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Orient $S^3$ as the boundary of $B^4$. Push the interior of an oriented Seifert surface $F$ slightly into $B^4$, and let $X_F$ be the double branched covering of $B^4$ over this pushed-in surface. Its boundary is the two-fold branched cover of a knot. Define the knot signature byThe signature of a possibly degenerate intersection form means the number of positive eigenvalues minus the number of negative eigenvalues; its radical contributes zero.

Here is why the definition is independent of $F$. Given two choices $F_0,F_1$, glue their branched covers along the common boundary to obtainThis is the double branched covering of $S^4$ over the closed oriented surface obtained by gluing the pushed-in $F_0$ to $-F_1$. Every closed oriented surface in $S^4$ admits a Seifert hypersurface: its complement has an integral meridional class, represent it by a map to the circle, and take a regular value. Near the surface choose the angular map in its trivial normal disk bundle. The closure of a suitable regular fiber is a compact oriented three-manifold bounding the surface. Its trivial normal framing can be chosen to match this construction.

Push the interior of this Seifert hypersurface into $B^5$. The double branched covering of $B^5$ over the resulting properly embedded three-manifold has boundary $X$. The existence of the cover follows from the meridian homomorphism to $\mathbb{Z}/2$; smoothness near the branch set is the local map $(z,u)\mapsto (z^2,u)$. By the hypothesis about boundaries of five-manifolds, $\mathrm{sign}X=0$. Novikov additivity for gluing along an entire closed three-dimensional boundary givesThus the knot signature is independent of the Seifert surface. An isotopy of the knot carries the construction to an equivalent one, so it is also a knot invariant.

To calculate its intersection form, express $F$ as a disk with $2g$ bands. The double branched covering of $B^4$ over the pushed-in disk is again $B^4$. Each band lifts to a two-handle. The capped lifted cores give a basis of $H_2(X_F;\mathbb{Z})$ corresponding to the band-core basis of $H_1(F;\mathbb{Z})$. The framing of the $i$th lifted attaching circle is $2\mathrm{lk}(a_i^{+},a_i)=2V_{ii}$. For distinct bands, the two sheets contribute the two push-off linkings $V_{ij}$ and $V_{ji}$, so the mutual linking number is $V_{ij}+V_{ji}$. HenceA different integral basis changes this matrix by matrix congruence and leaves its signature unchanged.

In Figure 7 the upper knot is a connected sum of knots $T\#T$, while the lower one is $T\#\overline{T}$, where $T$ is the positive trefoil knot. Both knot groups have the same presentationTo see this, the Seifert-van Kampen theorem expresses the knot group of a connected sum of knots as the amalgamated free product of the two summand knot groups, amalgamating their meridians of a knot. A trefoil knot has presentation $⟨a,b\mid aba=bab⟩$ with $a$ a meridian. Reflection gives the same abstract group for the mirror. If it reverses the chosen meridian, send both generators to their inverses; the inverse braid relation is again the same relation. Thus the meridian amalgamation yields the displayed group presentation in both cases.

A band basis for the positive trefoil knot giveswhose symmetrization is negative definite. Therefore $\sigma (T)=-2$ and $\sigma (\overline{T})=2$. Boundary-connected-summing Seifert surfaces makes the Seifert matrix block diagonal, so the knot signature is additive under connected sum of knots. ConsequentlyThe upper knot and lower knot are therefore not isotopic, even though their knot groups are isomorphic. With the opposite global knot signature convention the first value is $+4$, and the distinction is unchanged.

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

A useful closed-cover version of the Seifert-van Kampen theorem is the following. Let $X=A\cup B$ with $A,B,A\cap B$ path-connected closed subspaces and basepoint in their intersection. Assume there are open neighbourhoods $U,V$ covering $X$ that deformation retract onto $A,B$, while $U\cap V$ deformation retracts onto $A\cap B$, with the induced inclusions compatible up to based homotopy. ThenThe neighbourhood hypothesis, available for the present finite subcomplex gluing, matters: arbitrary closed covers need not satisfy this theorem.

A Möbius band retracts onto its core circle, with generator $m$; its boundary represents $m^2$. The torus has generators $u,v$ and relation $[u,v]=1$. The attaching circle is the $u$ circle. Reversing the attaching homeomorphism merely replaces $u$ by its inverse and changes no isomorphism type. The closed-cover theorem givesMapping $m\mapsto 0$, $v\mapsto 1$ surjects onto $\mathbb{Z}$, so the group is infinite. Imposing the extra relation $m^2=1$ gives the quotient $C_2\ast \mathbb{Z}$, a nonabelian free product, since alternating reduced words $mv$ and $vm$ differ. A quotient of an abelian group would be abelian, so the fundamental group is nonabelian.

The Seifert-van Kampen theorem states: if $X=U\cup V$, where $U,V$ are open and path connected, $U\cap V$ is path connected and contains a basepoint $x$, then ${\pi }_1(X,x)$ is the pushout of groups of ${\pi }_1(U,x)$ and ${\pi }_1(V,x)$ over ${\pi }_1(U\cap V,x)$. Thus the fundamental group is obtained by amalgamating the overlap relations. Openness can be replaced by standard neighborhood hypotheses, as in the wedge calculation.

The real projective plane has fundamental group $C_2=⟨a\mid a^2=1⟩$, while the circle has fundamental group $\mathbb{Z}=⟨b⟩$. Apply the Seifert-van Kampen theorem to open neighborhoods of the two wedge summands whose overlap is contractible. This givesChanging the basepoint within this path-connected space gives an isomorphic group.

Let $X=Y\cup Z$, where $Y$, $Z$, and $Y\cap Z$ are path-connected subcomplexes containing the basepoint; small open neighbourhoods may be used to reduce to the usual open-cover statement. The Seifert-van Kampen theorem says thatMore explicitly, this is the quotient of the free product ${\pi }_1(Y)\ast {\pi }_1(Z)$ by the normal closure ofEquivalently, ${\pi }_1(X)$ is the pushout of the two inclusion-induced homomorphisms.

The wedge intersection is simply connected after taking suitable neighbourhoods, so repeated application of the Seifert-van Kampen theorem gives the free product of $n$ copies of $\mathbb{Z}$ withThus, for any basepoint,This is the fundamental group of a wedge of circles and a real projective plane.

The symplectic neighborhood theorem says that a symplectomorphism between closed symplectic submanifolds which lifts to an isomorphism of their symplectic normal bundles extends to a symplectomorphism of neighborhoods.

Let $C_X\subset X$ and $C_Y\subset Y$ be the given copies of $C$. Their self-intersection numbers are the Euler classes of their oriented normal bundles, so square zero makes both normal bundles trivial. The neighborhood theorem identifies neighborhoods with $C\times D^2$. Remove their interiors and identify the boundary circle bundles by a map covering the chosen identification $C_X\cong C_Y$ and reversing the normal-circle orientation. On collars, the two forms have the modeland the radial coordinate can be reversed while the circle coordinate is reversed so that the forms glue. A collar application of Moser's trick removes any discrepancy. This proves that the symplectic fiber sum along a square-zero surface $X{\#}_{C}Y$ has a natural symplectic form.

The displayed relation is an ordinary product relation, sowhere the relation eliminates $b=ac{a}^{-1}$. The Gompf realization theorem constructs a closed symplectic four-manifold $M_{\Gamma }$ with this fundamental group. Concretely, its construction starts from a product of a sufficiently high-genus surface and a torus, represents the two surviving generators and the relations by loops, and crosses the relevant loops with circle factors to obtain square-zero tori. Symplectic sums with copies of the rational elliptic surface kill the unwanted generators and impose the relations: the complement of a regular elliptic fiber is simply connected, so the Seifert-van Kampen theorem gives exactly ${\pi }_1(M_{\Gamma })=\Gamma \cong F_2$.

Finally, ${\mathbb{CP}}^{48}$ is simply connected and has real dimension $96$. With the product symplectic form,is a closed symplectic manifold of real dimension $100$ and has fundamental group $\Gamma$.

Writing the square coordinates modulo one, the curve has the parametrizationIt is therefore the $(2,3)$ torus knot, or trefoil, on the common boundary of the two solid tori. Applying the Seifert-van Kampen theorem to the complements on the two sides gives one generator $a$ for the meridian of one solid torus and one generator $b$ for the other; the boundary slope identifies two turns of the first with three turns of the second. HenceThis group is nonabelian: imposing the additional relation $a^2=b^3=1$ gives a quotientwhich maps onto the nonabelian group $S_3$ by sending $a$ to a transposition and $b$ to a three-cycle. An abelian group cannot have a nonabelian quotient.

Cover the attached disc by a slightly enlarged interior and a collar of its boundary, and apply the Seifert-van Kampen theorem. The collar intersection has fundamental group generated by the boundary loop, whose image in $X$ is $[\alpha ]$, while that loop becomes trivial in the disc. Van Kampen therefore quotients by the smallest normal subgroup containing $[\alpha ]$:This is the general fundamental group after attaching a 2-cell.

Start with a wedge of two circles, whose fundamental group is the free groupAttach a disc along the loop represented by $a^3{b}^{-7}$. The displayed quotient gives a connected two-dimensional CW complex $X$ with

The Seifert-van Kampen theorem says that if a path-connected space is the union of path-connected open sets $U,V$ with path-connected intersection, thenUsing the usual polygonal cell decompositions, the torus attaches a two-cell to $S^1\vee S^1$ along $ab{a}^{-1}{b}^{-1}$, while the projective plane attaches one along $a^2$. Hence

Cut the square model of the Klein bottle along the midline parallel to the pair of sides whose identification reverses orientation. Each half becomes a Möbius band, and the two new boundary circles are identified. If $a$ and $b$ are the core loops of these two bands, each boundary travels twice around its core. The Seifert-van Kampen theorem therefore gives

Writing $x=a^{-1}b$ and $y=a$ converts this into the standard Klein-bottle presentationThe orientation homomorphism sends both $a$ and $b$ to the nonzero element of $C_2$. Its kernel is the index-two subgroupBy the classification of connected covering spaces, $H$ determines the orientation double cover of the Klein bottleChoosing torus generators $\alpha ,\beta$ along the two translation directions, we may takeThese commute: the standard relation says that $a$ conjugates $a^{-1}b=x$ to $x^{-1}$, so $a^2$ centralizes $x$.

The space $Y$ is the mapping cylinder of $p$ with its unused end omitted. Sliding $(x,t)$ to $(x,0)$ gives a deformation retraction of $Y$ onto the bottom quotient $T^2/\sim \cong K$. Hence

Finally suppose $U\subseteq X$ is open and homeomorphic to $Y$, and identify the bottom copy of $K$ inside $U$. This $K$ is compact and therefore closed in the Hausdorff space $X$. Apply van Kampen, in its groupoid form if intersections are disconnected, toThe intersection $U\setminus K$ deformation retracts to a positive-height torus, and its map to ${\pi }_1(U)\cong {\pi }_1(K)$ has image preciselyThe orientation homomorphismis zero on this intersection, so it is compatible with the trivial homomorphism from ${\pi }_1(X\setminus K)$ to $C_2$. The pushout property in van Kampen extends it to a surjectionThereforeThis is the Klein-bottle mapping-cylinder obstruction to simple connectivity.

Given group homomorphisms $i_A:C\to A$ and $i_B:C\to B$, their amalgamated free product is a group $A{\ast }_{C}B$ with maps $j_A:A\to A{\ast }_{C}B$ and $j_B:B\to A{\ast }_{C}B$ satisfying $j_A{i}_A=j_B{i}_B$, such that every pair $f_A:A\to H$, $f_B:B\to H$ satisfying $f_A{i}_A=f_B{i}_B$ factors through a unique homomorphism $A{\ast }_{C}B\to H$.

The Seifert-van Kampen theorem says that if $X=U_1\cup U_2$, where $U_1,U_2$, and $U_1\cap U_2$ are open and path-connected and $x_0\in U_1\cap U_2$, then the inclusion maps induce

Here is a direct proof of the requested generation statement. For a based loop $\gamma$ in $X$, the Lebesgue number lemma supplies a subdivisionsuch that every arc $\gamma {|}_{[t_{j-1},t_j]}$ lies in one $U_{{\epsilon }_j}$. Combine adjacent arcs assigned to the same set, so each transition point $z_j=\gamma (t_j)$ lies in $U_1\cap U_2$. Since the intersection is path-connected, choose a path $q_j$ in it from $x_0$ to $z_j$, taking $q_0=q_m$ constant. Thenis a loop in $U_{{\epsilon }_j}$. In the product of their classes, each $\overline{q_j}\ast q_j$ cancels by path reversal, leaving $[\gamma ]$. Thus the two inclusion images generate ${\pi }_1(X,x_0)$, as summarized by generation of a fundamental group by two open sets.

Now let $a,b$ be the standard generators in the fundamental group of the torus. A Möbius band retracts to a core circle, while its boundary traverses that core twice. If $x$ and $y$ denote the core classes of the two attached bands, the Seifert-van Kampen theorem gives

The two attaching classes $A=ab$ and $B=a^2{b}^3$ commute. Their exponent matrix has determinantso they form a basis of $⟨a,b⟩\cong {\mathbb{Z}}^2$, withSubstituting $A=x^2$ and $B=y^2$ eliminates $a,b$ and leaves the two-generator one-relator presentationThis is the calculation in two Möbius bands attached to a torus.

Finally send $x$ to $(12)$ and $y$ to $(23)$ in the symmetric group $S_3$. Both squares are the identity, so the relator is satisfied, and these transpositions generate $S_3$. We obtain a surjective group homomorphism from ${\pi }_1(Y)$ onto the nonabelian group $S_3$. Therefore ${\pi }_1(Y)$ is nonabelian.

Write a point of the band $X$ as $(u,t)$ with $u\in S^2$, $|t|\le 1/2$, after rescaling the radius of the first three coordinates. The antipodal identification isThe homotopy $(u,t)\mapsto (u,(1-s)t)$ is equivariant under this identification, so it descends to a deformation retraction of $Y$ onto the central sliceThusThe two boundary spheres of $X$ are exchanged by the antipodal map, so $\partial Y\cong S^2$. This is the mapping-cylinder model of punctured real projective three-space.

The standard cellular chain complex for ${\mathbb{RP}}^3$, with one cell in each dimension zero through three, has boundary maps alternating between multiplication by two and zero:Therefore the integral homology of real projective three-space isIn particular,

Apply the Mayer–Vietoris theorem toThe relevant pieces of the long exact sequence areandThe last map is injective, so

Since $\partial Y\cong S^2$ is simply connected and ${\pi }_1(Y)\cong {\pi }_1({\mathbb{RP}}^2)\cong \mathbb{Z}/2$, the Seifert-van Kampen theorem givesthe infinite dihedral group. Topologically, $Z$ is the double of punctured real projective three-space, namely ${\mathbb{RP}}^3\#{\mathbb{RP}}^3$.

The universal cover of each copy of $Y$ is $S^3$ with two disjoint open balls removed, homeomorphic to $S^2\times [0,1]$. The universal cover of the double strings infinitely many such cylinders together according to the Cayley line of $D_{\infty }$. Hence the familiar covering space is

The Seifert-van Kampen theorem says that if $X=U\cup V$, where $U,V,U\cap V$ are path connected open sets containing $x_0$, then

The cell attachment is the quotientwith the image of $x_0=f(\ast )$ as base point. For $n=2$, take one open set that deformation retracts onto $X$ together with a boundary collar, and another that consists of the interior of the disc with a collar and is contractible. Their intersection deformation retracts onto $S^1$. Its generator maps to $[f]$ in the first set and to the identity in the second. Van Kampen therefore proves the fundamental group after attaching a 2-cell formula

Applying van Kampen to the two circles givesAttach three discs along loops representing $a^2$, $b^3$, and $(ab{)}^2$. The resulting presentation complex for the symmetric group on three letters has groupSending $a$ to $(12)$ and $b$ to $(123)$ gives a surjection onto $S_3$. The relations imply $aba=b^{-1}$, so every word reduces to one ofThe presented group has at most six elements and surjects onto the six-element group $S_3$, so this map is an isomorphism.

One form of the Seifert-van Kampen theorem says that if $X=U\cup V$, with $U,V,U\cap V$ open and path-connected and containing the base point, then ${\pi }_1(X)$ is the pushout of

For the fundamental group after attaching a 2-cell, take $U$ to be $X$ together with a collar of the attached disc and $V$ to be a slightly enlarged open disc. Then $U\simeq X$, $V$ is simply connected, and $U\cap V\simeq S^1$; its generator maps to $[\alpha ]$ in ${\pi }_1(X)$. Van Kampen therefore gives

Take $X=S_a^1\vee S_b^1$, whose fundamental group is the free group $F(a,b)$, and attach one 2-cell along the loop represented by $a^2{b}^{-3}$. The resulting connected presentation complex $Y$ satisfies