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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