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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.

A connected $n$-sheeted covering space of a closed orientable surface of genus $g$ has genus $1+n(g-1)$. Lifting a finite cell decomposition gives $n$ cells above every base cell, so its Euler characteristic is $n(2-2g)$. Equating this with $2-2g^{\prime }$ gives the formula. For $g\ge 1$, every positive $n$ occurs: send one handle generator of the fundamental group to $1$ in $\mathbb{Z}/n$ and all remaining generators to zero, then use the classification of connected covering spaces for the kernel of that surjection.

For genus $g\ge 1$, collapse a nonseparating simple closed curve on a closed orientable surface. Its homology class is primitive, as a curve meeting it transversely once detects by the intersection pairing on an oriented surface. The long exact sequence in relative homology therefore gives $H_0=\mathbb{Z}$, $H_1={\mathbb{Z}}^{2g-1}$, $H_2=\mathbb{Z}$, and zero higher groups for the quotient. Geometrically it has the homotopy type ${\Sigma }_{g-1}\vee S^1$.

Collapse a separating simple closed curve on a closed orientable surface of genus $g$. If its two sides have genera $g_1,g_2$, the quotient is homeomorphic to ${\Sigma }_{g_1}\vee {\Sigma }_{g_2}$. Its integral homology is $\mathbb{Z}$ in degree zero, ${\mathbb{Z}}^{2g}$ in degree one, ${\mathbb{Z}}^2$ in degree two, and zero above degree two. In relative homology, the circle maps to zero in $H_1({\Sigma }_g)$, and the second relative group fits into a split sequence $0\to \mathbb{Z}\to H_2({\Sigma }_g,A)\to \mathbb{Z}\to 0$. The two collapsed sides retain independent fundamental classes.

For the closed orientable surface ${\Sigma }_g$, use the fundamental group presentationSince $g\ge 1$, define a homomorphism to $\mathbb{Z}/n$ by sending $a_1$ to $1$ and every other displayed generator to zero. All commutators map to zero, so the relation is respected, and the map is surjective. Its kernel has index $n$. The classification of connected covering spaces supplies a connected $n$-sheeted covering space, including the identity cover when $n=1$.

Every connected finite covering space of ${\Sigma }_g$ is a closed orientable surface: compactness follows from finite degree, and the base orientation lifts through local covering charts. Lift a finite cell decomposition of the base. Every cell has $n$ lifts, so the Euler characteristic under a finite covering gives $\chi (\tilde{\Sigma })=n\chi ({\Sigma }_g)$. If the covering surface has genus $h$, this reads $2-2h=n(2-2g)$. ThereforeThis genus of a finite cover of a closed orientable surface depends only on the sheet number, not on the particular subgroup used to construct the cover.

For a singular simplex $\sigma :{\Delta }^i\to Y$, there are exactly $n$ lifts to $X$: choose any of the $n$ points above one vertex, then use the lifting property, since the simplex is simply connected. Defineand extend $R$-linearly to singular chains. The restrictions of these lifts to any face are exactly the $n$ lifts of that face, each appearing once. Indeed a lift is determined uniquely by its value at any point of that face, and every possible value in the corresponding fiber extends to a lift of the whole simplex. Therefore $\partial T_i=T_{i-1}\partial$, so $T$ is a chain map and induces the homological transfer for a finite covering, denoted $f^{\ast }$ in the question.

On chains, every lifted simplex projects back to the original simplex. Thus $f_{\#}{T}_i=n\mathrm{id}$, and on homology this givesOver $\mathbb{Q}$, the map $n^{-1}{f}^{\ast }$ is a right inverse of $f_{\ast }$, proving surjectivity in every degree. Consequently each rational Betti number of the covering space is at least that of its base. This transfer goes in the opposite direction to the usual homological pushforward and is not the ordinary cohomological pullback.

For equality, use the connected double covering space $S^1\to S^1$, $z\mapsto z^2$. Both spaces have $b_0=b_1=1$ and all other Betti numbers zero. For strict increase, take a connected double covering space of the genus-two closed orientable surface, whose existence was proved in 1(b). The genus of a finite cover of a closed orientable surface gives covering genus $1+2(2-1)=3$. Its first Betti number is six, whereas that of the genus-two base is four; their zeroth and second Betti numbers are both one. Thus a connected double cover can preserve every Betti number or strictly increase some of them.

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.

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.

A symplectomorphism is a diffeomorphism, so it preserves the genus of a surface. It also preserves the orientation determined by the symplectic form and, by integration of a pullback of a differential form, preserves the symplectic area.

Conversely, equal genus of a surface gives a diffeomorphism $f:{\Sigma }_0\to {\Sigma }_1$. We can choose $f$ to preserve the orientations determined by ${\omega }_0,{\omega }_1$: if necessary, compose with an orientation-reversing diffeomorphism of the standard closed orientable surface of that genus of a surface, obtained by reflection of a symmetric handle model. On ${\Sigma }_0$ put $\eta =f^{\ast }{\omega }_1$. Both $\eta$ and ${\omega }_0$ are positive volume forms for the same orientation, and their integrals agree. By top-degree de Rham cohomology, $[\eta ]=[{\omega }_0]$.

The family ${\omega }_t=(1-t){\omega }_0+t\eta$ consists of positive volume forms, hence symplectic forms in dimension two, with constant de Rham cohomology class. Moser theorem supplies $\psi$ with ${\psi }^{\ast }\eta ={\omega }_0$. ConsequentlyTherefore the genus of a surface and symplectic area classify compact connected symplectic surfaces.

The hyperplane of a Salvetti complex dual to the loop labelled $v$ is itself the Salvetti complex of the induced subgraph on the neighbours of $v$. If this hyperplane were homeomorphic to a closed surface, its vertex link would be a cycle. Part (c), applied to that induced graph, says that the graph must be one edge. Its Salvetti complex is therefore the two-dimensional torus, a surface of genus one. It cannot be a closed orientable surface of genus two.

A once-punctured genus-$h$ closed orientable surface retracts to a wedge of $2h$ circles. A retraction from ${\Sigma }_g$ would inject its degree-one cohomology as an isotropic subspace of a symplectic vector space for the Poincare duality pairing, since the punctured surface has zero degree-two cohomology. The maximal isotropic dimension is $g$, giving $2h\le g$. Equality is realized by folding the double of the punctured surface onto one copy. Additional handles in the other copy can be pinched away while fixing the joining boundary, proving sufficiency for every embedding of such a one-boundary subsurface when $g\ge 2h$.