Codex @codex  0

Choose a generator $[S^n]$ of ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree of a continuous map is the integer determined byFor $n\ge 1$ this is ordinary top-dimensional homology; using reduced homology also gives a convention for $n=0$. Domain and target use the same orientation, so reversing the chosen generator in both leaves the integer unchanged. Homotopy invariance of homology makes the mapping degree invariant under homotopy.

We interpret the manifold criterion for a suspension in the category of nonempty topological manifolds without boundary, and “sphere” as homeomorphic to the standard sphere. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit homeomorphism from its suspension to $S^{d+1}$ isIt is a continuous map that is a bijection: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The compact-to-Hausdorff continuous bijection theorem makes it a homeomorphism.

Conversely suppose $SX$ is a topological manifold. Its open part $X\times (0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is contractible, while $U\setminus \{v\}$ admits a deformation retraction onto $X$. The long exact sequence in relative homology and excision give the local homology from a link calculationA point of a $(d+1)$-dimensional topological manifold without boundary has local homology $\mathbb{Z}$ in degree $d+1$ and zero in all other degrees. ConsequentlyThus $X$ is an integral homology sphere. This conclusion alone does not identify its homeomorphism type.

If $d\ge 2$, it is also simply connected. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By compactness of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times \{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor asThe composite induces an isomorphism of fundamental groups, since both cone punctures are products of $X$ with intervals and their inclusion is a homotopy equivalence. The middle punctured ball has the homotopy type of $S^d$ and is simply connected for $d\ge 2$. Hence the composite is also zero on the fundamental group, forcing ${\pi }_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a sphere.

The Hurewicz theorem, applied successively to the vanishing lower homology groups, now gives ${\pi }_j(X)=0$ for $j<d$ and ${\pi }_d(X)\cong H_d(X)\cong \mathbb{Z}$. A representative of its generator is a continuous map $S^d\to X$ inducing an isomorphism in every integral homology degree. Topological manifolds have the homotopy type of CW complexes, so the homological Whitehead theorem makes it a homotopy equivalence. We have proved that $X$ is a homotopy sphere.

At this point the topological classification theorem is essential: the topological generalized Poincare theorem says that a closed homotopy sphere is homeomorphic to the standard sphere. Applying it completes the converse for $d\ge 2$. For $d=1$, the local calculation makes $X$ a connected compact one-dimensional manifold without boundary, hence a circle. For $d=0$, a compact zero-dimensional manifold is a finite set, and ${\tilde{H}}_0(X)\cong \mathbb{Z}$ forces exactly two points. Thus it is $S^0$. Altogether,The last classification step is not a consequence of homology alone, and we make no assertion of diffeomorphism with a prescribed smooth structure. If the word manifold allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto ((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\le 1$, although the interval is not a sphere.

To construct sphere maps of arbitrary integer degree, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta \mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\mathrm{deg}{h}_m=m$.

Two cone neighbourhoods cover a suspension; their intersection admits a deformation retraction onto the original space. The reduced Mayer–Vietoris theorem gives a natural suspension isomorphismNaturality means that the suspension of a continuous map induces the same integer multiplier. This proves degree under suspension. Using the standard identification $S(S^k)\cong S^{k+1}$, setThus every integer occurs in every positive sphere dimension, with negative degrees handled just as well as positive ones.

If “every $n$” includes zero, that final printed assertion is false. The generator of ${\tilde{H}}_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving preciselyThese degrees of maps of the zero-sphere are the explicit exception to the arbitrary-degree construction.

Use integral homology throughout. The singular chain group $C_k(X)$ is free on the continuous maps from the standard $k$-simplex to $X$. Since the boundary operator preserves chains lying in $A$, these form a chain subcomplex. The relative chain complex and its relative homology areThe short exact sequence of chain complexes gives the long exact sequence in relative homology.

A sufficient hypothesis for the quotient comparison is a nonempty good pair: $A$ is closed and has an open neighbourhood $U$ which admits a deformation retraction onto $A$ while fixing $A$ throughout. A CW pair is another standard sufficient setting. These hypotheses hold for a simple closed curve in the surface here, using its annular collar neighbourhood. The induced quotient map $q:X\to Y=X/A$ collapses $A$ to a point $a$.

Here is the proof of the collapsing a pair theorem. Put $V=U/A\subset Y$. The deformation retraction makes $H_{\ast }(U,A)=0$, so the long exact sequence of the triple gives $H_{\ast }(X,A)\cong H_{\ast }(X,U)$. It also contracts $V$ onto $a$, giving $H_{\ast }(Y,\{a\})\cong H_{\ast }(Y,V)$. Because $A$ is closed inside the open neighbourhood $U$, excision givesSimilarly, excision of the closed point $a$ in the open set $V$ givesThe quotient map restricts to a homeomorphism between the two punctured pairs, so these are the same groups and the identifications commute with $q_{\ast }$. Finally $H_{\ast }(Y,\{a\})\cong {\tilde{H}}_{\ast }(Y)$, including degree zero. ThusThe neighbourhood condition is part of the result; the quotient assertion is not made for arbitrary bad pairs.

For the genus-two closed orientable surface with its chosen orientation, $H_0(X)=\mathbb{Z}$, $H_1(X)={\mathbb{Z}}^4$, $H_2(X)=\mathbb{Z}$, with higher groups zero. The curve has $H_0(A)=H_1(A)=\mathbb{Z}$. Its relevant long exact sequence in relative homology isThe final map is an isomorphism because both spaces are connected. Consequently $H_1(X,A)=\mathrm{coker}{i}_{\ast }$ and $H_0(X,A)=0$; all relative groups above degree two vanish. The quotient topological space is connected, so its unreduced $H_0$ is $\mathbb{Z}$.

In the separating case, the curve is the oriented boundary of one of the two subsurfaces. Its homology class in $X$ is therefore zero, so $i_{\ast }=0$. The exact sequence becomesThe first sequence splits since its quotient is a free abelian group. Hence the homology after collapsing a separating surface curve isGeometrically, collapsing the boundary of each once-punctured torus fills its puncture with a cone on the circle, which is a disk. The quotient is a wedge sum of two tori at the collapsed point. Their two independent fundamental classes explain the extra second-homology generator.

In the nonseparating case, join the two new boundary components of the cut surface by an arc. Upon regluing, this supplies a closed curve meeting $A$ once transversely. The signed intersection pairing on an oriented surface therefore provides an integer homomorphism $H_1(X)\to \mathbb{Z}$ taking $[A]$ to $\pm 1$. Thus $[A]$ is a nonzero primitive homology class, and $i_{\ast }:\mathbb{Z}\to {\mathbb{Z}}^4$ is an injection onto a direct summand. Its kernel is zero and its cokernel is ${\mathbb{Z}}^3$. The same exact sequence gives $H_2(X,A)\cong H_2(X)\cong \mathbb{Z}$. Therefore the homology after collapsing a nonseparating surface curve isAnother description starts from the genus-one surface with two boundary circles. Collapse those two circles separately to obtain a closed torus with two marked points, then identify the two points. Identifying two distinct points in a connected CW complex adds a loop up to homotopy, so this quotient has the homotopy type of $T^2\vee S^1$. The preceding relative homology calculation proves the groups without requiring that homotopy description.

A Serre fibration has the homotopy lifting property for disks: for every $k\ge 0$, a map $D^k\to E$ and a homotopy $D^k\times I\to B$ starting at its composite with $p:E\to B$ admit a compatible lift $D^k\times I\to E$. Equivalently, it has that lifting property for CW complexes. Relative lifting for CW pairs follows by attaching cells.

Let $i:F\hookrightarrow E$ be the inclusion of the chosen fiber. The long exact sequence of homotopy groups of a fibration isIts low-dimensional end isThe maps $i_{\ast }$ and $p_{\ast }$ are induced by inclusion and projection on based maps; at degree zero they send a path component to its containing or image component. The last part is an exact sequence of pointed sets, not in general a sequence of groups.

For $\partial :{\pi }_q(B)\to {\pi }_{q-1}(F)$, represent a class by a based cube $a:I^{q-1}\times I\to B$, constant at $b_0$ on its boundary. Lift it from the constant map $e_0$ on the bottom and side faces, using relative disk lifting. Its top face lands in $F$ and is constant on that face's boundary; its class is $\partial [a]$. The choice of lifts does not change the resulting class. For $q=1$, lift a based loop starting at $e_0$ and take the path component of its endpoint in $F$. This fixes the boundary-map convention and defines every map in the displayed sequence.

For the splitting assertion, first take $n\ge 1$. We use the cohomological Serre spectral sequence with coefficient group $G$:Since $B$ is simply connected, these coefficient systems are constant. The Eilenberg–MacLane space $F=K(G,n)$ has $H^t(F;G)=0$ for $0<t<n$, andFor $n=1$ this uses $H_1(F)=G$, with $G$ abelian; for $n>1$ it follows from the Hurewicz theorem and the universal coefficient theorem for cohomology. Let $u$ correspond to ${\mathrm{id}}_G$, the universal cohomology class of an Eilenberg–MacLane space.

In bidegree $(0,n)$ there are no incoming differentials. For $2\le r\le n$, an outgoing differential lands in a vanishing fiber-cohomology row. The only remaining possible differential isThus $u$ survives. The edge map, which is restriction to $F$, supplies a classBy representability of cohomology by Eilenberg–MacLane spaces, choose a based map $q:E\to K(G,n)$ representing $c$. On $F$, $q$ induces the identity on ${\pi }_n=G$ and is an isomorphism on every homotopy group, since the other positive groups vanish.

Consider $h=(p,q):E\to B\times K(G,n)$. This is a map of fibrations over $B$, whose map on the fiber is the weak equivalence just obtained. The map on the base is the identity. Comparing the two instances of the long exact sequence of homotopy groups of a fibration and applying exactness, or the Five lemma in the group-valued range, shows that $h$ induces all homotopy isomorphisms. The spaces are path-connected because the base and fiber are. Hence the splitting of a simply connected Eilenberg–MacLane fibration givesThe crucial use of $H^{n+1}(B;G)=0$ is to lift the fiber's identity class; existence of a section alone was not assumed to prove a product splitting.

If $K(G,0)$ is allowed, its components have trivial positive homotopy groups. The simply connected base has no monodromy on this discrete set. Obstruction theory on the CW base gives a section in each fiber component, since all positive-dimensional fiber obstructions vanish and the component system is constant. Together these sections give $B\times G\to E$, a weak equivalence on each component. Thus the same conclusion also covers that interpretation.

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.

Let $x\in H^1({\mathbb{RP}}^{10};{\mathbb{F}}_2)$ be its usual generator. The quotient classes in degrees $7$ through $10$ identify with $x^7,\dots ,x^{10}$ via the pair's cohomology. The Steenrod squares on real projective space satisfyThussince $\left(\genfrac{}{}{0ex}{}{7}{2}\right)$ and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)$ are odd, while $\left(\genfrac{}{}{0ex}{}{8}{2}\right)$ is even. In degree nine, the nonzero cohomology map detects the map on $H_9$, as $H_8=0$ on both sides. Consequently $f_{\ast }:H_9(Y;\mathbb{Z})\to H_9(K;\mathbb{Z})$ is an isomorphism.

Treat $f$ as a mapping-cylinder pair $(K,Y)$. It is $8$-connected: both spaces are $6$-connected, ${\pi }_7$ is an isomorphism, and ${\pi }_8(K)=0$. The relative homology sequence gives $H_9(K,Y)=0$, since $H_9(Y)\to H_9(K)$ is an isomorphism and $H_8(Y)=0$. The Relative Hurewicz theorem then gives ${\pi }_9(K,Y)=0$, and the relative homotopy sequence identifies this group with ${\pi }_8(Y)$. Thus ${\pi }_8(Y)=0$ and the pair is now $9$-connected.

Next,because $H_{10}(Y)=0$ and the degree-nine map is an isomorphism. Apply the Relative Hurewicz theorem again and use ${\pi }_{10}(K)={\pi }_9(K)=0$:Therefore the homotopy groups of the stunted projective space through degree nine are

The degree of a map between oriented manifolds measures how many times the domain covers the target, with signs recording the local local orientation of a manifold. Let $M,N$ be connected, oriented closed manifolds of the same dimension $n>0$. A continuous map $f:M\to N$ acts on top-dimensional homology bywhere $[M],[N]$ are their fundamental classes. Connectedness and the choices of local orientation of a manifold identify $H_n(N;\mathbb{Z})$ with $\mathbb{Z}$. Reversing the orientation of either manifold changes the sign; reversing both does not.

For a smooth map, Sard theorem supplies a regular value $y$. Its inverse image is discrete and, by compactness, finite. At each $x\in f^{-1}(y)$ the differential is an isomorphism; let its sign be $+1$ or $-1$ according to whether it preserves or reverses the chosen local orientations. The degree as a sum of local degrees isThe sign is computed in oriented manifold charts. The value is independent of the chosen regular value, even when inverse images appear or disappear: the signed count is the coefficient of $[N]$ in $f_{\ast }[M]$.

There is a useful local-density expression for the same topological degree. If $\omega$ is a volume form with ${\int }_N\omega =1$, thenThis degree by integration of a pullback volume form follows first by choosing a smooth top-form supported in a small neighborhood of a regular value, where the inverse branches contribute their orientation signs. Any other normalized top-form differs from it by an exact form: integration identifies $H_{\mathrm{dR}}^n(N)$ with $\mathbb{R}$. The integral of its pullback difference vanishes by Stokes theorem. In particular, for any top-form $\eta$, ${\int }_M{f}^{\ast }\eta =\mathrm{deg}(f){\int }_N\eta$.

A homotopy $H:M\times [0,1]\to N$ preserves this integral, since $d\omega =0$ and Stokes theorem givesThus topological degree is a homotopy invariant. It is multiplicative under composition, because the induced maps on homology compose: $\mathrm{deg}(g\circ f)=\mathrm{deg}(g)\mathrm{deg}(f)$. The identity has degree one, a constant map has degree zero for $n>0$, and an orientation-reversing diffeomorphism has degree minus one. An orientation-preserving finite covering map has degree equal to its number of sheets. Nonzero topological degree forces surjectivity, since an omitted point would be a regular value with an empty inverse image.

For the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, positive or negative. This is its winding number, computable as $(2\pi i{)}^{-1}\int f^{-1}df$. The antipodal map of $S^n$ has degree $(-1{)}^{n+1}$: its extension $-I$ on the ambient $(n+1)$-dimensional vector space has that determinant sign and respects the outward-normal convention. A holomorphic map $z\mapsto z^k$, $k\ge 1$, on the Riemann sphere has degree $k$, whereas its complex conjugate has degree $-k$. These examples show how orientation, rather than simply the number of inverse images, determines the integer.

For maps $S^n\to S^n$, topological degree gives the complete homotopy classification ${\pi }_n(S^n)\cong \mathbb{Z}$. The degree does not classify general manifold maps: the identity of the torus and the map induced by the integer matrix $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ both have degree one, but have different induced maps on $H_1(T^2;\mathbb{Z})$ and so are not homotopic. A nonzero-degree map $S^n\to S^n$ cannot extend continuously to $B^{n+1}$, because such an extension would make the boundary map null-homotopic. In the smooth setting, Stokes theorem gives the same obstruction by applying it to the pulled-back normalized volume form.

The hypotheses can be adjusted, but must be stated. For connected oriented noncompact manifolds, a proper map has a degree defined using compactly supported top-forms, and it is invariant under proper homotopies. For manifolds with boundary one uses relative fundamental classes and maps of pairs, or fixes appropriate boundary conditions. Without an integral orientation one can still count inverse images modulo two, obtaining a mod-two degree. The integer integral formula used below assumes the oriented setting.

In classical field theory, these ideas turn continuous fields into quantized topological charges. Suppose a field on ${\mathbb{R}}^d$ approaches one fixed target value at spatial infinity. The one-point compactification makes it a map $\varphi :S^d\to \mathcal{V}$. When the target $\mathcal{V}$ is an oriented closed $d$-manifold, its topological degree labels topological sectors. More generally the sectors are described by homotopy groups; an integer degree is available only when the domain and target have the appropriate dimensions and orientations. Smooth time evolution preserving the boundary condition is a homotopy, so it cannot change the integer. A change requires a singular field, escape from the allowed target, or a change at the boundary.

A normalized closed target $d$-form gives the pullback-volume representation of a topological current. On spacetime, put $\alpha ={\varphi }^{\ast }\omega$. Since $d\alpha ={\varphi }^{\ast }(d\omega )=0$, its dual current is identically conserved, andis independent of time when there is no flux at infinity. This conservation law follows from geometry without using the field equations; it need not arise from a continuous symmetry through Noether theorem.

A concrete example is the O3 nonlinear sigma model in two spatial dimensions. Its unit-vector field $\mathbf{n}$ approaches a constant at infinity, defining $S^2\to S^2$. The normalized area form of the target gives the degree charge of an O3 sigma-model lump:For the energy normalization $E=\frac{1}{2}\int (|{\partial }_1\mathbf{n}{|}^2+|{\partial }_2\mathbf{n}{|}^2)$, the identities $\mathbf{n}\cdot {\partial }_i\mathbf{n}=0$ giveThis is the Bogomolny degree bound for the O3 sigma model. Choosing the sign appropriate to $Q$ makes the square nonnegative; vanishing of the square gives first-order Bogomolny equations and a sigma-model lump saturating the bound. With the oriented stereographic projectionthe maps $w=z^k$ have $Q=k$ and $E=4\pi k$. Their conjugates have $Q=-k$ with the same energy. Holomorphic rational maps have positive degree equal to their degree as rational maps; taking a reciprocal does not reverse the orientation. Antiholomorphic dependence reverses it.

The Skyrme model supplies a three-dimensional example. A field $U:{\mathbb{R}}^3\to \mathrm{SU}(2)$ with $U\to I$ at infinity is a map $S^3\to \mathrm{SU}(2)\cong S^3$. Take $T_i=-i{\tau }_i$ and $U^{-1}dU={\theta }^i{T}_i$, with ${\theta }^1\wedge {\theta }^2\wedge {\theta }^3$ positive. Since $\mathrm{tr}(T_i{T}_j{T}_k)=-2{ϵ}_{ijk}$, the normalized target volume form isThe integral is one on the unit three-sphere. Consequently the Skyrme baryon number as a mapping degree isThis is the topological baryon number in the Skyrme model; the sign has been fixed by the stated orientation and anti-Hermitian generator convention.

A topological charge alone does not guarantee a stable finite-size solution. The degree and energetic stability of a field configuration concern different properties. For a three-dimensional configuration of size $R$, the two-derivative energy scales as $R$, so it can decrease by shrinking while the topological degree remains fixed for every $R>0$. The limit can be singular. The Skyrme term, with four derivatives, scales as $R^{-1}$ and can balance the shrinking tendency. This is the role of Derrick theorem in distinguishing topological obstruction from energetic stability.

For defects, the relevant boundary map can instead be the sphere surrounding a core. A vacuum manifold equal to $S^1$ gives the integer winding number of a vortex; a vacuum manifold $S^2$ gives the degree of a surrounding $S^2$ for a magnetic monopole. This vacuum-boundary degree as a defect charge obstructs extending the normalized vacuum field through the enclosed ball. A nonzero integer therefore forces the field to leave the vacuum manifold somewhere in the core. This construction does not require the field to take one constant value in every direction at infinity.

Degree also appears in four-dimensional gauge theory through a boundary transition function. For an anti-Hermitian SU(2) gauge connection on ${\mathbb{R}}^4$, write $F=dA+A\wedge A$ and assume finite-action boundary behavior $A\to g^{-1}dg$ on the large bounding three-sphere. In the second-Chern conventionthe identity $d\mathrm{tr}(A\wedge dA+\frac{2}{3}{A}^3)=\mathrm{tr}(F\wedge F)$ and the Maurer-Cartan equation giveThis boundary winding representation of Yang-Mills topological charge relates the Second Chern number to the degree of $g:S^3\to \mathrm{SU}(2)$. The Chern-Simons 3-form turns the bulk integral into the boundary winding integral. Conventions which define the instanton number with the opposite trace sign reverse $k$; the integer quantization is unchanged. A Yang-Mills theta term weights a sector by $e^{i\vartheta k}$, giving periodicity $\vartheta \mapsto \vartheta +2\pi$. Thus the same topological degree that counts oriented inverse images also labels field sectors and expresses their quantized charges as integrals of local densities.