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The topological degree measures an oriented net number of sheets of a map. Its simplest integer-valued setting is a map $f:M\to N$ between connected oriented closed manifolds of the same positive dimension $n$. The hypotheses matter: an integer sign count needs orientations; compactness prevents preimages from escaping; and the same dimension makes a regular fiber discrete. A fundamental class gives the definition for continuous maps:Here $H_n(M;\mathbb{Z})$ and $H_n(N;\mathbb{Z})$ are generated by their chosen fundamental classes. Reversing either orientation changes the sign of the degree of a map between oriented manifolds. Disconnected sources contribute a sum over components; disconnected targets require a degree for each target component rather than one universal integer.

For a smooth map, choose a regular value $y$. The Sard theorem guarantees plentiful choices. Each point of $f^{-1}(y)$ has an invertible derivative, so the inverse function theorem makes it an isolated point. The fiber is finite by compactness of $M$. An oriented coordinate chart defines $\mathrm{sgn}\det d{f}_x$, and the degree as a sum of local degrees becomesAn empty fiber contributes zero. The sign is independent of the chosen oriented charts. Thus the mapping degree counts sheets with signs, not simply the number of preimages: orientation-preserving and orientation-reversing sheets cancel.

A differential-form argument both proves that this count is independent of $y$ and relates it to the topological definition. Around a regular value, take a small neighborhood $V$ whose full preimage is a disjoint union of neighborhoods mapped diffeomorphically onto $V$. Compactness rules out additional preimages approaching $V$ from elsewhere. Choose a smooth top-degree form $\eta$ supported in $V$, with ${\int }_N\eta =1$. The change of variables formula givesAlthough $\eta$ need not be everywhere positive, it represents the same normalized top cohomology class as any normalized volume form. The top de Rham cohomology of a compact connected oriented manifold says that two top forms with equal integrals differ by $d\beta$. Their pullback integrals agree, because the Generalized Stokes theorem gives ${\int }_{M}d(f^{\ast }\beta )=0$. Therefore the degree by integration of a pullback volume form isfor every smooth top form $\eta$. Pairing de Rham cohomology with fundamental classes identifies this integer with the homological definition. This proof also shows that every regular value gives the same signed sum.

The homotopy invariance of mapping degree makes it stable under continuous deformation. Homologically this follows from homotopic maps inducing the same map on homology. For a smooth homotopy $F:M\times [0,1]\to N$, a direct proof uses $d\eta =0$ and the Generalized Stokes theorem on the cylinder to getwith the cylinder orientation chosen to give the indicated boundary signs. Smooth approximation extends the geometric description to continuous maps, while the fundamental class definition already applies without differentiability. The multiplicativity of mapping degree follows from functoriality and givesA constant map has degree zero when $n>0$. If the mapping degree is nonzero, the map is onto: a point outside its image is a regular value with empty preimage. An orientation-preserving diffeomorphism has degree $1$, and an orientation-reversing one has degree $-1$. An orientation-preserving $k$-sheeted covering map has degree $k$; the signs of sheets must be included for other orientation conventions.

Examples show both the strength and the limits of the invariant. On the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, including negative integers and zero; it is the winding number. On the sphere $S^n$, an ambient reflection has degree $-1$, and the antipodal map has degree $(-1{)}^{n+1}$, because the determinant of $-I$ on the ambient ${\mathbb{R}}^{n+1}$ is $(-1{)}^{n+1}$. On the torus, an integer matrix $A$ induces a map of degree $\det A$, obtained by pulling back the constant top form. However, degree does not classify general manifold maps: the identity and an integer shear on the two-torus both have degree one but induce different maps on $H_1$, so are not homotopic.

The mapping degree gives useful obstruction and existence arguments. There is no retraction of a closed ball onto its boundary sphere: such a retraction would make the degree-one identity of the boundary null-homotopic, since it would extend across a contractible ball, contradicting homotopy invariance and the degree-zero constant map. This is the degree obstruction underlying the Brouwer fixed-point theorem. Also an everywhere nonzero tangent vector field on an even-dimensional sphere would, after normalization, produce the homotopy $\cos (\pi t)x+\sin (\pi t)v(x)$ from the identity to the antipodal map. Orthogonality makes every value a unit vector, but the endpoint degrees are $1$ and $-1$, impossible. This proves the corresponding Hairy ball theorem obstruction.

There are appropriate extensions rather than an unrestricted integer degree for every pair of manifolds. For a proper map between connected oriented manifolds without boundary of equal dimension, the regular-value count remains finite and is invariant; compactly supported top forms or locally finite homology replace the compact fundamental-class description. Without orientations, degree modulo two counts preimages modulo two and uses mod-two fundamental classes. For manifolds with boundary one uses relative mapping degree for maps of pairs, keeping the boundary away from the chosen target value. Maps of unequal dimensions, or nonproper maps where preimages can escape to infinity, do not in general admit this same signed sheet-count invariant. The central principle is an integer signed count stable under homotopy, with its hypotheses explicitly preserved.

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.

For $n\ge 1$, let $\#$ denote concatenation in the first cube coordinate and let $\cdot$ denote pointwise multiplication of maps into the topological group. Both operations descend to the homotopy group, and the constant map is their common identity. They satisfy the interchange rulewhich follows by considering the two halves of the first coordinate. Using the identity class $0$ givesTherefore $[\alpha \beta ]=[\alpha ]+[\beta ]$. The equality is on based homotopy classes; the usual reparametrization homotopies justify the unit identities for concatenation. This is the Eckmann-Hilton argument, and proves homotopy-group addition in a topological group even for $n=1$.

For the unit quaternions, the explicit inverse isBoth compositions cancel in the displayed order, without commuting the quaternions. Multiplication, inversion and all integer powers are continuous, so this proves that ${\Phi }_{k,l}$ is a homeomorphism for all integers $k,l$.

Put $m=k+l$. Let $a,b$ be the standard generators of $H_3(S^3\times S^3;\mathbb{Z})$, represented by the first and second factors. The previous pointwise-multiplication result and the Hurewicz theorem imply that the mapping degree of $q\mapsto q^m$ on $S^3$ is $m$, including negative integers. Restricting $\Phi$ to the two factors therefore givesThe map on $H_0$ is the identity. For the top homology group, take the dual degree-three cohomology classes $x,y$. We have ${\Phi }^{\ast }x=mx+y$, ${\Phi }^{\ast }y=x$. Since these classes have odd degree, graded commutativity of the cup product givesThus ${\Phi }_{\ast }$ is multiplication by $-1$ on $H_6$, and all other homology groups of the product are zero.

For the gluing, regard $\Phi$ as the attaching identification from $\partial (D^4\times S^3)$ to $\partial (S^3\times D^4)$. In these coordinates its first input is the boundary coordinate of $D^4$, its second input the fibre coordinate; its second output is the boundary coordinate of the other $D^4$. This makes the two pieces the usual two trivializations of a three-sphere bundle over the four-sphere.

Let $A=S^3\times D^4$ and $B=D^4\times S^3$, and parametrize their common boundary $T=S^3\times S^3$ using the $B$ coordinates. Collar neighbourhoods give an open cover with the same homotopy types, so the Mayer–Vietoris sequence applies. Both pieces retract onto $S^3$. In degree three, the map into the homology of the pieces isIndeed, inclusion into $A$ retains the first output coordinate, while inclusion into $B$ retains the second input coordinate. Exactness now givesThe second relation eliminates the second generator of the cokernel, leaving a single generator with relation $m$ times that generator equal to zero. Thus $\mathrm{coker}L=\mathbb{Z}/m\mathbb{Z}$. The kernel is zero if $m\ne 0$ and is generated by $(1,0)$ if $m=0$.

The degree-six boundary class gives $H_7(X_{k,l})\cong H_6(T)=\mathbb{Z}$. All remaining positive-degree groups outside degrees $3,4,7$ vanish by the same Mayer–Vietoris sequence; connectedness gives $H_0=\mathbb{Z}$. Consequently the complete integral answer isHere $\mathbb{Z}/0\mathbb{Z}$ means $\mathbb{Z}$, and a negative $m$ gives the same cyclic group as $|m|$. In particular, $m=\pm 1$ gives the integral homology of a seven-sphere, while $m=0$ gives the integral homology of $S^3\times S^4$.

The degree and its local signs. Orient $S^n$ and let $[S^n]$ generate its top reduced homology ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree is the integer characterized byFor $n\ge 1$ this is the usual top homology definition using the fundamental class; using reduced homology also covers $S^0$.

Suppose $f$ is smooth and $p$ is a regular value. Each $x\in f^{-1}(p)$ has invertible tangent map $D{f}_x$, so the inverse function theorem makes $f^{-1}(p)$ discrete. It is also closed in the compact sphere, hence finite. Choose disjoint small neighborhoods of these points on which $f$ is a local diffeomorphism. The Excision theorem identifies the source local homology with the direct sum of one copy of $\mathbb{Z}$ for each inverse image. The induced local map is multiplication by $+1$ or $-1$ according as $D{f}_x$ preserves or reverses orientation. The map from the global fundamental class to these local orientation classes therefore proves the degree as a sum of local degrees formula:Here the determinant is computed in positively oriented tangent bases. Thus the mapping degree counts inverse images with signs, rather than just their cardinality.

The quotient map. Put $m=n+1$ and write $f=F{|}_{S^{m-1}}$. Give $D^m$ its standard orientation and its boundary the induced orientation. The connecting homomorphismis an isomorphism: the disk has zero positive reduced homology. It sends the relative fundamental class $[D^m,S^{m-1}]$ to $[S^{m-1}]$. If the map on relative homology induced by $F$ multiplies this class by $d$, naturality givesThus $d=\mathrm{deg}f$. Collapsing the boundary gives a sphere $D^m/S^{m-1}\cong S^m$, and the quotient map identifies its top reduced homology with the top relative homology of the disk pair. Give this quotient sphere the orientation determined by that identification. The relation $qF=\tilde{F}q$ now showsThis is the quotient-sphere degree identity.

The graph intersection. Orient $D^m\times D^m$ by the product orientation, orient $N=D^m\times \{0\}$ by its first factor, and orient the graph of a function $\Gamma$ by $x\mapsto (x,F(x))$. An intersection is precisely a zero of $F$, and no such zero lies on the boundary because $F(S^{m-1})\subset S^{m-1}$. At an intersection, transverse intersection means that $D{F}_x$ is surjective, hence invertible. The zeros are consequently isolated and finite.

For the smooth intersection number use the ordered tangent spaces $T_{x}N$ first and $T_{(x,F(x))}\Gamma$ second. Relative to the product basis their concatenated basis has matrixIts determinant is $\det D{F}_x$, so the intersection sign is ${ϵ}_x=\mathrm{sgn}\det D{F}_x$. By the same local Excision theorem argument, now in relative homology at the interior point $0$, the sum of these signs equals the multiplier of $F_{\ast }$ on $H_m(D^m,S^{m-1};\mathbb{Z})$. Therefore the graph intersection formula for mapping degree isIf the tangent spaces are ordered $\Gamma$ first and $N$ second, every sign changes by $(-1{)}^m$; the order above specifies the appropriate convention.

The one-dimensional model $F(x)=2x^3-x$ on $[-1,1]$ illustrates the graph intersection formula for mapping degree: its three zeros have signs $+1,-1,+1$, and its endpoint map has mapping degree $1$ on reduced homology.

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.

Euler number and self-intersection. Orient the rank-two normal bundle ${\nu }_S$ by the orientations of $S$ and $M$: the ordered tangent and normal spaces have the ambient orientation. Choose a smooth normal section $s$ transverse to the zero section. Its zeros are isolated, and their signed number is the evaluation $⟨e({\nu }_S),[S]⟩$ of the Euler class on the fundamental class. Scale the section sufficiently small to lie in a tubular neighborhood. Its graph is a push-off $S^{\prime }$ isotopic to $S$.

The intersections of $S$ with $S^{\prime }$ occur exactly at the zeros of $s$. In a positively oriented local splitting $TS\oplus {\nu }_S$, the basis formed from the tangent spaces to $S$ and to the graph has block matrixIts determinant is $\det Ds$, so the local intersection sign is precisely the local zero index of the section. Summing proves the Euler number equals self-intersection identity

The conic. The given map is the degree-two Veronese map $[u:v]\mapsto [u^2:uv:v^2]$ followed by the projective linear transformationThe Veronese map is injective with nonvanishing differential: on the chart $u\ne 0$ its coordinates are $[1:t:t^2]$, and on $v\ne 0$ they are $[s^2:s:1]$. It is therefore a smooth embedding, and $S$ is a smooth plane conic diffeomorphic to ${\mathbb{CP}}^1\cong S^2$. Its equation in the given coordinates is $z_0^2-z_2^2=4z_1^2$.

Let $L$ be a projective line, with its complex orientation. The cohomology ring of complex projective space gives $[L]\cdot [L]=1$ in ${\mathbb{CP}}^2$. The hyperplane $z_1=0$ meets $S$ at $[u:v]=[1:0]$ and $[0:1]$. The zeros of $uv$ are simple in the respective local coordinates, and complex intersections have positive signs. Hence $[S]\cdot [L]=2$, so $[S]=2[L]$. The self-intersection number and the normal Euler class arewhere $u\in H^2(S^2;\mathbb{Z})$ has $⟨u,[S^2]⟩=1$.

The boundary of the tubular neighborhood. Write $N=\nu (S)$ for the closed disk tubular neighborhood and $E=\partial N$ for its boundary. This distinguishes the disk neighborhood $N$ from the vector normal bundle ${\nu }_S$. The space $E$ is the oriented circle bundle of ${\nu }_S$, with Euler class $4u$. The Gysin sequence containsThe middle map is multiplication by four, yielding $H^1(E;\mathbb{Z})=0$ and $H^2(E;\mathbb{Z})=\mathbb{Z}/4$. The same Gysin sequence gives $H^0(E;\mathbb{Z})=H^3(E;\mathbb{Z})=\mathbb{Z}$. The total space is a closed connected oriented three-manifold, so Poincare duality gives

The exterior. Work first with $W={\mathbb{CP}}^2\setminus \mathrm{int}N$. A collar neighborhood of $\partial W$ shows that its interior, the requested ${\mathbb{CP}}^2\setminus N$, has the same homotopy equivalence type: push the boundary a small positive distance into the collar. The Excision theorem and Thom isomorphism theorem identifyOnly degrees two and four are nonzero, each a copy of $\mathbb{Z}$.

In degree four the map $H_4({\mathbb{CP}}^2;\mathbb{Z})\to H_4(N,E;\mathbb{Z})$ sends the ambient fundamental class to the relative fundamental class of $N$. Under the Thom isomorphism theorem this becomes $[S]$. Thus this map is multiplication by one with compatible orientations. The long exact sequence in homology givesso $H_4(W)=H_3(W)=0$.

In degree two, the Thom isomorphism theorem identifies the map $H_2({\mathbb{CP}}^2)\to H_0(S^2)$ with intersection against $[S]$. A projective line intersects the conic twice, so this map is multiplication by two. The long exact sequence in homology isIt yields $H_2(W)=0$ and $H_1(W)=\mathbb{Z}/2$. In degree zero the relative groups vanish, so $H_0(W)\cong H_0({\mathbb{CP}}^2)\cong \mathbb{Z}$. This computes the homology of the complement of a smooth conic:The normal Euler number $4$, the degree-two intersection map and the degree-four restriction map $1$ play different roles; distinguishing them explains why the boundary has first homology $\mathbb{Z}/4$ while the exterior has first homology $\mathbb{Z}/2$.

We use the James reduced product theorem and the Bott–Samelson theorem: if $X$ is a connected based CW complex with free integral homology, then $JX\to \Omega \Sigma X$ is a weak equivalence, and its induced Pontryagin ring isHere $\Sigma$ is the reduced suspension, $T$ is the tensor algebra, and the multiplication is induced by concatenating James words, hence by concatenating loops. Taking $X=S^{2n}$ supplies a single generator $x$ of degree $2n$, sois free, with one generator in every degree $2nk$ and zero in the other degrees.

The diagonal makes this a homology coalgebra. The generator $x$ is a primitive homology class:There are no nontrivial lower positive degrees in which its reduced diagonal could land. Compatibility of the diagonal with loop multiplication givesThe degree of $x$ is even, so the two tensor factors commute without a Koszul sign rule.

The universal coefficient theorem for cohomology has no Ext terms here because homology is free. Let $a_k$ be the cohomology class dual to $x^k$, with $a_0=1$ and $|a_k|=2nk$. The cup product is dual to the diagonal, henceTherefore the integral cohomology ring is the divided power algebraExplicitly $H^{2nk}=\mathbb{Z}{a}_k$ for $k\ge 0$, and all other cohomology groups vanish. In particular $a_1^k=k!a_k$: over the integers this is not a polynomial ring on $a_1$. This is the integral cohomology of an odd-sphere loop space.

For the requested homology multiplication, use the homology cross product followed by concatenation:The constant loop gives the degree-zero unit. Loop concatenation is associative up to homotopy, which suffices for associativity on homology; Moore loops can make the space-level operation strictly associative. The Künneth theorem identifies the tensor-product homology because its groups are free abelian groups. The Bott–Samelson theorem identifies this product with word multiplication, so $x^i\ast x^j=x^{i+j}$. The Pontryagin ring of an odd-sphere loop space has presentationWith only one generator the tensor algebra has the indicated polynomial presentation. The homology product and the divided-power cohomology product are different operations.

Let ${\mathcal{C}}_{\mathrm{fg}}$ be the Serre class of finitely generated abelian groups. A homomorphism is an isomorphism modulo this class when its kernel and cokernel are finitely generated. The Hurewicz theorem modulo a Serre class says that, for a simply connected space and $r\ge 2$, if ${\pi }_i(X)\in {\mathcal{C}}_{\mathrm{fg}}$ for $2\le i<r$, thenThe corresponding lower-dimensional homology condition is equivalent to the lower homotopy condition. In particular, degreewise finite generation of integral homology and homotopy are equivalent for simply connected spaces. The Serre class is closed under subgroups, quotients and extensions; these closure properties are what make the modulo-class formulation useful.

Here is the consequence needed for construction. Inductively, if the lower homotopy groups are finitely generated, the theorem gives a finitely generated kernel for $h_r$. Its image is a subgroup of the finitely generated $H_r(X)$, so the image is also finitely generated. The resulting extension proves ${\pi }_r(X)$ finitely generated. Starting with the ordinary isomorphism ${\pi }_2(X)\cong H_2(X)$ yields this for every positive degree.

We now construct a finite type CW approximation. Choose finitely many maps $S^2\to X$ generating ${\pi }_2(X)$, and let $C_2$ be their wedge. The resulting map $C_2\to X$ is $2$-connected: it is an isomorphism below degree two and a surjection in degree two. In general, call a map $r$-connected when its mapping-cylinder pair has relative homotopy groups zero through degree $r$.

Suppose $f_r:C_r\to X$ is $r$-connected and $C_r$ is a finite simply connected CW complex of dimension at most $r$. The homotopy groups of $C_r$ are finitely generated by the same modulo-class theorem, since a finite CW complex has finitely generated homology. The long exact sequence of relative homotopy groups shows thatis finitely generated: it lies between a quotient of ${\pi }_{r+1}(X)$ and a subgroup of ${\pi }_r(C_r)$. Relative groups here refer to the mapping cylinder of $f_r$.

Represent a finite set of generators by relative disks. Attach their boundary spheres to $C_r$, and extend the map over the disks by their chosen maps into $X$. This adds finitely many $(r+1)$-cells and kills the relative group in that dimension without changing lower relative groups. The new map $C_{r+1}\to X$ is $(r+1)$-connected. Iterating givesThere are finitely many cells of each dimension. Every fixed homotopy degree stabilizes to an isomorphism once sufficiently high-dimensional cells have been added, so this is the required weak equivalence:

For the bounded-homology assertion, assume first $n\ge 2$ and take the finite $n$-dimensional $C_n$ just constructed. The Relative Hurewicz theorem for its $n$-connected map givesThe relative homology sequence and $H_{n+1}(X)=0$ identify the latter withBecause $C_n$ is $n$-dimensional, $H_n(C_n)$ is a subgroup of its free cellular $n$-chain group. It is finite free, and so is $K$.

Choose a basis of $K$, lift it using the Relative Hurewicz theorem, and attach exactly those finitely many $(n+1)$-cells to $C_n$. Denote the resulting complex by $C^{\prime }$. Its new cellular boundary has image $K$ and is injective: the selected cycles are linearly independent in $H_n(C_n)$, and there are no old $(n+1)$-boundaries. ThusLower homology remains unchanged, and all higher homology is zero on both sides. The map $C^{\prime }\to X$ is an integral homology isomorphism between simply connected spaces. The homological Whitehead theorem consequently makes it a weak homotopy equivalence. Hence the finite CW approximation from bounded homology hasIf $n=0$ or $1$, simple connectivity and the homology hypothesis make all reduced homology zero. The Hurewicz theorem, applied at the first possible nonzero homotopy degree, shows that $X$ is weakly contractible, so a point suffices. If the wording requires dimension exactly $n+1$ rather than at most $n+1$, add a contractible cancelling pair of $n$- and $(n+1)$-cells, mapping constantly to the basepoint. This does not change the weak homotopy type.

Let $u_n$ be the rationalized universal class of degree $n$. The rational cohomology of an integral Eilenberg–MacLane space isHere $\Lambda$ is an exterior algebra. For $n=1$, $K(\mathbb{Z},1)\simeq S^1$. The standard path-loop spectral-sequence calculation supplies the induction: inthe total space is contractible and the fundamental fiber class transgresses to $u_n$. An odd exterior fiber generator gives an even polynomial base generator. An even polynomial fiber generator gives an odd exterior base generator; the differential on its $k$th power has coefficient $k$, which is invertible over $\mathbb{Q}$. The multiplicative spectral sequence then has no remaining positive-degree classes in the total space. This is the rational transgression calculation for Eilenberg–MacLane spaces; it includes the absence of additional base generators.

For $m=2n+1\ge 3$, choose $S^m\to K(\mathbb{Z},m)$ representing the integral fundamental class. The ring calculation shows that this map is a rational homology equivalence: both spaces have rational cohomology only in degrees zero and $m$. The rational Whitehead theorem for simply connected spaces identifies their rational homotopy groups. Since the target has only ${\pi }_m=\mathbb{Z}$, the rational homotopy groups of a sphere in odd dimension areFor $n=0$, this follows directly from ${\pi }_1(S^1)=\mathbb{Z}$ and the contractible universal cover of the circle, which makes every higher homotopy group zero.

Let $M={\mathbb{CP}}^2\#{\mathbb{CP}}^2$, with the standard complex orientations. It is simply connected by the Seifert-van Kampen theorem applied to the punctured summands. Classes $a,b\in H^2(M;\mathbb{Q})$ can be chosen from the two summands. Their cross product vanishes, while their squares equal the oriented top class:Thus the cohomology ring of the connected sum of two complex projective planes isThe two relations also kill all cubic monomials, so its dimensions are $1,2,1$ in degrees $0,2,4$ and zero otherwise.

We use the Sullivan minimal model dictionary: for a simply connected finite-type space, the dual of its degree-$i$ generator space is ${\pi }_i(M)\otimes \mathbb{Q}$. The following free graded-commutative differential algebra is the Sullivan model of the connected sum of two complex projective planes:It is minimal because all differentials of generators are decomposable.

To verify that no further generators are required, observe that $ab,a^2-b^2$ is a regular sequence in $\mathbb{Q}[a,b]$. The first polynomial is a nonzerodivisor. If $(a^2-b^2)h$ is divisible by $ab$, restricting to each coordinate axis forces $h$ to vanish on both axes, hence to be divisible by $ab$. The second polynomial is therefore a nonzerodivisor modulo the first. The Koszul complex of this regular sequence is exactly $\mathcal{M}$, so its cohomology is the quotient ring above, with no additional odd cohomology.

For completeness, choose rational polynomial forms representing $a,b$ on $M$. Their product and the difference of their squares are exact; choose degree-three primitives for them. Sending $r,s$ to those primitives defines a differential-algebra map from $\mathcal{M}$ to the rational polynomial forms on $M$. It induces the specified cohomology-ring isomorphism and hence is a quasi-isomorphism. This verifies the model directly, rather than assuming that a cohomology presentation alone automatically determines all rational homotopy.

The model has exactly two degree-two and two degree-three generators. Consequently the rational homotopy groups of the connected sum of two complex projective planes are

Write $u=u_n\in H^n(K(\mathbb{Z}/2,n);{\mathbb{F}}_2)$ for the fundamental class. We use the path-loop Serre spectral sequence, its multiplicative structure, and the Kudo transgression theorem: transgression of the fundamental classes and their compatible Steenrod squares is natural, equivalently cohomology suspension commutes with these stable operations. The initial ring is $H^{\ast }(K(\mathbb{Z}/2,1);{\mathbb{F}}_2)={\mathbb{F}}_2[t_1]$.

Here is the low-degree calculation. In the first path fibration, $t$, $t^2$ and $t^4$ transgress respectively to $u_2$, ${\mathrm{Sq}}^1{u}_2$ and ${\mathrm{Sq}}^2{\mathrm{Sq}}^1{u}_2$, in degrees $2,3,5$. These supply all indecomposable base classes through degree five; the other classes there are products. Repeating the path-loop calculation raises the degree of the transgressive classes by one. For $n=3$, the class ${\mathrm{Sq}}^2{u}_3$ transgresses from $u_2^2$, and ${\mathrm{Sq}}^2{\mathrm{Sq}}^1{u}_3$ from the degree-five generator. For $n=4$, ${\mathrm{Sq}}^3{u}_4$ transgresses from $u_3^2$. For larger $n$ all four displayed low-degree operations are transgressive indecomposables. Acyclicity of the path-space total cohomology forces these transgressions and excludes additional classes in this range.

More systematically, this is the range up to $n+3$ of the Serre polynomial generators for mod-two Eilenberg–MacLane cohomology:For an admissible sequence of Steenrod squares, $I=(i_1,\dots ,i_r)$ has $i_j\ge 2i_{j+1}$; its degree increment is $\sum i_j$, and its Steenrod excess is $e(I)=i_1-i_2-\cdots -i_r$. Include the empty sequence. Up to increment three the nonempty possibilities are $(1),(2),(3),(2,1)$; the strict excess bound and possible products explain precisely the small-$n$ exceptions.

The low-degree mod-two cohomology of an Eilenberg–MacLane space isNegative-degree cohomology is zero. For $n=2$, ${\mathrm{Sq}}^{2}u=u^2$ and ${\mathrm{Sq}}^{3}u=0$ by instability, so the product in degree five must not be omitted. For $n=3$, ${\mathrm{Sq}}^{3}u=u^2$ is a product, still independent from ${\mathrm{Sq}}^2{\mathrm{Sq}}^{1}u$. For $n\ge 4$ the listed classes are indecomposable. The Adem relations include ${\mathrm{Sq}}^1{\mathrm{Sq}}^1=0$ and ${\mathrm{Sq}}^1{\mathrm{Sq}}^2={\mathrm{Sq}}^3$, so no further increment-three class comes from reversing the two squares.

Now use the space actually printed in the PDF,the converted TeX dropped the projective-space $P$'s. This stunted real projective space has one cell in dimensions $7,8,9,10$, besides its basepoint, and is $6$-connected. The integral cellular boundary is $2$ in even dimensions and zero in odd dimensions. HenceThe Hurewicz theorem gives ${\pi }_7(Y)=\mathbb{Z}/2$.

For the next two groups, let $K=K(\mathbb{Z}/2,7)$ and choose $f:Y\to K$ representing the generator of $H^7(Y;{\mathbb{F}}_2)$. It induces an isomorphism on ${\pi }_7$, and the target's higher homotopy groups vanish.

We need the low-degree integral homology of a mod-two Eilenberg–MacLane space. Here it isTo justify the torsion orders, the Serre class theorem first makes every positive-degree integral homology group of $K$ a finite 2-group. The universal coefficient theorem for cohomology with ${\mathbb{F}}_2$ and the low-degree dimensions give one cyclic summand in degrees seven, nine and ten and none in degree eight. The mod-two Bockstein homomorphism is ${\mathrm{Sq}}^1$. Its relevant nonzero actions areThe last target is nonzero: $(3,1)$ is admissible with excess two, below seven. It also follows by transgressing the nonzero square $({\mathrm{Sq}}^1{u}_2{)}^2$ through successive path fibrations. Each nonzero Bockstein pairs the mod-two classes associated with a cyclic integral summand of order exactly two; for a summand of order $2^r$ with $r>1$, this first Bockstein would be zero. In degree ten, ${\mathrm{Sq}}^{3}u$ is already the Ext class from $H_9$, so the independent class ${\mathrm{Sq}}^2{\mathrm{Sq}}^{1}u$ detects $H_{10}$. This proves all four integral groups without confusing them with mod-two Betti numbers.