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Algebraic topology translates questions about topological spaces and continuous maps into questions about groups and other structures from algebra. A topological invariant compresses the enormous amount of data in a space into something one can calculate. For example, the fundamental group records which based loops can be continuously deformed into one another. A filled disk has trivial fundamental group, whereas a circle has fundamental group $\mathbb{Z}$: an integer remembers how many times a loop goes around.

This compression also respects maps. By functoriality of the fundamental group, a continuous map preserving the chosen basepoints induces a group homomorphism; successive maps induce successive group homomorphisms. Thus geometry supplies arithmetic constraints. A proposed homotopy equivalence must induce an isomorphism, so incompatible fundamental groups rule it out before anyone attempts to construct the deformation.

The beauty is that a whole annulus can retain exactly the same homotopy type as its inner circle. Its thickness disappears under a deformation retraction, while the obstruction to shrinking a loop survives. Equal fundamental groups alone do not establish a homeomorphism or a homotopy equivalence between arbitrary topological spaces.

This turns a visual question into a reusable test. Suppose a continuous map sends the inner circle of an annulus once around another circle. Could it extend over a filled disk spanning that boundary? Such an extension would supply a homotopy shrinking the image loop to a point. Its winding number would have to change from one to zero, which homotopy invariance of winding number forbids. Algebra rules out the extension even when no particular formula for it has been proposed.

In the quotient space description of the Möbius band, follow $v=+1$ once along the rectangle. The seam sends its endpoint to the start of $v=-1$, which must also be traversed before returning to the original point. Under the deformation retraction onto the core circle of a Möbius band, these two stretches each travel once around the same circle. Orienting both circles by increasing longitudinal parameter gives a map of topological degree two. Equivalently, the group homomorphism on fundamental groups induced by inclusion of the boundary sends a chosen generator to twice the core generator, or to $c^2$ in multiplicative notation. Reversing the chosen boundary generator changes the sign, not the fact that it winds twice.

The core circle is the image of the centre line in the rectangle defining a Möbius band. Its endpoints are identified, so it is a circle. The formuladefines a deformation retraction onto that circle: it respects $(0,v)\sim (1,-v)$, starts at the identity map, and fixes the core throughout. Consequently the Möbius band has fundamental group isomorphic to $\mathbb{Z}$, just like a circle; this alone cannot distinguish it from an annulus.

A deformation retraction of $X$ onto a subspace $A$ is a homotopy $H:X\times [0,1]\to X$ such that $H(x,0)=x$, $H(x,1)=r(x)\in A$, and $H(a,t)=a$ for every $a\in A$ and every $t$. Thus the retract is fixed pointwise throughout. With inclusion $i:A\hookrightarrow X$, the equations $r\circ i={\mathrm{id}}_A$ and $i\circ r\simeq {\mathrm{id}}_X$ exhibit a homotopy equivalence. See Hatcher, Algebraic Topology, Chapter 0, pp. 2–3.

Take the closed annulusAn exact deformation retraction onto its inner circle isThe denominator is nonzero everywhere on the annulus. If the starting radius is $\rho =\Vert x\Vert$, its radius at time $t$ is $(1-t)\rho +t$. This lies between $1$ and $\rho \le 2$, verifying that the entire homotopy stays in the original annulus. When $\rho =1$, the point is stationary, as required.

For $x=(0,2)$, times $0,1/2,1$ give $(0,2),(0,3/2),(0,1)$. A radius-$3/2$ loop similarly becomes a radius-$5/4$ loop halfway through, then the unit circle. Its angle and winding number are preserved. This is useful because computing the fundamental group of the annulus reduces to computing that of the circle; the explicit homotopy equivalence establishes why the answer transfers.

Build a closed annulus from two nested $n$-sided polygons, $n\ge 3$, with $n$ radial edges joining corresponding vertices. There are $2n$ vertices, $3n$ edges and $n$ quadrilateral topological cells, giving $\chi =2n-3n+n=0$. The central missing region and the exterior are not faces of the annulus. A deformation retraction onto a circle gives $(b_0,b_1,b_2)=(1,1,0)$, agreeing with the cell count. This is a useful reduction: the annulus is two-dimensional, yet its homology is that of a one-dimensional cycle graph.

The fundamental group ${\pi }_1(X,x_0)$ consists of based loops modulo based homotopy. Multiplication follows one based loop and then the other; the constant based loop represents the identity, and path reversal gives the inverse. Passing to homotopy classes makes this operation associative, as explained by well-defined loop concatenation.

Identify the circle with $S^1=\{z\in \mathbb{C}:|z|=1\}$ and choose basepoint $1$. Its based loopshave winding number $n$: positive $n$ counts counterclockwise turns, negative $n$ counts clockwise turns, and $n=0$ gives the constant based loop. Every based loop on the circle is joined by a based homotopy to exactly one ${\gamma }_n$. ConsequentlyThis is Hatcher's Theorem 1.7 in Algebraic Topology, §1.1, pp. 29–30. For example, two counterclockwise turns followed by one clockwise turn represent $2+(-1)=1$, rather than three turns. The winding graph in the figure above makes the signed arithmetic visible.

A proof outline uses the covering map $q:\mathbb{R}\to S^1$, $q(u)=e^{2\pi iu}$. By the path lifting theorem, a based loop has a unique lift starting at zero and ending at an integer $n$. Interpolating its lift to $s\mapsto ns$ produces a based homotopy to ${\gamma }_n$. The homotopy lifting theorem for a covering map keeps that endpoint unchanged during a based homotopy, proving uniqueness.

In the filled disk, every based loop $\gamma$ at $x_0$ has a based homotopy to the constant based loop:Both endpoints remain $x_0$, and the convex set property of the disk keeps every intermediate point inside. Thus ${\pi }_1(D^2,x_0)$ is the trivial group. Together with the deformation retraction above, this computes three concrete answers: the filled disk has trivial fundamental group, while the circle and the annulus have infinite cyclic groups.

Maps have computable algebraic shadows too. The continuous map $p_n:S^1\to S^1$, $p_n(z)=z^n$, fixes $1$ and sends ${\gamma }_k$ to ${\gamma }_{nk}$. Its induced group homomorphism is therefore $k\mapsto nk$, and its mapping degree is $n$. For $n=2$ the image consists of even integers, so this map cannot be a homotopy equivalence. Composing $p_2$ with $p_{-3}$ gives $p_{-6}$: map composition multiplies degrees, whereas concatenating based loops adds winding numbers. Distinct degrees distinguish these maps up to homotopy; this calculation does not classify arbitrary topological spaces.

Historically, Poincaré's 1895 Analysis Situs introduced the fundamental group; its original §12 explicitly names it. The CNRS transcription of Poincaré's §12, Groupe fondamental provides a direct view of this early formulation.

For $C_2\ast \mathbb{Z}\to S_3$ taking the two generators to a transposition and a $3$-cycle, the kernel's six-sheeted regular covering space consists of three spheres and two circles. Each sphere has two marked points, and each circle meets one point on each sphere. It retracts onto a rank-four graph by replacing each sphere by a meridian arc, but admits no deformation retraction onto any graph because its degree-two homology is ${\mathbb{Z}}^3$.

The Möbius band is the topological surface obtained by gluing the ends of a rectangle with a half-turn. Precisely, it is the quotient spaceThe twist identifies the two apparent long edges into one boundary circle. Following the core circle of a Möbius band reverses local handedness, so this surface is not orientable. An untwisted untwisted band has two boundary circles and is orientable: the distinction survives stretching and bending because orientability is a topological invariant.

To explore the construction with paper, mark the two long edges in different colours, join the short ends after a half-turn, and trace an edge. The trace passes from one marked edge to the other without ever leaving the boundary. Separately, trace the centre line and track a local transverse arrow: it comes back reversed after one circuit. The boundary closes after twice the travel around the centre, while the transverse arrow needs two circuits to return to its original direction.

This original mathematical model has central radius $R=2$ and half-width $a=0.55$:Its seam obeys $F(2\pi ,w)=F(0,-w)$. The dotted circle is the core circle of a Möbius band; the teal curve follows the single boundary through $0\le \theta \le 4\pi$. At the same point on the core, the arrows for $\theta =0$ and $\theta =2\pi$ are opposite. The graph compares a continuously followed unit normal with its starting value in the ambient space; the endpoint reversal, rather than any intermediate zero of the graph, establishes the obstruction. It is not a statement about parallel transport.

The Möbius band has the homotopy type of a circle: a deformation retraction shrinks each width coordinate to zero. Nevertheless its boundary represents twice a generator of its fundamental group. The boundary of a Möbius band winds twice around its core, so with compatible generator choices the boundary generator maps to $c^2$ in ${\pi }_1(M)\cong \mathbb{Z}$.

The historical account by Julyan H. E. Cartwright and Diego L. González places the mathematical discovery by August Möbius and Johann Listing in 1858, while investigating representations of twisted bands predating that work. A familiar-looking picture and a recognition of the underlying topological invariant are different kinds of evidence. ^[ref]

The total space of the orientation line bundle is orientable even when its base is not: an orientation reversal in a base coordinate transition is matched by a reversal in the line coordinate, so their product has positive sign. Fiberwise multiplication by $1-t$ gives a deformation retraction onto the zero section. For the real projective plane this total space is $(S^2\times \mathbb{R})/((v,t)\sim (-v,-t))$, an oriented noncompact three-manifold homotopy equivalent to the plane.

A rank-$k$ complex vector bundle is a smooth map $\pi :E\to B$ whose fibers are $k$-dimensional complex vector spaces, with local diffeomorphisms ${\Phi }_i:{\pi }^{-1}(U_i)\to U_i\times {\mathbb{C}}^k$ commuting with projection and linear on every fiber. Define transition functions of a vector bundle byThey are smooth and obeyon the appropriate double/triple overlaps. The structure group of a vector bundle is a Lie subgroup $G\subset GL(k,\mathbb{C})$ in which a chosen set of transitions takes values. It records allowed changes of fiber frame; different trivializations may exhibit different reductions of that group.

Using the same cocycle, glue the disjoint union of $U_i\times G$ byThe cocycle makes this an equivalence relation and makes its local charts compatible. Right multiplication $(b,h{)}_i\cdot a=(b,ha{)}_i$ is well-defined because it commutes with the left gluing matrices. It is free and transitive on each fiber. These charts therefore construct the associated right principal bundle $P\to B$ with group $G$.

For the natural line realization, take the complex tautological line bundle $L=\{(\ell ,v):\ell \in {\mathbb{CP}}^1,v\in \ell \subset {\mathbb{C}}^2\}$. Its nonzero vectors identify with ${\mathbb{C}}^2\setminus \{0\}$ by $(\ell ,v)\mapsto v$, with projection $v\mapsto [v]$. On the two standard charts, use $\zeta =z_2/z_1$ and $\eta =z_1/z_2=1/\zeta$. The local frames are $e_0=(1,\zeta )$ and $e_1=(\eta ,1)$, so $e_1=e_0/\zeta$. Normalize them using the ambient Hermitian norm:If $v=t_0{u}_0=t_1{u}_1$, then $t_0=(|\zeta |/\zeta )t_1$. Thus the unitary transitions of the tautological line over the projective line areThese are smooth on the overlap $\zeta \ne 0$. A Hermitian metric on a smooth complex vector bundle generally permits precisely this unitary reduction; here the ambient norm supplies it explicitly.

The corresponding principal circle bundle consists of unit vectors in the tautological line. Sending $(\ell ,v)$ to $v$ identifies its total space with $S^3\subset {\mathbb{C}}^2$, and the right circle action is $v\cdot e^{i\theta }=e^{i\theta }v$. Identify ${\mathbb{CP}}^1$ with $S^2$ usingIts affine-coordinate inverse on the chart $Z\ne -1$ is $\zeta =(X-iY)/(1+Z)$; the other chart handles the missing pole. This gives a diffeomorphism, and the projection becomes the Hopf fibration

There is a distinction if the stated punctured-bundle identification is interpreted only as a projection-preserving diffeomorphism, rather than a fiber-linear identification. It does not determine whether the complex line is tautological or dual: for a nonzero $\varphi \in L_{\ell }^{\ast }$, the unique $v\in \ell$ with $\varphi (v)=1$ gives a smooth map $L^{\ast }\setminus 0\to L\setminus 0$. Locally it is inversion of a nonzero complex number. Thus punctured line bundles do not determine their duality sign.

For completeness, these are the only possible signs here. Complex line bundles over $S^2$ are described by the clutching construction with a loop in ${\mathbb{C}}^{\ast }$, reducible to $S^1$, and its integer winding number determines the bundle: a zero-degree ratio has a logarithm and can be removed by changes of trivializations extending over the two disks. For winding number $d$, the unit circle bundle has fundamental group $\mathbb{Z}/d\mathbb{Z}$, as the two solid-torus trivializations identify the common fiber generator and impose its $d$th power as the equatorial gluing relation. There is a fiberwise radial deformation retraction from the punctured line bundle onto its unit circle bundle. Since ${\mathbb{C}}^2\setminus 0$ retracts to simply connected $S^3$, the hypothesis forces $|d|=1$.

Consequently the allowed unitary transitions are the displayed $g_{01}$ or its reciprocal, according to the bundle's sign. For the dual, the same underlying Hopf projection has the inverse principal action $v\cdot e^{i\theta }=e^{-i\theta }v$. In either case the principal total space and projection are $S^3\to S^2$; the purely diffeomorphic hypothesis alone leaves the circle-action sign unspecified.

Suppose, for a contradiction, that $X=M\times \mathbb{R}$ has a complete metric with zero Ricci curvature. Since $M$ is closed and connected, the complement of $M\times [-A,A]$ has two unbounded components in any complete metric on $X$: their closures are noncompact, while closed bounded sets are compact by the Hopf-Rinow theorem. Thus $X$ is disconnected at infinity, and part (a) supplies a line. The Cheeger-Gromoll splitting theorem gives an isometryThe Riemannian product curvature formula makes $N$ complete and Ricci-flat. A three-dimensional Ricci-flat Riemannian manifold is flat: for any orthonormal triple, the three trace equations are ${\mathrm{Ric}}_{11}=K_{12}+K_{13}$ and their cyclic companions. Solving gives $K_{ij}={\mathrm{Ric}}_{ii}+{\mathrm{Ric}}_{jj}-\mathrm{Scal}/2=0$. Every plane can occur in such a triple, so every sectional curvature vanishes, and hence so does the Riemann curvature tensor.

The complete simply connected universal cover of $N$ is therefore Euclidean ${\mathbb{R}}^3$, and that of $X$ is Euclidean ${\mathbb{R}}^4$. But the topological universal cover of $X$ is $\tilde{M}\times \mathbb{R}$. Its deformation retraction onto $\tilde{M}\times \{0\}$ shows that contractibility of this cover would imply contractibility of $\tilde{M}$, contrary to the hypothesis. Therefore no complete Ricci-flat metric exists on $M\times \mathbb{R}$. Notice that no product assumption was made about the hypothetical original metric.

The Excision theorem for singular homology states that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq {\mathrm{int}}_{X}A$, inclusion induces isomorphisms $H_q(X\setminus Z,A\setminus Z;\mathbb{Z})\cong H_q(X,A;\mathbb{Z})$ for all $q$. For an open cover $X=U\cup V$, its small-chain form says that the subcomplex of singular chains whose individual simplices lie in $U$ or $V$ computes $H_{\ast }(X)$. Concretely, repeated barycentric subdivision puts each finite chain in that subcomplex, and subdivision is chain-homotopic to the identity. There is then a short exact sequence of chain complexesIts long exact sequence in homology, with excision identifying the last complex's homology with $H_{\ast }(X)$, is the Mayer–Vietoris sequence:

The Real projective space ${\mathbb{RP}}^1$ is a circle. The projective plane has one cell in each dimension $0,1,2$, with cellular homology boundary maps $d_2=2$ and $d_1=0$. Thus, writing $A_q=H_q({\mathbb{RP}}^2;\mathbb{Z})$, we have $A_0=\mathbb{Z}$, $A_1=\mathbb{Z}/2$ and $A_q=0$ for $q\ge 2$. Cover the circle by two open arcs with a two-component intersection and multiply this cover by the projective plane. In the Mayer–Vietoris sequence the overlap-to-cover map isIts kernel and cokernel are each $A_q$. Consequently $0\to A_q\to H_q(X)\to A_{q-1}\to 0$ for $q\ge 1$. In degree one the quotient is free, so this sequence splits; in degree two it identifies the group directly with $A_1$. The integral homology of a circle times a real projective plane is therefore

Without a compactness restriction, the answer to the final existence question is yes: there is a noncompact oriented four-manifold. Explicitly, takeThe involution is free and reverses orientation on both $S^2$ and $\mathbb{R}$, hence preserves their product orientation. Its quotient is consequently an oriented three-manifold, the orientable total space of the orientation line bundle of the projective plane. The contraction $[v,t]\mapsto [v,(1-s)t]$ is a deformation retraction onto ${\mathbb{RP}}^2$, so $Y\simeq X$.

If a closed oriented manifold is intended, the answer instead is no. A closed connected oriented $d$-manifold has $H_d\cong \mathbb{Z}$ and no homology above $d$. Here the highest nonzero positive-dimensional homology is the torsion group $H_2=\mathbb{Z}/2$, so no dimension can supply that required top fundamental class. The distinction is necessary because the printed wording does not require the oriented manifold to be closed or three-dimensional.

Choose open neighborhoods of the sphere and circle in the wedge, each enlarged by a small contractible neighborhood of the attaching point in the other piece. They deformation retraction to $S^2$ and $S^1$ and have contractible intersection. Van Kampen therefore givesThe generator traverses the circle once; the sphere contributes no fundamental-group generator.

Write a point of the larger sphere as $(u,v)\in {\mathbb{R}}^{m+1}\times {\mathbb{R}}^{n-m}$. In the complement, $v\ne 0$. There is a homeomorphismwhose inverse sends $(u,w)$ to $(u,\sqrt{1-\Vert u{\Vert }^2}w)$. In particular the complement has a deformation retraction onto its transverse sphere. Explicitly,The denominator stays positive, the homotopy stays in the complement, and points with $u=0$ remain fixed. This establishes the homology of an equatorial sphere complement. By homotopy invariance of homology, its homology is therefore the homology of a sphere of dimension $d=n-m-1$:In the codimension-one case, the complement has two contractible components; this is why degree-zero homology has rank two. Equivalently, in every case its only nonzero reduced homology is ${\tilde{H}}_d(X;\mathbb{Z})=\mathbb{Z}$.

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.

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

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