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A simple closed curve in a closed orientable surface has an annular collar neighbourhood, so its collapse is governed by the collapsing a pair theorem. In the long exact sequence in relative homology, the decisive map is $H_1(A;\mathbb{Z})\to H_1({\Sigma }_g;\mathbb{Z})$, sending the circle generator to its homology class. That class is zero for a separating curve and a primitive homology class for a nonseparating curve. The two possibilities lead to homology after collapsing a separating surface curve and homology after collapsing a nonseparating surface curve.

Represent one-dimensional homology classes on a closed oriented topological surface by transverse oriented curves and sum their signed crossing points. This gives a bilinear form that is skew-symmetric. A symplectic handle basis has matrix $\left(\begin{array}{cc}0& I_g\\ -I_g& 0\end{array}\right)$, which is unimodular. In particular, a class detected with intersection $1$ or $-1$ by another integral class is a primitive homology class: the detecting homomorphism splits its rank-one subgroup.

A simple closed curve in a connected oriented surface is nonseparating when cutting along it leaves a connected surface. A separating curve bounds a subsurface and represents zero in first homology. For a nonseparating curve, an arc between the two new boundary components closes to a curve meeting it once. Its algebraic intersection number of curves on an oriented surface is $\pm 1$, proving that the original curve represents a primitive homology class.

A chain complex consists of abelian groups $C_q$ and homomorphisms ${\partial }_q:C_q\to C_{q-1}$ with ${\partial }_{q-1}{\partial }_q=0$. Its groups of chain cycles and chain boundaries are $Z_q=\ker {\partial }_q$ and $B_q=\mathrm{im}{\partial }_{q+1}$; the identity ensures $B_q\subseteq Z_q$. Its homology groups areFor a nonnegative complex, take $C_q=0$ for $q<0$.

The singular chain group $C_q(X)$ is the free abelian group on continuous maps $\sigma :{\Delta }^q\to X$, called singular simplices. With ${\delta }_i$ the inclusion of the face opposite vertex $i$, defineDeleting any two vertices in the two possible orders gives the same face with opposite signs, proving ${\partial }^2=0$. This defines the singular chain complex and its singular homology. A continuous map $f$ induces the chain map $f_{\#}(\sigma )=f\circ \sigma$; it commutes with the boundary because composition commutes with restriction to faces.

To prove homotopy invariance of homology, let $F:X\times I\to Y$ have endpoints $f,g$. Triangulate ${\Delta }^q\times I$ by the $q+1$ simplices with ordered verticesLet $p_i:{\Delta }^{q+1}\to {\Delta }^q\times I$ be their affine parameterizations. The singular prism operator isOn taking boundaries, shared internal faces of consecutive prism simplices cancel. The top and bottom faces give $g_{\#}\sigma -f_{\#}\sigma$, and the remaining faces are $-P_{q-1}(\partial \sigma )$ with their induced signs. ThusFor a chain cycle $z$, $g_{\#}z-f_{\#}z=\partial Pz$ is a chain boundary, so $f_{\ast }=g_{\ast }$ on homology. If $f$ is a homotopy equivalence, its homotopy inverse therefore induces an inverse to $f_{\ast }$, proving that singular homology depends only on homotopy type.

Use the standard CW complex for an orientable genus-three surface: one vertex, six edges $a_1,b_1,a_2,b_2,a_3,b_3$, and a two-cell whose attaching word is ${\prod }_{j=1}^3[a_j,b_j]$. The cellular homology theorem identifies this cellular calculation with singular homology. The boundary of every edge is zero, and each edge's total exponent in the attaching word is zero, so the two-cell boundary is zero. HenceChoose $C_1$ to be the nonseparating curve represented by $a_1$, and choose $C_2$ as a small circle bounding a disc in a surface patch. Then $[C_1]$ is a nonzero primitive homology class, whereas the inclusion of $C_2$ induces zero on $H_1$. If a homotopy equivalence carried $C_1$ onto $C_2$, its restriction would factor through $C_2$ and kill $[C_1]$, contradicting injectivity on first homology. This is the homology obstruction to carrying surface curves onto one another.

There is nevertheless such a continuous map. Map the edge $a_1$ homeomorphically onto $S^1$ and all other edges to its base point. The two-cell attaching word then has winding number zero in $S^1$, hence is null-homotopic and the map extends to $r:{\Sigma }_3\to S^1$. Compose $r$ with a homeomorphism $S^1\to C_2$ and the inclusion of $C_2$ in the surface. This gives $g(C_1)=C_2$. Thus an arbitrary map can carry the first curve onto the second, but a homotopy equivalence cannot.

For an unambiguous drawing, realize ${\Sigma }_3$ as the double of a manifold formed from a disc with three holes. Its two copies are glued along all four boundary circles. The red curve below follows an arc from the outer boundary to the first hole on the front copy and returns along the matching arc on the back copy; it is a nonseparating curve. The blue circle is entirely in a disc patch on the front copy and is contractible.

For a framed embedding $S^{\lambda -1}\times D^{m+1-\lambda }\hookrightarrow M^m$, index-$\lambda$ surgery on a smooth manifold removes the interior of that region and replaces it by $D^{\lambda }\times S^{m-\lambda }$. Their common boundary is $S^{\lambda -1}\times S^{m-\lambda }$, along which the framing specifies the gluing. This is a $(\lambda ,m+1-\lambda )$ surgery. Its trace is an $(m+1)$-dimensional cobordism obtained by attaching the handle $D^{\lambda }\times D^{m+1-\lambda }$ to $M\times [0,1]$ along the outgoing boundary.

In the present surface case, $\lambda =2$ and $m=2$. The operation removes an annulus around the embedded curve $\gamma$ and caps its two boundary circles by closed disks. Since $[\gamma ]=pa+qb\ne 0$, the curve cannot separate the oriented torus: a separating curve is the oriented boundary of one of the complementary subsurfaces and is therefore null-homologous. Thus it is a nonseparating curve, and the cut-open surface $F$ is connected with two boundary components. Its Euler characteristic is zero, since both the removed annulus and the gluing circles have Euler characteristic zero. Capping adds two, so the resulting oriented closed connected surface has $\chi (M)=2$. By the allowed surface classification it is a sphere. ThereforeEquivalently the classification shows that $F$ is a cylinder. The argument is conditional on the asserted embedding existing; it does not assign a surgery manifold to a nonexistent embedded curve.

Choose an arc in this connected cylinder joining its two boundary circles and close it by an arc crossing the removed annulus. After smoothing the joints, its closed curve $\delta$ meets $\gamma$ transversely exactly once. If $[\delta ]=ra+sb$, then their algebraic intersection number of curves on an oriented surface isThis proves that $[\gamma ]$ is a primitive homology class and forces $\gcd (p,q)=1$.

Conversely, if $\gcd (p,q)=1$, the mapis an immersion with the required homology class. If its values at $t,s$ coincide, then $p(t-s)$ and $q(t-s)$ are integers; Bézout's identity gives $t-s\in \mathbb{Z}$. It is therefore injective and, by compactness, an embedded circle. An explicit annular thickening isThe determinant of its coordinate derivatives is $\delta (p^2+q^2)\ne 0$. If two image points coincide, solving for the difference in the normal coordinate shows that $\delta (u-u^{\prime })$ is an integer multiple of $1/(p^2+q^2)$. The strict choice of $\delta$ forces that integer to be zero, so $u=u^{\prime }$ and then $t=t^{\prime }$ modulo one. This supplies the actual required embedding, not just a homology class. Hence

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.

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.

A nonzero element of a free abelian group is primitive when it is part of an integral basis, equivalently when some homomorphism to $\mathbb{Z}$ takes it to one. Its coordinates have greatest common divisor one. This lattice meaning differs from a primitive homology class in the coalgebra sense.