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The monodromy theorem states that a germ admitting analytic continuation along every path in a simply connected domain extends to a single-valued holomorphic function there. Equivalently continuations along homotopic paths with fixed endpoints agree. For a biholomorphism between the two tori, lift its composition with $\mathbb{C}\to \mathbb{C}/{\Lambda }_1$ through the second covering. Simple connectivity, or monodromy applied to local inverse charts, yields a holomorphic lift $F:\mathbb{C}\to \mathbb{C}$.

Lift the inverse similarly. The compositions of the lifts differ from the identity by lattice translations; adjusting the lifts shows that $F$ is bijective. By the allowed classification, $F(z)=az+b$ with $a\ne 0$. For $\lambda \in {\Lambda }_1$, its two arguments $z,z+\lambda$ represent the same torus point, so $a\lambda =F(z+\lambda )-F(z)\in {\Lambda }_2$. Thus $a{\Lambda }_1\subseteq {\Lambda }_2$. Apply the inverse lift to obtain the reverse inclusion. ThereforeThe translation $b$ affects the chosen origin, not the period lattice.

Label the six vertices $A,B,C,D,E,F$ clockwise, starting at the upper-left vertex. The arrows identify $AB$ with $ED$, $BC$ with $FE$, and $FA$ with $DC$, preserving the indicated directions. Consequently the quotient space has exactly two vertices,Orient its three topological cells by $a:A\to B$, $b:B\to C$, and $c:F\to A$. The clockwise boundary of the hexagonal topological cell traversesEvery topological cell of dimension one therefore occurs once with each sign. The cellular chain complex, with integral coefficients, isHere the rows correspond to $P,Q$. The image of ${\partial }_1$ is the primitive subgroup generated by $Q-P$, and its kernel is freely generated by $a+b$ and $a+c$. Thus the homology groups areThe two vertex classes are important: identifying all six vertices would give an incorrect cellular chain complex. The result agrees with the homology of a torus.

The torus map acts on $H_1(T^2;\mathbb{Z})$ byIts determinant is one, so it is an orientation-preserving homeomorphism and acts as the identity on both $H_0(T^2)$ and $H_2(T^2)$.

To calculate the homology of its mapping torus, cover the time circle by two arcs. Each inverse image has the homotopy type of $T^2$, while their intersection has two components of that homotopy type. The map in the Mayer–Vietoris sequence can be writtenAdding the output coordinates and using the input coordinate $a+b$ separates an identity map from $f_{\ast }-I$. Canceling the identity summands gives the Wang sequence, and henceIn degree one,Its image and kernel are both generated by $(1,-1)$; its cokernel is $\mathbb{Z}$ since this image is primitive. Its Smith normal form is $\mathrm{diag}(1,0)$, so its torsion subgroup is zero. On the other two nonzero homology groups of the torus, $f_{\ast }-I=0$. The displayed short exact sequences split because their right-hand groups are free. ThereforeIn particular, the mapping torus contributes an additional degree-one class from the time circle, as well as the invariant degree-one class in the fiber.

For a left $\mathbb{Z}G$-module, group cohomology iswith $\mathbb{Z}$ the trivial module. Equivalently take a projective resolution $P_{\bullet }\to \mathbb{Z}$ and the cohomology of ${\mathrm{Hom}}_{\mathbb{Z}G}(P_{\bullet },M)$. This also describes the right-derived functors of the invariant-module functor $M\mapsto M^G$.

For an explicit model, the homogeneous bar free resolution has $P_n=\mathbb{Z}[G^{n+1}]$, with diagonal $G$-action and boundary the alternating sum of coordinate deletions. It is free over $\mathbb{Z}G$: each orbit has a unique representative with first coordinate $1$. The augmented complex is exact because inserting $1$ as a first coordinate supplies a contracting homotopy of underlying abelian groups; its lack of $G$-equivariance does not affect exactness. The associated inhomogeneous group cochains are functions $f:G^n\to M$, withFor group cochains of degree zero, $(dm)(g)=gm-m$. Adjacent deletions cancel in pairs, giving $d^2=0$ and hence $H^n=\ker d/\mathrm{im}d$.

In degree zero this givesIt is also ${\mathrm{Hom}}_{\mathbb{Z}G}(\mathbb{Z},M)$, since a map is determined by the invariant image of $1$.

In degree one the cocycle condition is $f(gh)=f(g)+gf(h)$, soThese are crossed homomorphisms modulo principal crossed homomorphisms. For trivial action they reduce to ordinary group homomorphisms, so $H^1(G,M)=\mathrm{Hom}(G_{\mathrm{ab}},M)$.

There are two further useful descriptions. If $I=\ker (\mathbb{Z}G\to \mathbb{Z})$ is the augmentation ideal, a crossed homomorphism gives an equivariant map $I\to M$ by $g-1\mapsto f(g)$, because $gh-1=(g-1)+g(h-1)$. Applying $\mathrm{Hom}(-,M)$ to $0\to I\to \mathbb{Z}G\to \mathbb{Z}\to 0$ therefore givesAlso a section of $M⋊G\to G$ has the form $g\mapsto (f(g),g)$ and is a group homomorphism exactly when $f$ is a crossed homomorphism. Conjugation of sections by an element of $M$ changes $f$ by a principal crossed homomorphism. Thus $H^1$ classifies the $M$-conjugacy classes of such sections, or equivalently of complements to $M$ in the semidirect product.

In degree two, a normalized two-cocycle $c:G\times G\to M$ satisfiesDefine multiplication on $M\times G$ byThe displayed cocycle identity is precisely associativity; normalization gives the identity and inverses exist. This produces a group extension $1\to M\to E_c\to G\to 1$ with the prescribed action on $M$.

Conversely a set-theoretic section $s:G\to E$, $s(1)=1$, gives $c(g,h)=s(g)s(h)s(gh{)}^{-1}$ in the additive kernel of a group homomorphism. Associativity gives the cocycle condition. Replacing $s(g)$ by $b(g)s(g)$ changes $c$ byThe resulting extensions are equivalent through a change of coordinates in $M\times G$. Conversely any extension equivalence preserving both kernel of a group homomorphism and quotient yields such a change of section. HenceThe zero class corresponds exactly to a split extension. With trivial action these are central group extensions; addition of classes corresponds to the Baer sum. This proves the second group cohomology classifies group extensions interpretation.

For $G={\mathbb{Z}}^2$ and trivial coefficients $\mathbb{Z}$, put $R=\mathbb{Z}[x^{\pm 1},y^{\pm 1}]$. The Koszul resolution for a rank-two free abelian group iswhere $d_1(a,b)=(x-1)a+(y-1)b$ and $d_2(c)=(-(y-1)c,(x-1)c)$. To check exactness at $R^2$, reduce $d_1(a,b)=0$ modulo $x-1$. Since $y-1$ is not a zero divisor in $\mathbb{Z}[y^{\pm 1}]$, write $b=(x-1)c$, and cancellation gives $a=-(y-1)c$. Injectivity of $d_2$ follows because $R$ is a domain. The augmentation kernel is generated by $x-1,y-1$, proving the other exactness statement.

Applying ${\mathrm{Hom}}_R(-,\mathbb{Z})$ makes both differentials zero, since $x,y$ act trivially. ThusDegree-one classes assign arbitrary integers to the two generators. Geometrically these are also the cohomology groups of the torus, a classifying space for ${\mathbb{Z}}^2$; its cohomology ring is the exterior algebra on the two degree-one classes.

The integer $n$ in degree two has an explicit extension representative. On triples $(a,b,m)\in {\mathbb{Z}}^3$, setThe bilinear term is a two-cocycle. With generators $x=(1,0,0)$, $y=(0,1,0)$ and central $z=(0,0,1)$,Every central extension of ${\mathbb{Z}}^2$ by $\mathbb{Z}$ has this form: choose lifts of the two generators and the fixed generator of the kernel of a group homomorphism, then their group commutator determines $n$ and every element has a unique form $z^m{y}^b{x}^a$. Changing the lifts by central elements leaves that commutator unchanged. Thus distinct integers give inequivalent extensions, and the cocycles $nad$ add with $n$. The case $n=0$ is ${\mathbb{Z}}^3$; $n=1$ is the integer Heisenberg group. This realizes the integral central extensions of a rank-two free abelian group concretely.

Use ${\mathbb{T}}^n={\mathbb{R}}^n/(2\pi \mathbb{Z}{)}^n$ and the periodic Sobolev space conventionThe Cauchy-Schwarz inequality givesTo check the last sum, partition the lattice into $|k|<1$ and dyadic shells $2^j\le |k|<2^{j+1}$. A shell contains at most $C_n{2}^{jn}$ points and contributes at most $C_n{2}^{j(n-2s)}$. This is a convergent geometric series exactly when $s>n/2$.

Thus the Fourier series ${\sum }_k\hat{f}(k)e^{ik\cdot x}$ converges absolutely and uniformly on the torus. Its sum is a continuous function and has the original Fourier coefficients, so it represents the original periodic distribution. In particular the representative is unique andThis proves the requested inclusion, with its continuous Sobolev embedding estimate.

We use a closed manifold of positive dimension $n$. The nonnegative positive Laplace-Beltrami operator on functions isIt satisfies $⟨\Delta f,f⟩={\int }_M|df{|}_g^{2}d{\mathrm{vol}}_g$. Its domain as a self-adjoint operator on $L^2(M)$ is $H^2(M)$. The analytic argument in Question 4(d), with $q=0$, gives a smooth orthonormal eigenbasis and ordered eigenvalues $0={\lambda }_0\le {\lambda }_1\le \cdots \to \infty$.

We prove a quantitative counting bound from local Fourier series, rather than assume a spectral asymptotic formula. Choose a finite coordinate cover and smooth functions ${\chi }_{\alpha }$ compactly supported in its charts with ${\sum }_{\alpha }{\chi }_{\alpha }^2=1$. This can be obtained by normalizing an ordinary nonnegative partition of unity by the square root of the sum of its squares. Embed each chart support in a fixed coordinate torus, and extend $v_{\alpha }={\chi }_{\alpha }f$ by zero there. Uniform comparison of the coordinate and metric norms, and differentiation of the cutoffs, give constants independent of $f$ such thatLet $P_R$ retain the Fourier modes $|k|\le R$. Parseval identity givesThe map $f\mapsto (P_R{v}_{\alpha }{)}_{\alpha }$ has target dimension at most $C_0{R}^n$, since there are at most $C_n{R}^n$ lattice points in each cutoff ball.

Let $E_{\Lambda }$ span all eigenfunctions with eigenvalues at most $\Lambda$. For $f\in E_{\Lambda }$, $\Vert df{\Vert }_2^2\le \Lambda \Vert f{\Vert }_2^2$. If every retained local coefficient vanishes, the preceding estimates implyChoose $R=b\sqrt{1+\Lambda }$ with fixed $b$ sufficiently large. Then this is impossible unless $f=0$, so the finite-dimensional coefficient map is injective on $E_{\Lambda }$. Therefore the eigenvalue counting function satisfiesTaking $\Lambda ={\lambda }_k$ gives $k+1\le C_1(1+{\lambda }_k{)}^{n/2}$, so ${\lambda }_k\ge c_1(k+1{)}^{2/n}-1$. For all sufficiently large $k$, this implies ${\lambda }_k\ge A_0{k}^{2/n}$.

If $M$ is connected, the zero eigenvalue is simple: $\Delta f=0$ implies $df=0$, hence $f$ is constant. Thus every ${\lambda }_k$ for $k\ge 1$ is positive. Taking the minimum of $A_0$ and the finitely many positive ratios ${\lambda }_k/k^{2/n}$ below the large-$k$ threshold provesConnectedness is implicit in the printed claim. If $M$ has $b>1$ connected components, zero has multiplicity $b$, and the claim fails already at $k=1$. The same proof gives ${\lambda }_k\ge A\max \{k-b+1,0{\}}^{2/n}$ for all $k$, as well as the universal counting bound above. These finite initial zero modes do not affect part (d).

Use ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$ and the uniform metric $d(f,g)=\Vert f-g{\Vert }_{\infty }$. Let $N(\epsilon ,E)$ be the least number of radius-$\epsilon$ balls in this metric covering $E$. The Kolmogorov entropy, equivalently metric entropy, isIt is infinite if no finite cover exists; set the entropy of the empty set to zero. Changing the logarithm base or requiring centres in $E$ changes only the constants and resolution factors in the estimates below.

For an explicit appropriate norm, takeAn equivalent standard norm gives the same exponents, with different constants. We prove the two bounds for the entropy of a smooth periodic function ball separately.

For the lower bound, choose a nonnegative bump function $\phi \in C_c^{\infty }((-1,1{)}^n)$ with $0\le \phi \le 1$ and $\phi (0)=1$. Let $K$ bound all its derivatives of orders at most $p$, and choose $c>0$ with $cK\le 1$. For sufficiently small $h\le 1$, place $L\ge c_n{h}^{-n}$ translates of its scaled support disjointly on the torus, for instance by taking grid centres separated by $3h$ in a fundamental cube. Define the real functionsusing periodic extension. Disjointness means that at any point only one summand contributes. For every $|\alpha |\le p$, its derivative is bounded by $c{h}^{p-|\alpha |}K\le 1$. Thus every $f_{\sigma }$ lies in $B_{n,p}$. If $\sigma \ne \tau$, evaluate at a centre where the binary coefficients differ to obtainChoose $h=(4\epsilon /c{)}^{1/p}$ for sufficiently small $\epsilon$. The $2^L$ functions are more than $2\epsilon$ apart, so no radius-$\epsilon$ ball covers two of them. The metric covering number consequently satisfiesTo include the remaining $\epsilon <1/2$, observe that the two constant functions $1,-1$ belong to the ball and require distinct covering balls. Reducing the positive lower-bound constant extends the estimate over that fixed remaining interval.

For the upper bound, use this precise version of multivariable Jackson approximation: for every $f\in C^p({\mathbb{T}}^n)$ and integer $N\ge 1$, there exists a real trigonometric polynomial $P_N$ with $|k_j|\le N$ in every coordinate andChoose $N=\lceil (4J_{n,p}/\epsilon {)}^{1/p}\rceil$. Then the approximation error is at most $\epsilon /4$, and $\Vert P_N{\Vert }_{\infty }\le 2$. Write $P_N(x)={\sum }_{k\in \{-N,\dots ,N{\}}^n}{c}_k{e}^{ik\cdot x}$. There are $D=(2N+1{)}^n$ coefficients, and normalized integration gives $|c_k|\le \Vert P_N{\Vert }_{\infty }\le 2$.

Round their real and imaginary parts with mesh $\delta =\epsilon /(8D)$, respecting $c_{-k}=\overline{c_k}$ so the rounded polynomial remains real. Each coefficient changes by at most $2\delta$, and hence the uniform change in the polynomial is at most $2D\delta =\epsilon /4$. At most $(C/\delta {)}^{2D}$ choices are needed; allowing redundant choices only enlarges this bound. These rounded polynomials form an $\epsilon$-net for the ball, andHere $D\le C_{n,p}{\epsilon }^{-n/p}$, and the last logarithm absorbs the constant and $\log D$ for $\epsilon <1/2$. Together the bounds give

For the unit ball of $C({\mathbb{T}}^n)$ alone, take infinitely many disjoint small neighbourhoods and a height-one continuous bump supported in each. Distinct such functions have uniform distance one. Thus no finite $\epsilon$-net exists when $\epsilon <1/2$, andThe derivative bound was essential for total boundedness; mere continuity gives no common small-scale oscillation bound.

For the final composition estimate, interpret outer periodic functions on ${\mathbb{T}}^m$ by their $2\pi$-periodic extensions to ${\mathbb{R}}^m$. Since $p\ge 1$, each $f\in B_{m,p}$ satisfiesby integrating its first partial derivatives along coordinate segments. This uniform Lipschitz continuity controls errors in the inner functions.

Choose an outer net of radius $\eta =\epsilon /2$ for $B_{m,p}$ and an inner net of radius $\delta =\epsilon /(2m)$ for $E$. For each $g=f(u_1,\dots ,u_m)$ choose corresponding approximants $f_0,v_1,\dots ,v_m$. No derivative bound for $f_0$ is needed: use the derivative bound of $f$ before replacing the outer function. Pointwise,All composed centres are continuous. Multiplying the numbers of outer and inner choices gives the entropy of Lipschitz compositions inequalityApply the proved upper bound with dimension $m$, and the assumed estimate for $E$. Fixed rescalings of $\epsilon$ preserve its exponent and change $\log (1/\epsilon )$ by only an additive constant. Therefore

The Riemannian volume form of an oriented $n$-dimensional Riemannian manifold is the unique positive $n$-form taking value one on every positively oriented orthonormal frame. In positive coordinates,The metric induces an inner product on exterior powers of the cotangent bundle, with wedges of distinct orthonormal covectors forming an orthonormal basis. The Hodge star is the pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ specified byIt is an isometry and satisfies $\ast \ast \alpha =(-1{)}^{p(n-p)}\alpha$. With the positive-sign convention, the Hodge Laplacian on differential forms isHere $d$ is the exterior derivative and $\delta$ its formal adjoint; $\delta$ on functions is zero. On functions this is the Laplace-Beltrami operatorThe convention $\mathrm{div}\mathrm{grad}$ gives the negative of this operator. All subsequent Laplacians use the nonnegative convention above.

On a compact oriented manifold without boundary, the Hodge decomposition theorem gives an orthogonal decomposition of smooth real forms,The three component forms are uniquely determined, although the potentials in the last two summands need not be unique. The space ${\mathcal{H}}^p$ is finite-dimensional, and each de Rham cohomology class has a unique harmonic representative. Integration by parts gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic forms are exactly the closed and coclosed forms. Set ${\Omega }^k=0$ outside $0\le k\le n$. Compactness without boundary is a hypothesis of this version of the theorem; orientation alone does not supply it.

A connection on $TM$ induces a dual connection on $T^{\ast }M$ and then connections on its tensor and exterior powers by the Leibniz rule. Explicitly, for a $p$-form $\alpha$,To prove the parallelism of the Riemannian volume form, use a local positively oriented orthonormal frame $E_i$, its dual coframe ${\theta }^i$, and write ${\nabla }_X{E}_j={\sum }_i{a}_{ij}(X)E_i$. Metric compatibility makes the matrix $a(X)$ skew-symmetric. Since ${\omega }_g={\theta }^1\wedge \cdots \wedge {\theta }^n$, differentiating the wedge yieldsOff-diagonal terms vanish because they repeat a coframe factor. Thus $\nabla {\omega }_g=0$; equivalently parallel transport preserves both the metric and the continuously chosen orientation.

Now let $M$ be compact, oriented and without boundary. For representing functionals on harmonic forms by wedge pairing, choose an $L^2$-orthonormal basis $h_1,\dots ,h_b$ of ${\mathcal{H}}^p$ and defineIf $\phi ={\sum }_a{c}_a{h}_a$, linearity givesThis also works when $b=0$, using the empty sum. Since the Hodge star commutes with the Hodge Laplacian, ${\psi }_0$ is harmonic of degree $n-p$.

For the full ambiguity of harmonic wedge-pairing representatives, write another representative as ${\psi }_0+\chi$. Because the star is an isometry,The forms $\ast \phi$ range over all of ${\mathcal{H}}^{n-p}$. Hence $\chi$ gives zero functional exactly when its harmonic projection is zero. Hodge decomposition proves that the complete answer isBoth summands in this ambiguity are permitted; it is not only an exact-form ambiguity. The exact and coexact component forms are orthogonal to every harmonic form, so every displayed choice works. Conversely a nonzero harmonic component would pair nontrivially with some $\ast \phi$, and therefore cannot occur. The harmonic representative $\ast h_f$ is unique; the arbitrary smooth representative generally is not. The formula also covers the endpoint degrees and any exceptional case where the ambiguity space is zero.

For a concrete coexact ambiguity, take the torus $(\mathbb{R}/2\pi \mathbb{Z}{)}^2$ with metric $d{x}^2+d{y}^2$. The form $\chi =\delta (\sin ydx\wedge dy)=\cos ydx$ annihilates every harmonic one-form by orthogonality, yet $d\chi =\sin ydx\wedge dy$ is nonzero. Thus an annihilating representative need not even be closed, let alone exact.

Use generators $a,b,c$ for the three one-handles in the generalized Heegaard diagram. Read a downward crossing of the cut from a left-hand hole to its right-hand partner as the corresponding generator. Traverse the attaching curve around the left-hand $A$ hole, starting at its outgoing downward strand, and then traverse the attaching curve around the left-hand $B$ hole in the same way. Their cut-crossing words areThese are the two relators imposed by the two handle attachments. The remaining boundary is a torus; no third relator should be added. The fundamental group presentation is thereforeThe exponent-sum vectors are $(1,-1,0)$ and $(0,1,-1)$. Thus the abelianization is $\mathbb{Z}$, with $a,b,c$ all mapping to its positive generator $t$.

Apply the Fox derivatives and then this abelianization. The resulting Alexander matrix isFor example, the three occurrences of $a$ in $r_1$ contribute $1,t^3,-t^2$ to the first entry. Every row sums to zero, as the Fox calculus identity predicts.

The cellular boundary in degree one for the infinite cyclic cover is $(t-1,t-1,t-1)$. Its kernel is freely generated by $e_a-e_c,e_b-e_c$. Expressing the two relation boundaries in this basis gives the two-by-two matrix obtained by deleting column $c$ of $F(t)$. Its determinant is the order of the Alexander module of a space. Equivalently, all three maximal matrix minors of $F(t)$ agree up to sign. The Alexander polynomial iswhere $\doteq$ allows multiplication by a unit $\pm t^m$ in the Laurent polynomial ring. Its value at $1$ is $1$.

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.

The topological degree measures an oriented net number of sheets of a map. Its simplest integer-valued setting is a map $f:M\to N$ between connected oriented closed manifolds of the same positive dimension $n$. The hypotheses matter: an integer sign count needs orientations; compactness prevents preimages from escaping; and the same dimension makes a regular fiber discrete. A fundamental class gives the definition for continuous maps:Here $H_n(M;\mathbb{Z})$ and $H_n(N;\mathbb{Z})$ are generated by their chosen fundamental classes. Reversing either orientation changes the sign of the degree of a map between oriented manifolds. Disconnected sources contribute a sum over components; disconnected targets require a degree for each target component rather than one universal integer.

For a smooth map, choose a regular value $y$. The Sard theorem guarantees plentiful choices. Each point of $f^{-1}(y)$ has an invertible derivative, so the inverse function theorem makes it an isolated point. The fiber is finite by compactness of $M$. An oriented coordinate chart defines $\mathrm{sgn}\det d{f}_x$, and the degree as a sum of local degrees becomesAn empty fiber contributes zero. The sign is independent of the chosen oriented charts. Thus the mapping degree counts sheets with signs, not simply the number of preimages: orientation-preserving and orientation-reversing sheets cancel.

A differential-form argument both proves that this count is independent of $y$ and relates it to the topological definition. Around a regular value, take a small neighborhood $V$ whose full preimage is a disjoint union of neighborhoods mapped diffeomorphically onto $V$. Compactness rules out additional preimages approaching $V$ from elsewhere. Choose a smooth top-degree form $\eta$ supported in $V$, with ${\int }_N\eta =1$. The change of variables formula givesAlthough $\eta$ need not be everywhere positive, it represents the same normalized top cohomology class as any normalized volume form. The top de Rham cohomology of a compact connected oriented manifold says that two top forms with equal integrals differ by $d\beta$. Their pullback integrals agree, because the Generalized Stokes theorem gives ${\int }_{M}d(f^{\ast }\beta )=0$. Therefore the degree by integration of a pullback volume form isfor every smooth top form $\eta$. Pairing de Rham cohomology with fundamental classes identifies this integer with the homological definition. This proof also shows that every regular value gives the same signed sum.

The homotopy invariance of mapping degree makes it stable under continuous deformation. Homologically this follows from homotopic maps inducing the same map on homology. For a smooth homotopy $F:M\times [0,1]\to N$, a direct proof uses $d\eta =0$ and the Generalized Stokes theorem on the cylinder to getwith the cylinder orientation chosen to give the indicated boundary signs. Smooth approximation extends the geometric description to continuous maps, while the fundamental class definition already applies without differentiability. The multiplicativity of mapping degree follows from functoriality and givesA constant map has degree zero when $n>0$. If the mapping degree is nonzero, the map is onto: a point outside its image is a regular value with empty preimage. An orientation-preserving diffeomorphism has degree $1$, and an orientation-reversing one has degree $-1$. An orientation-preserving $k$-sheeted covering map has degree $k$; the signs of sheets must be included for other orientation conventions.

Examples show both the strength and the limits of the invariant. On the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, including negative integers and zero; it is the winding number. On the sphere $S^n$, an ambient reflection has degree $-1$, and the antipodal map has degree $(-1{)}^{n+1}$, because the determinant of $-I$ on the ambient ${\mathbb{R}}^{n+1}$ is $(-1{)}^{n+1}$. On the torus, an integer matrix $A$ induces a map of degree $\det A$, obtained by pulling back the constant top form. However, degree does not classify general manifold maps: the identity and an integer shear on the two-torus both have degree one but induce different maps on $H_1$, so are not homotopic.

The mapping degree gives useful obstruction and existence arguments. There is no retraction of a closed ball onto its boundary sphere: such a retraction would make the degree-one identity of the boundary null-homotopic, since it would extend across a contractible ball, contradicting homotopy invariance and the degree-zero constant map. This is the degree obstruction underlying the Brouwer fixed-point theorem. Also an everywhere nonzero tangent vector field on an even-dimensional sphere would, after normalization, produce the homotopy $\cos (\pi t)x+\sin (\pi t)v(x)$ from the identity to the antipodal map. Orthogonality makes every value a unit vector, but the endpoint degrees are $1$ and $-1$, impossible. This proves the corresponding Hairy ball theorem obstruction.

There are appropriate extensions rather than an unrestricted integer degree for every pair of manifolds. For a proper map between connected oriented manifolds without boundary of equal dimension, the regular-value count remains finite and is invariant; compactly supported top forms or locally finite homology replace the compact fundamental-class description. Without orientations, degree modulo two counts preimages modulo two and uses mod-two fundamental classes. For manifolds with boundary one uses relative mapping degree for maps of pairs, keeping the boundary away from the chosen target value. Maps of unequal dimensions, or nonproper maps where preimages can escape to infinity, do not in general admit this same signed sheet-count invariant. The central principle is an integer signed count stable under homotopy, with its hypotheses explicitly preserved.

Write $R=a+b\cos u>0$. The tangent vectors areTheir squared lengths and inner product give the first fundamental formThe unit normal determined by this ordered parametrization isTaking inner products of $\nu$ with ${\sigma }_{uu}$, ${\sigma }_{uv}$ and ${\sigma }_{vv}$ yields $e=b$, $f=0$, $g=R\cos u$. Consequently the second fundamental form of a ring torus isThis normal points into the tube. Choosing the outward unit normal reverses both coefficients and gives $-II$; the choice must therefore accompany an answer for the second fundamental form.

The Gaussian curvature is unchanged by that reversal:Since the denominator is positive, $K>0$ on the outside of the torus, where $\cos u>0$, and $K<0$ on its inner side, where $\cos u<0$. It vanishes on the two circles $u=\pi /2,3\pi /2$. This is the Gaussian curvature of a ring torus.