions zero, one and two, and cellular chain complexThe degree-two boundary is injective and its cokernel is $\mathbb{Z}/3$. ThereforeThe attaching word is $a^3$, so the Seifert-van Kampen theorem also gives ${\pi }_1(X)\cong \mathbb{Z}/3$. This is the degree-three circle-attachment model of a Moore space.

Let $a$ be the projective-line loop and $b$ the core loop of the Möbius band. The boundary of the Möbius band goes twice around its core. Its boundary inclusion is a cofibration, so replacing the gluing map by a homotopic cellular degree-$n$ map preserves the homotopy type. More precisely, the band is a mapping cylinder of the boundary double covering onto the core circle. After attaching its other end and collapsing the connecting tree, the resulting space has a homotopy-equivalent CW complex with one vertex, edges $a,b$, and two two-cells with attaching wordsThe first is the projective-plane face; the second records equality of the attached boundary loop $a^n$ and the band boundary loop $b^2$. Taking exponent sums gives the cellular chain complexSince $\det D=4$, its kernel is zero. The first invariant factor in the Smith normal form is the greatest common divisor of its entries, $d=\gcd (2,n)$, and the second is $4/d$. HenceThis includes $n=0$ and negative degrees. The fundamental group after attaching a Möbius band is $⟨a,b\mid a^2=1,b^2=a^n⟩$, consistent with the computed abelianization. The calculation is an example of homology after attaching a Möbius band to the real projective plane.

Label the six vertices $A,B,C,D,E,F$ clockwise, starting at the upper-left vertex. The arrows identify $AB$ with $ED$, $BC$ with $FE$, and $FA$ with $DC$, preserving the indicated directions. Consequently the quotient space has exactly two vertices,Orient its three topological cells by $a:A\to B$, $b:B\to C$, and $c:F\to A$. The clockwise boundary of the hexagonal topological cell traversesEvery topological cell of dimension one therefore occurs once with each sign. The cellular chain complex, with integral coefficients, isHere the rows correspond to $P,Q$. The image of ${\partial }_1$ is the primitive subgroup generated by $Q-P$, and its kernel is freely generated by $a+b$ and $a+c$. Thus the homology groups areThe two vertex classes are important: identifying all six vertices would give an incorrect cellular chain complex. The result agrees with the homology of a torus.

For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.

The filtration ${\mathbb{CP}}^0\subset {\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ gives one cell in each of dimensions zero, two and four, since ${\mathbb{CP}}^k\setminus {\mathbb{CP}}^{k-1}\cong {\mathbb{C}}^k$. There are no odd-dimensional cells, so every cellular differential vanishes. ThusThe complex projective line has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its cellular chain complex again has zero differentials, givingA homeomorphism induces isomorphisms on homology groups. The second groups have different ranks, and therefore

Now orient the Complex projective plane by its complex coordinates. The inclusion of its two-skeleton ${\mathbb{CP}}^1$ identifies the fundamental class of a projective line with a generator $h$ of $H_2({\mathbb{CP}}^2;\mathbb{Z})\cong \mathbb{Z}$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.

Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented homology classes are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their tangent spaces are the complex $w$-axis and complex $v$-axis, so the intersection is a transverse intersection. The ordered real basesconcatenate to the complex orientation of the ambient four-manifold. Thus the local sign is $+1$, andThis computes the intersection form using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now givesEquivalently, the map $H_2\to \mathrm{Hom}(H_2,\mathbb{Z})$ sending $a$ to $\lambda (a,-)$ is an isomorphism. This is exactly a unimodular intersection pairing, and proves projective lines generate a unimodular intersection form. Reversing the ambient orientation would change the matrix to $(-1)$ and would still be unimodular.

Write $A=S^2\times S^2$ and let $i:A\to X$ be its inclusion after the attachment. The cohomology ring of a product of two spheres iswhere $a,b$ come from the first and second factors and $ab$ is the chosen orientation class. The diagonal pulls both $a$ and $b$ back to the same generator of $H^2(S^2;\mathbb{Z})$.

There is one new three-cell. In the cellular chain complex, its boundary has coordinates $(1,1)$ in the two-dimensional cells. Thus the relevant differential isand the original four-cell still has zero boundary. This gives $H_2(X;\mathbb{Z})=\mathbb{Z}$, $H_3(X;\mathbb{Z})=0$, $H_4(X;\mathbb{Z})=\mathbb{Z}$, and no other positive homology groups. Equivalently, the relative cohomology sequence of $(X,A)$ gives an injective restriction in degree two with image $\mathbb{Z}(a-b)$, and an isomorphism in degree four.

Choose $u\in H^2(X;\mathbb{Z})$ and $v\in H^4(X;\mathbb{Z})$ by $i^{\ast }u=a-b$, $i^{\ast }v=ab$. Naturality of the cup product givesSince restriction is injective in degree four, $u^2=-2v$. All further positive-degree cup products vanish by dimension. ThereforeChanging the sign of $v$ would instead give $u^2=2v$; the displayed sign uses the product orientation $ab$. The factor $2$ is the characteristic feature of a cup square after a diagonal sphere attachment.

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.

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.

Both relators have zero exponent sums, so abelianization gives $H_1(Y)={\mathbb{Z}}^3$, with meridian variables $x,y,z$ corresponding to $a,b,c$. The universal abelian cover has deck transformation group ${\mathbb{Z}}^3$ and coefficient group ringThe generalized Heegaard diagram gives a two-dimensional spine with one vertex, three edges and two faces. Its lifted cellular chain complex iswhere chosen lifts of the cells giveThe two columns are the abelianized Fox derivatives of $[a,bc]$ and $[ab,c]$. The Fox calculus identity gives $d_1{d}_2=0$; this can also be checked by multiplying the displayed matrices. There is no three-cell in this spine. One may use the lifted spine because its deformation retraction from $Y$ lifts to the universal abelian cover.

The maximal minors of the Alexander matrix, in row-pair order $(1,2),(1,3),(2,3)$, areTheir greatest common divisor in $R$ is $xyz-1$, since $x-1,y-1,z-1$ have no common nonunit divisor. Accordingly the multivariable Alexander polynomial isThe allowed units are $\pm x^i{y}^j{z}^k$. The single-variable specialization convention can introduce extra $(t-1)$ factors; the answer here is the genuinely multivariable Alexander polynomial.

Cells and attaching maps. Regard ${\mathbb{RP}}^k$ as the lines in ${\mathbb{R}}^{k+1}$ and include ${\mathbb{RP}}^{k-1}$ as the lines whose last coordinate is zero. Its complement consists of lines with a unique representative $(x_1,\dots ,x_k,1)$, and is therefore an open $k$-cell. Inductively this makes Real projective space a CW complex with one cell in each dimension from zero to $n$.

A characteristic map is obtained from the northern closed hemisphere of $S^k$: send a unit vector to the line it spans. Its interior maps homeomorphically onto the open cell, while its equator has the antipodal identification. Thus the $k$-cell is attached by the quotient mapwhich is the antipodal two-sheeted covering. This includes the two endpoints of the one-cell attaching to the zero-cell.

The cellular chain complex has $C_k^{\mathrm{cell}}=\mathbb{Z}$ for $0\le k\le n$ and zero otherwise. To compute its differential, follow the attaching map by collapse of the $(k-2)$-skeleton. The resulting map to ${\mathbb{RP}}^{k-1}/{\mathbb{RP}}^{k-2}\cong S^{k-1}$ has two local contributions. They differ by the mapping degree $(-1{)}^k$ of the antipodal map on $S^{k-1}$. With compatible cell orientations,For $k=1$ the two oriented endpoints cancel, giving the same formula. Consecutive differentials compose to zero, as required.

The mod-two cup products. Modulo two every cellular differential vanishes, so cellular cohomology gives a one-dimensional group in each degree $0,\dots ,n$. Let $a\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$ be the Poincare dual of a projective hyperplane. This class is nonzero: a projective line transverse to that hyperplane meets it once. Intersecting $j$ generic projective hyperplanes produces ${\mathbb{RP}}^{n-j}$, and the cup product of their Poincare duals is the Poincare dual of that intersection. In particular, $a^n$ evaluates to one on the mod-two fundamental class. Therefore every $a^j$, $0\le j\le n$, is nonzero, since otherwise multiplying it by $a^{n-j}$ would contradict $a^n\ne 0$. Dimension makes $a^{n+1}=0$. The mod-two cohomology ring of real projective space isFor $n=0$ this simply means the cohomology of a point.

The product with integral coefficients. The final product's coefficients are unstated; take $\mathbb{Z}$ as the default. Dualizing the cellular chain complex above giveswith all unlisted groups zero. The integral Künneth theorem has tensor terms with $i+j=q$ and Tor functor terms with $i+j=q+1$:For these finite free cellular complexes it splits additively, though not canonically. Both the tensor and Tor functor of two $\mathbb{Z}/2$ summands give $\mathbb{Z}/2$, so no order-four summands occur.

For an efficient count, write $F_A(t)$ for the free-rank polynomial and $T_A(t)$ for the number of $\mathbb{Z}/2$ summands in each degree. The integral Künneth torsion polynomial rule isThe last two factors record respectively the tensor contribution in summed degree and the Tor functor contribution one degree lower. The three factors haveThe first two give $F_{23}=1+t^3$ and $T_{23}=2t^2+t^3+t^4+t^5$. Multiplying by the third givesConsequently the integral cohomology of a product of finite real projective spaces in this case isIf the intended coefficients were instead ${\mathbb{F}}_2$, the Künneth theorem over a field gives the dimension polynomialThus the mod-two groups in degrees zero through nine are respectivelyand all other degrees vanish.

First account for the unlabelled coefficient-sequence construction. The two short exact sequences of abelian groups areThe singular chain groups of $X$ are free abelian. Applying $\mathrm{Hom}(C_q(X),-)$ therefore preserves these exact sequences, degree by degree, giving short exact sequences of cochain complexes. The associated long exact sequence from a coefficient sequence gives the displayed maps in cohomology; the connecting maps are the integral Bockstein homomorphism $\hat{\beta }$ and the modulo-$n$ Bockstein homomorphism $\beta$. The first omitted map is multiplication by $n$, and the second is induced by $j$.

For the requested example, attach an $(i+1)$-cell to $S^i$ using a map of degree $n$. The resulting Moore space $X=M(\mathbb{Z}/n,i)$ has positive-degree cellular chain complexin degrees $i+1,i$. This construction also works for $i=1$, using the degree-$n$ map of the circle. In cellular cohomology with coefficients $\mathbb{Z}/n$, the differential is zero, so both $H^i$ and $H^{i+1}$ are $\mathbb{Z}/n$.

Lift the cochain taking value $1$ on the $i$-cell to a cochain with coefficients $\mathbb{Z}/n^2$. Its coboundary takes value $n$ on the $(i+1)$-cell, which is $j(1)$. The definition of the connecting homomorphism therefore sends the degree-$i$ generator to the degree-$(i+1)$ generator. HenceIt is nonzero for every $i\ge 1$ and $n\ge 2$, including composite $n$. This is the Bockstein on a cyclic Moore space.

No, it cannot deformation retract onto a graph. Each sphere contributes a degree-$2$ homology class independent of the other two. Attaching circles at points introduces only vertices and edges, soFor example, the cellular chain complex obtained by giving each sphere one $2$-cell and a tree of arcs has three zero-boundary $2$-cells and no $3$-cells. Every graph has zero degree-$2$ homology. A deformation retraction would induce a homotopy equivalence and an isomorphism on homology, which is impossible. An ordinary retraction has no such requirement.

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.

Put $F={\mathbb{F}}_2$. Under the usual bundle convention of a paracompact base, choose a fiber metric and let $D(E),S(E)$ be the disk bundle and sphere bundle of the rank-$n$ real vector bundle. A mod-two Thom class is a relative cohomology classwhose restriction to every fiber pair $(D^n,S^{n-1})$ is the nonzero generator. Equivalently it is a class in $H^n(E,E\setminus B;F)$, where $B$ is identified with the zero section. No orientation choices are needed over $F$.

The Thom isomorphism theorem asserts that this class exists and that, for every $r$, the mapis an isomorphism. The disk bundle retracts to the zero section $s:B\to D(E)$, so $H^r(D(E);F)\cong H^r(B;F)$. Define the mod-two Euler class $e_2(E)=s^{\ast }u\in H^n(B;F)$, where the pullback includes the passage from relative to absolute cohomology. Under the Thom identification, the map from relative to absolute cohomology sends $a$ to $a⌣e_2(E)$. Substituting into the pair's long exact sequence in cohomology gives the Gysin sequence of a sphere bundle:Here ${\pi }_{!}$ is the connecting homomorphism followed by ${\Phi }^{-1}$. Thus the sequence and the cup-product map have been derived from the Thom theorem, not just stated. The Thom/Gysin constructions are also discussed in Hatcher's Vector Bundles and K-Theory, §3.2.