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If $\gamma \cdot u$ uses the ordered-tangent orientation convention for algebraic intersection number of curves on an oriented surface, a right-handed Dehn twist takes $u$ to $u+([\gamma ]\cdot u)[\gamma ]$ on first homology. In particular a separating-curve twist acts trivially there.

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.

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

Choose an orientation for each curve and let $c_i=[C_i]\in H_1(X;\mathbb{Q})$. The algebraic intersection number of curves on an oriented surface gives a nondegenerate alternating bilinear form on the $2g$-dimensional space $H_1(X;\mathbb{Q})$, by Poincare duality. Distinct curves are disjoint, so their intersection numbers are zero; each self-pairing is zero by alternation. Thus the span $W=\mathrm{span}\{c_1,\dots ,c_n\}$ is an isotropic subspace of a symplectic vector space. The assumed linear independence gives $\dim W=n$.

Nondegeneracy identifies $H_1(X;\mathbb{Q})$ with its dual. Restricting functionals to $W$ is surjective, so the symplectic orthogonal complement has dimension $\dim W^{\perp }=2g-n$. Isotropy says $W\subseteq W^{\perp }$, giving $n\le 2g-n$. HenceThe bound is sharp: one meridian on each handle of a genus-$g$ surface gives $g$ pairwise disjoint curves with independent homology classes. This is an algebraic bound on independent classes, not on the number of disjoint separating or mutually homologous curves.

Fix the algebraic intersection number of curves on an oriented surface by declaring $u\cdot v=+1$ when the ordered tangent pair $(u,v)$ agrees with the surface orientation. Orient an annular neighbourhood of $\gamma$ with coordinates $(\theta ,r)$ and orientation $d\theta \wedge dr$. A right-handed Dehn twist is represented there by $(\theta ,r)\mapsto (\theta +2\pi f(r),r)$, where $f$ increases from zero to one and is constant near the two boundary circles; it is the identity outside this annulus. A transverse arc gains one oriented copy of $\gamma$ per signed crossing. Consequently its homology action of a Dehn twist isReversing the orientation of $\gamma$ changes both factors' signs and leaves this expression unchanged. A separating curve has $[\gamma ]=0$, so its Dehn twist acts trivially on first homology.

The first assertion is true. Choose a basis $(a,b)$ for the first homology group of the torus with $a\cdot b=1$. The positive Dehn twists about these curves have matricesThe mapping class group of the oriented closed torus is $S{L}_2(\mathbb{Z})$, generated by $A,B$ and their inverses; the Euclidean algorithm on a primitive column gives this generation. The displayed relation rewrites those inverses as positive words:Replacing every inverse in a generating word proves the assertion. The closed torus hypothesis matters: a boundary twist is retained when a boundary circle must be fixed pointwise.

Give the closed orientable surface its standard CW complex structure: one zero-cell, $2g$ one-cells, and one two-cell attached by the product of $g$ commutators. The cellular boundary of the two-cell is zero, since every edge occurs once with each orientation in that word; the one-cell boundaries are also zero. HenceHere we use the cellular homology theorem, identifying cellular and singular homology, and the universal coefficient theorem for cohomology: its exact sequence has terms $\mathrm{Ext}(H_{q-1},\mathbb{Z})$ and $\mathrm{Hom}(H_q,\mathbb{Z})$. All the homology groups here are free, so the Ext terms vanish.

The ring structure comes from Poincare duality and algebraic intersection number of curves on an oriented surface. For a closed oriented surface, cap product with its fundamental class identifies degree-one cohomology with degree-one homology; evaluating the cup product of two such classes equals the signed intersection number of their dual one-cycles. Choose the usual $g$ pairs of handle curves, each pair meeting positively once, and distinct pairs disjoint. Their dual classes can accordingly be named $a_1,b_1,\dots ,a_g,b_g$ so that, for the positive orientation class $\omega$,These formulas include squares. More generally, graded commutativity of the cup product kills every degree-one square here because $H^2$ is torsion-free. The unit generates $H^0$, and products involving $\omega$ and any positive-degree class vanish for dimensional reasons. These additive groups and multiplication rules completely describe the cohomology ring of a closed oriented surface, including $g=0$, when there are no degree-one generators. The intersection pairing is integral and unimodular, rather than merely nondegenerate over a field.

Use the meridian of a knot $\mu$ and Seifert longitude $\lambda$ of the unknot on its boundary torus. The unknot's knot exterior is a solid torus whose disk-bounding curve is $\lambda$. A genus-one Heegaard diagram is thereforeHere the horizontal coordinate is $\mu$ and the vertical coordinate is $\lambda$. The drawing uses a small translate of $\alpha$ to avoid intersections on the edge of the square; opposite edges are identified. The algebraic intersection number of curves on an oriented surface has absolute value five.

Starting with $\Sigma \times [0,1]$, attach a three-dimensional two-handle along $\alpha \times \{0\}$, using its surface framing, and cap the resulting sphere with a three-handle. This gives the solid torus on that side. Attach a two-handle along $\beta \times \{1\}$, again using the surface framing, and cap its sphere with a three-handle. The second solid torus has its disk-bounding curve identified with $5\mu +3\lambda$, so the resulting oriented lens space is precisely the specified $5/3$ Dehn filling.

Write $V=M_{S^2}(p^{\ast }/q,\ast )$. Filling one end of $S^1$ times an annulus gives a solid torus, so $V$ is a solid torus. The prescribed algebraic intersection number of curves on an oriented surface convention gives ${\tilde{f}}_2\cdot {\tilde{h}}_2=1$. HenceIn $H_1(V;\mathbb{Z})$, the first Dehn filling imposes $p^{\ast }f-q{h}_1=0$, while $h_1+h_2=0$. ThereforeThe class ${\lambda }_2$ is primitive because $\gcd (p^{\ast },q)=1$, a consequence of $p{p}^{\ast }-q{q}^{\ast }=1$. It is therefore the disk-bounding meridian of a solid torus of $V$. Since ${\mu }_2$ meets it once, Dehn filling $V$ along ${\mu }_2$ glues two solid torus pieces with disk boundaries meeting once. This is the standard genus-one Heegaard splitting of $S^3$, soTo identify the regular fiber geometrically, put it on their common boundary torus. Invert the displayed basis change:Here ${\mu }_2$ bounds a disk in the newly attached solid torus and ${\lambda }_2$ bounds a disk in $V$. Thus the curve has winding numbers $p$ and $q$ in the two complementary solid torus pieces. Both coefficients are positive and $\gcd (p,q)=1$, so it is the positive $(p,q)$ torus knot. This identifies the torus knot directly from the genus-one splitting, without invoking a theorem specifically about torus knots.

Whenever $\alpha$ and $\beta$ bound a bigon, isotope one side across the bigon. The two removed crossings have opposite local signs: the induced directions around the two corners of an oriented disc are opposite. Thus the move reduces the geometric intersection number by two while leaving the algebraic intersection number of curves on an oriented surface unchanged.

Repeatedly remove bigons. The process terminates because the intersection count is a nonnegative integer, and the bigon criterion says that the resulting curves ${\alpha }_0$ and ${\beta }_0$ are in minimal position. Since every removal cancelled one positive and one negative crossing,