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For a continuous map $f:M\to N$ of closed connected oriented $d$-dimensional manifolds, suppose $f^{-1}(y)=\{x_1,\dots ,x_s\}$ is finite. ThenBy excision, the degree-$d$ relative homology group $H_d(M,M\setminus f^{-1}(y);\mathbb{Z})$ is the direct sum of the local groups at $x_j$. The fundamental class maps to the tuple of local orientation generators. Mapping this tuple to $H_d(N,N\setminus \{y\};\mathbb{Z})$ adds their local degrees. Naturality identifies the result with the image of $(\mathrm{deg}f)[N]$. If the fiber is empty, the same argument gives degree zero.

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.

The integral homology of the loop space of a sphere implies that the unit $S^{r-1}\to \Omega S^r$ has vanishing relative homology through degree $2r-3$. Both spaces are simply connected for $r\ge 3$. The absolute Hurewicz theorem first identifies their degree-two homotopy groups with homology; the relative Hurewicz theorem then rules out a first nonzero relative homotopy group in degrees three through $2r-3$. The relative homotopy sequence makes the unit an isomorphism on ${\pi }_i$ through degree $2r-4$. Loop-suspension adjunction identifies this map with suspension, yielding ${\pi }_i(S^{r-1})\cong {\pi }_{i+1}(S^r)$ in that range.

Use integral singular chains. The inclusion $Y\subseteq X$ identifies $C_k(Y)$ with the subgroup of $C_k(X)$ generated by the singular simplices whose images lie in $Y$. It is a chain subcomplex. DefineThis boundary is well defined because $\partial C_k(Y)\subseteq C_{k-1}(Y)$, and its square is zero because ${\partial }^2=0$. Its homology is, by definition, $H_k(X,Y)$. A relative cycle has a representative $c$ with $\partial c\in C_{k-1}(Y)$; representatives of the same relative homology class differ by $\partial b+y$, where $b\in C_{k+1}(X)$ and $y\in C_k(Y)$.

Inclusion and quotient induce homomorphisms $i_{\ast }:H_k(Y)\to H_k(X)$ and $j_{\ast }:H_k(X)\to H_k(X,Y)$. Define the relative homology connecting homomorphism byHere $\partial c$ is a cycle in $Y$ since ${\partial }^{2}c=0$. If $c$ is changed to $c+\partial b+y$, its boundary changes by $\partial y$, a boundary in $Y$, proving independence of representatives. The formula is additive, so $\delta$ is a homomorphism.

The long exact sequence in relative homology isand ends inWe prove all three types of exactness directly, so no unproved abstract lemma about chain complexes is required.

At $H_k(X)$, $j_{\ast }{i}_{\ast }=0$ because a chain in $Y$ is zero in the relative chain group. Conversely, let an absolute cycle $c$ have $j_{\ast }[c]=0$. The definition of a relative boundary gives $c=\partial b+y$ with $y\in C_k(Y)$. Since $c$ is a cycle, $\partial y=0$. Thus $[c]=i_{\ast }[y]$, proving $\ker j_{\ast }=\mathrm{im}{i}_{\ast }$.

At $H_k(X,Y)$, $\delta j_{\ast }=0$ because an absolute cycle has zero boundary. Conversely, suppose $\delta [c]=0$. Then $\partial c=\partial y$ for some $y\in C_k(Y)$. Therefore $c-y$ is an absolute cycle and $j_{\ast }[c-y]=[c]$. Hence $\ker \delta =\mathrm{im}{j}_{\ast }$.

At $H_{k-1}(Y)$, $i_{\ast }\delta =0$ because $\partial c$ is a boundary in $X$. Conversely, if a cycle $y\in C_{k-1}(Y)$ becomes zero in $H_{k-1}(X)$, then $y=\partial c$ for some $c\in C_k(X)$. That $c$ is a relative cycle, and $\delta [c]=[y]$. Thus $\ker i_{\ast }=\mathrm{im}\delta$.

These arguments also prove the assertions at shifted indices. Finally, every zero-dimensional relative cycle lifts to a chain in $C_0(X)$, and every zero-chain is an absolute cycle; this proves surjectivity onto $H_0(X,Y)$. All maps, representative changes, compositions and exactness assertions have consequently been justified from the chain definitions.

Let $C_{\ast }(X)$ be the integral singular chain complex. Augment it by the homomorphism $C_0(X)\to \mathbb{Z}$ sending every point simplex to one. The homology of this augmented complex is reduced homology: for a nonempty space, it agrees with ordinary homology in positive degrees and ${\tilde{H}}_0(X)$ is the kernel of $H_0(X)\to \mathbb{Z}$. The relative homology $H_{\ast }(X,A)$ is the homology of $C_{\ast }(X)/C_{\ast }(A)$; a relative cycle has boundary in $A$, and chains differing by a boundary or a chain in $A$ represent the same class.

Excision theorem: if $U\subset A\subset X$ and $\overline{U}\subset {\mathrm{int}}_{X}A$, the inclusion $(X\setminus U,A\setminus U)\hookrightarrow (X,A)$ induces isomorphisms on all relative homology groups. In particular, for a nonempty CW subcomplex $A$, choose a neighborhood $N$ deforming onto $A$, with the deformation fixing $A$. Such neighborhoods exist for CW pairs. Then $N/A$ is contractible. The long exact sequence of a pair gives $H_{\ast }(X,A)\cong H_{\ast }(X,N)$ and $H_{\ast }(X/A,\{\ast \})\cong H_{\ast }(X/A,N/A)$. Excision removes $A$ and the collapsed point respectively; the resulting pairs $(X\setminus A,N\setminus A)$ and $((X/A)\setminus \{\ast \},(N/A)\setminus \{\ast \})$ are homeomorphic. ConsequentlyHere the nonempty-subcomplex condition is the usual quotient convention in this statement. If $A=\varnothing$, use the pointed cofiber $X_{+}$ with a disjoint basepoint; the literal unpointed quotient $X/\varnothing =X$ would not give the asserted degree-zero identity.

For one embedded circle, its long exact sequence of a pair reduces toA separating circle bounds a subsurface, so its homology class is zero. A nonseparating circle has a transverse closed curve meeting it once, showing that its class is nonzero and primitive. Thus $i_{\ast }$ is either zero or an injection with torsion-free cokernel. The resulting possibilities arewith all other relative groups zero. These ranks determine whether the circle is null-homologous.

For $g\ge 2$, a disk boundary and a separating circle bounding a once-punctured torus have the same relative groups. The first is null-homotopic; the second is not. For a direct verification, use the surface-group presentation $⟨a_1,b_1,\dots ,a_g,b_g\mid {\prod }_i[a_i,b_i]=1⟩$. Send $a_1,b_1$ to free generators $u,v$, send $a_2,b_2$ to $v,u$, and all remaining generators to one. The relation maps to $[u,v][v,u]=1$, so this defines a homomorphism to the free group. The separating circle represents $[a_1,b_1]$ and maps to the nonempty reduced word $uv{u}^{-1}{v}^{-1}$; it is therefore not null-homotopic. This proves the requested failure to detect nullhomotopy. There is a necessary low-genus qualification: on ${\Sigma }_0$ or ${\Sigma }_1$, an embedded circle is nullhomologous exactly when it is nullhomotopic. On the sphere every circle bounds a disk; on the torus every essential embedded circle is nonseparating. The printed nondetection assertion therefore needs $g\ge 2$.

For two disjoint embedded circles let $r$ be the rank of their image in $H_1({\Sigma }_g)$, and let $c$ be the number of connected components of the cut surface. Removing open annular neighborhoods and applying excision gives the compact cut surface $W$ relative to its boundary. Poincare-Lefschetz duality gives $H_2(W,\partial W)\cong H^0(W)={\mathbb{Z}}^c$ and $H_1(W,\partial W)\cong H^1(W)$, which is free. Thus there is no hidden torsion. The long exact sequence gives $c=3-r$, and the Euler characteristic of the pair is $2-2g$, givingThe possibilities are $r=0,1,2$: two separating circles give zero; a nonseparating circle with a disk boundary, or two parallel essential circles, give one; circles around two different handles give two. For genus zero only $r=0$ occurs; for genus one $r=0,1$ occur; for every genus at least two all three occur. This is relative homology of a surface modulo disjoint circles.

Let $p:E\to S^r$ be the Serre fibration. For $r\ge 1$, cover the base by $U=S^r\setminus \{s\}$ and a small open ball $V$ about $s$. Write $A=p^{-1}U$ and $B=p^{-1}V$. The contractible-base assumption gives $A\simeq F$ and identifies the pair $(B,A\cap B)$, up to fiberwise homotopy, with $(F\times V,F\times (V\setminus \{s\}))$.

Excision for the open cover $E=A\cup B$ gives $H_i(E,A)\cong H_i(B,A\cap B)$. The relative base pair hasby the deformation retraction of the punctured ball onto $S^{r-1}$ and its relative homology sequence. The relative Künneth theorem therefore givesThere is no Tor term because the only nonzero base group is free. The long exact sequence of a pair, replacing $A$ by its homotopy-equivalent fiber, now becomesThe last arrow is the relative boundary transported through these identifications; it can encode twisting, so it has not been set to zero. This proves the relative homology of a fibration over a sphere sequence. In the dimension-zero case, the total space is the disjoint union of the two fibers and the same formula is the split sequence for that union.

The smooth h-cobordism theorem states: if $(W;M_0,M_1)$ is a compact smooth h-cobordism, $\dim W\ge 6$, and its connected boundary manifolds are simply connected, then there is a diffeomorphism $W\cong M_0\times [0,1]$ restricting to the prescribed identification on $M_0$. In particular $M_0$ and $M_1$ are diffeomorphic. Being an h-cobordism means that both inclusions $M_i\hookrightarrow W$ are homotopy equivalences; their fundamental groups then also identify with that of $W$.

The homotopy-equivalence condition is necessary for a trivial cobordism: each end of a cylinder is a deformation retract, by $(x,t)\mapsto (x,(1-s)t)$ for the incoming end and $(x,t)\mapsto (x,(1-s)t+s)$ for the outgoing end. Its relative homology with respect to either end is consequently zero. Compactness is also necessary for a product with a closed compact end. The dimensional bound and simple connectivity are not necessary for an individual cobordism to be a product. The cylinder $S^2\times [0,1]$ is a simply connected trivial cobordism of dimension three; and, in the theorem's dimension range, $(S^1\times S^4)\times [0,1]$ is a six-dimensional trivial cobordism with fundamental group $\mathbb{Z}$. These prove the two non-necessity assertions explicitly, while the inclusion argument proves necessity of the h-condition.

For the remaining claims, use a cobordism Morse function: its two regular boundary levels are zero and one, and all critical points are interior. An arbitrary function with unrelated boundary values would not have the relative handle-count property. The relative Morse chain complex for $(W,M_0)$ has one generator per critical point, in its Morse index. ThusA trivial cobordism has zero relative homology, hence zero relative Euler characteristic. Modulo two the displayed sum is the number of critical points. An odd number makes it odd, and therefore nonzero. This proves an odd-critical-point cobordism is not trivial.

With precisely two critical points of indices $\lambda \le \mu$, a trivial cobordism requires the two-generator relative complex to be acyclic over $\mathbb{Z}$. Equal indices leave a rank-two homology group; a gap of more than one leaves both generators with zero possible differential. Therefore $\mu =\lambda +1$ is necessary. The remaining two-term complex isChoose a transverse gradient-like vector field. A nonzero differential requires the index-$\mu$ point to be above the index-$\lambda$ point; otherwise no descending connecting trajectories exist. In a common regular level between them, the belt sphere $S_U(p)$ has dimension $d-\lambda -1$, and the attaching sphere $S_L(q)$ has dimension $\mu -1=\lambda$, where $d=\dim W$. Their dimensions add to $d-1$, the level's dimension, so their transverse intersection consists of points. Each intersection is one descending trajectory from $q$ to $p$, modulo time translation. Orient the attaching sphere and use an orientation of a vector space on the handle core to coorient the belt sphere; the resulting signed count is the cellular boundary coefficientThe two homology groups are the kernel and cokernel of multiplication by $a$. They vanish precisely when $a$ is a unit of $\mathbb{Z}$. Consequentlyare necessary. This proves the two-critical-point obstruction to a trivial cobordism; it makes an algebraic incidence statement and does not assume that an algebraic count of one automatically supplies geometric handle cancellation.

For an explicit nontrivial cobordism from $S^3$ to itself, attach two zero-framed two-handles to the outgoing end of $S^3\times [0,1]$ along the standard Hopf link. Its two components are unknots with linking number one. This realizes two successive $(2,2)$ surgeries, as suggested by the hint. The outgoing boundary is again $S^3$, and this can be checked geometrically: the Hopf-link exterior is $T^2\times [0,1]$, and each component's zero-framing longitude is the other's meridian. Filling along these two slopes glues two solid tori with meridians intersecting once, the standard genus-one Heegaard decomposition of $S^3$. Thus zero surgery on the Hopf link does return the three-sphere, not just an unspecified homology sphere.

The trace has only the two added index-two handles. Its relative cellular complex has $C_2(W,S^3)={\mathbb{Z}}^2$ and all other groups zero, soIt cannot be a product cylinder. Equivalently this trace is $S^2\times S^2$ with two disjoint open four-balls removed: the usual handle presentation of $S^2\times S^2$ has one zero-handle, the two zero-framed Hopf-linked two-handles, and one four-handle. Removing the bottom and top balls exposes the two three-sphere ends. The relative-homology calculation alone already proves the requested cobordism is nontrivial.

Use relative homology. Every simplex of $X=B\cup C$ lies in $B$ or $C$, and those lying in both are exactly the simplices of $A$. Consequently inclusion induces a chain-complex isomorphismIt is bijective on the respective bases of simplices and commutes with boundary because it comes from inclusion. Thus

For any pair $(Y,Z)$, the long exact sequence in homology shows that inclusion induces isomorphisms in every degree exactly when $H_j(Y,Z)=0$ for every $j$. To verify both directions, if all relative groups vanish, exactness makes each $H_j(Z)\to H_j(Y)$ injective and surjective. Conversely, if the inclusion maps are isomorphisms, the map from $H_j(Y)$ into the relative group is zero, while the relative connecting map is injective and has zero image because the next inclusion is injective; hence the relative group itself is zero. This includes degree zero, with the negative-degree ordinary group zero.

Apply the criterion first to $(B,A)$ and then to $(X,C)$ using the chain isomorphism. It proves

For closed connected oriented $d$-manifolds, the degree of a map between oriented manifolds is the integer defined by $f_{\ast }[M]=(\mathrm{deg}f)[N]$ in $H_d(N;\mathbb{Z})$. If $y\in N$ has finite fiber $\{x_1,\dots ,x_r\}$, choose disjoint small neighbourhoods $U_i$ containing no other points of that fiber. The map of local relative homology groupssends the local orientation generator at $x_i$ to $d_i$ times the one at $y$. This integer is its local degree of a continuous map, independent of the small neighbourhood by excision. Excision also identifies $H_d(M,M\setminus f^{-1}(y))$ with the direct sum of the local groups, and the global fundamental class restricts to their orientation generators. Naturality of the maps to relative homology therefore proves the degree as a sum of local degrees:The empty fiber gives degree zero. For smooth maps at a regular value, a local degree is the sign of the derivative determinant, but general continuous maps can have larger local multiplicities.

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.

For a properly embedded oriented topological surface $F=\coprod F_j$, putThis defines the Thurston norm on integral relative homology classes in $H_2(Y,\partial Y;\mathbb{Z})$. Homogeneity extends it to rational classes, and continuity extends it to real classes. It is a seminorm: nonzero classes carried by spheres, disks, annuli or tori can have zero value.

Not every integral class has a connected embedded representative. In $Y=S^1\times S^2$, the class $2[\{\ast \}\times S^2]$ is a counterexample. A connected embedded oriented surface either separates, in which case it is null-homologous, or has connected complement. In the latter case join its two local sides through its complement to obtain a loop intersecting it exactly once. Its homology class is consequently a primitive lattice element. The proposed double has all intersection numbers even, so cannot have such a representative. Two disjoint parallel spheres do represent it.

For the knot exterior in an integral homology sphere, $H_2(Y,\partial Y;\mathbb{Z})=\mathbb{Z}$ is generated by a Seifert surface. Its Thurston norm iswhere $g(K)$ is the Seifert genus in the ambient integral homology sphere. To justify using a one-boundary-component Seifert surface, simplify a minimizing representative's boundary to parallel essential longitudes. The algebraic sum of these longitudes is one. Join oppositely oriented pairs by boundary annuli and push inward; this preserves Euler characteristic and does not increase ${\chi }_{-}$. Discard closed components, which represent zero because the ambient integral homology sphere has $H_2=0$. The component retaining the single boundary is a Seifert surface. This proves the formula, including the disk case.

When $\Vert y{\Vert }_T>0$, $g(K)\ge 1$. In zero Dehn surgery the longitude bounds a meridional disk in the filling solid torus. Cap a minimizing Seifert surface with that disk. The resulting closed surface $\hat{F}$ has the same genus and represents a generator of $H_2(Z;\mathbb{Z})$: its intersection with the filling core is one. ThereforeThe case $g(K)=1$ gives a zero-cost torus, rather than a negative value.

Here is a rigorous family of strict examples with $Z=S^1\times S^2$. Use the excellent knot representative theorem to choose a knot $J$ representing the generator of ${\pi }_1(Z)$ with an excellent three-manifold as exterior $E_0$. Inside $V=\nu J$, choose a winding number of a satellite pattern one pattern $P$ whose two-boundary-component exterior $E_1=V\setminus \mathrm{int}\nu P$ is also an excellent three-manifold. Both choices are available because neither ambient manifold has a spherical boundary component. Put $\Gamma =P(J)$ and $E=Z\setminus \mathrm{int}\nu \Gamma =E_0{\cup }_{\partial V}{E}_1$.

The interface is an incompressible surface. On a Thurston norm minimizing surface in $E$ with boundary the meridian ${\mu }_{\Gamma }$, arrange all interface intersections to be essential. The winding number of a satellite pattern one condition forces the oriented boundary on that interface to have net class ${\mu }_J$. The piece in $E_0$ has odd meridional boundary sum and costs at least one: a zero-cost representative would require an essential disk or annulus, forbidden by excellence. The piece in $E_1$ has opposite net meridians on its two boundary tori. It costs at least two: zero-cost components are boundary-parallel annuli or tori and carry no such boundary class, and its total meridional boundary count is even, so its nonzero negative Euler characteristic has even magnitude. Cutting along the interface adds the costs, since the essential pieces have no disk or sphere components. ThusHere the notation denotes the relative class whose boundary is ${\mu }_{\Gamma }$, not the meridian curve itself. This is the usual Thurston norm gluing along an incompressible torus argument.

Because $\Gamma$ represents the generator of $H_1(Z)$, $H_1(E)=\mathbb{Z}$, its meridian is null-homologous in $E$, and a longitudinal curve ${\lambda }_{\Gamma }$ generates $H_1(E)$. Fill $E$ along ${\lambda }_{\Gamma }$ to obtain $X$. The Mayer–Vietoris sequence gives $H_1(X)=0$, so $X$ is an integral homology sphere. Let $K$ be the filling core. Its preferred longitude is ${\mu }_{\Gamma }$; zero Dehn surgery on $K$ consequently recovers $Z$. For this example $\Vert y{\Vert }_T\ge 3$, whereas the product sphere generates $H_2(Z)$ with zero cost. HenceThe construction needs a nontrivial winding-one pattern; merely tying a local knot into the product core would not provide this lower bound.

The dual Thurston polytope is the polar of the Thurston norm unit ball. More intrinsically, in the real dual of $H_2(Y,\partial Y)$ it isThe pairing can be regarded as evaluation of $H^2(Y,\partial Y;\mathbb{R})$ on relative homology. Its definition remains valid when the Thurston norm has a kernel: the polytope then lies in the annihilator of that kernel.

Identify $Y$ with the pair of pants product $P\times S^1$. A regular Seifert fiber has homology class $h={\mu }_1+{\mu }_2+{\mu }_3$. Let $e_i$ be the relative homology class corresponding by Poincare-Lefschetz duality to the homomorphism taking the $i$th meridian to one and the other two to zero. A spanning disk for $L_1$ punctured once by each of $L_2,L_3$ is an embedded pair of pants $F$ representing $e_1$, with $c(F)=1$.

Take two arcs in $P$, one joining boundary one to boundary two, the other joining boundary one to boundary three. Their products with $S^1$ are embedded vertical surfaces in a Seifert fibered space, namely annuli $A_{12},A_{13}$. Orient them so that their relative classes are $e_1-e_2$ and $e_1-e_3$. They cost zero. Thus the three required inequalities, with ${\alpha }_i=\alpha (e_i)$, areFor completeness they give the whole polytope. Oriented cut-and-paste of copies of $F,A_{12},A_{13}$ gives the upper bound $\Vert a{e}_1+b{e}_2+c{e}_3{\Vert }_T\le |a+b+c|$. For the reverse bound, compress a minimizing surface and use the classification of incompressible surfaces in Seifert fibered spaces. Its horizontal components cover $P$ and have negative Euler characteristic equal to their unsigned covering degree; its vertical components have zero cost and zero intersection with a regular Seifert fiber. The total signed horizontal degree is $a+b+c$, so its cost is at least $|a+b+c|$. HenceIt is a line segment, because this Thurston norm has a two-dimensional kernel.

The degree and its local signs. Orient $S^n$ and let $[S^n]$ generate its top reduced homology ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree is the integer characterized byFor $n\ge 1$ this is the usual top homology definition using the fundamental class; using reduced homology also covers $S^0$.

Suppose $f$ is smooth and $p$ is a regular value. Each $x\in f^{-1}(p)$ has invertible tangent map $D{f}_x$, so the inverse function theorem makes $f^{-1}(p)$ discrete. It is also closed in the compact sphere, hence finite. Choose disjoint small neighborhoods of these points on which $f$ is a local diffeomorphism. The Excision theorem identifies the source local homology with the direct sum of one copy of $\mathbb{Z}$ for each inverse image. The induced local map is multiplication by $+1$ or $-1$ according as $D{f}_x$ preserves or reverses orientation. The map from the global fundamental class to these local orientation classes therefore proves the degree as a sum of local degrees formula:Here the determinant is computed in positively oriented tangent bases. Thus the mapping degree counts inverse images with signs, rather than just their cardinality.

The quotient map. Put $m=n+1$ and write $f=F{|}_{S^{m-1}}$. Give $D^m$ its standard orientation and its boundary the induced orientation. The connecting homomorphismis an isomorphism: the disk has zero positive reduced homology. It sends the relative fundamental class $[D^m,S^{m-1}]$ to $[S^{m-1}]$. If the map on relative homology induced by $F$ multiplies this class by $d$, naturality givesThus $d=\mathrm{deg}f$. Collapsing the boundary gives a sphere $D^m/S^{m-1}\cong S^m$, and the quotient map identifies its top reduced homology with the top relative homology of the disk pair. Give this quotient sphere the orientation determined by that identification. The relation $qF=\tilde{F}q$ now showsThis is the quotient-sphere degree identity.

The graph intersection. Orient $D^m\times D^m$ by the product orientation, orient $N=D^m\times \{0\}$ by its first factor, and orient the graph of a function $\Gamma$ by $x\mapsto (x,F(x))$. An intersection is precisely a zero of $F$, and no such zero lies on the boundary because $F(S^{m-1})\subset S^{m-1}$. At an intersection, transverse intersection means that $D{F}_x$ is surjective, hence invertible. The zeros are consequently isolated and finite.

For the smooth intersection number use the ordered tangent spaces $T_{x}N$ first and $T_{(x,F(x))}\Gamma$ second. Relative to the product basis their concatenated basis has matrixIts determinant is $\det D{F}_x$, so the intersection sign is ${ϵ}_x=\mathrm{sgn}\det D{F}_x$. By the same local Excision theorem argument, now in relative homology at the interior point $0$, the sum of these signs equals the multiplier of $F_{\ast }$ on $H_m(D^m,S^{m-1};\mathbb{Z})$. Therefore the graph intersection formula for mapping degree isIf the tangent spaces are ordered $\Gamma$ first and $N$ second, every sign changes by $(-1{)}^m$; the order above specifies the appropriate convention.

The one-dimensional model $F(x)=2x^3-x$ on $[-1,1]$ illustrates the graph intersection formula for mapping degree: its three zeros have signs $+1,-1,+1$, and its endpoint map has mapping degree $1$ on reduced homology.

For the sphere homology calculation use singular homology, homotopy invariance of homology, and the reduced Mayer–Vietoris sequence, not cellular homology. For a point, its chain group for singular homology is $\mathbb{Z}$ in every nonnegative degree, and its boundary operator is multiplication by ${\sum }_{i=0}^q(-1{)}^i$, hence alternately zero and one in positive degrees. Its homology is therefore $\mathbb{Z}$ in degree zero and zero above. Singular chains of a disjoint union split into the summand chain complexes. Start with $S^0$, the disjoint union of two points: $H_0(S^0;\mathbb{Z})={\mathbb{Z}}^2$, all positive groups vanish, and ${\tilde{H}}_0(S^0;\mathbb{Z})=\mathbb{Z}$. For $n\ge 1$, remove the north and south poles to obtain an open cover $U,V$ of $S^n$. Each member is contractible, while $U\cap V$ has a deformation retraction to $S^{n-1}$, including the disconnected intersection when $n=1$. The reduced exact sequence therefore suppliesAlso $S^n$ is connected, so its reduced degree-zero group vanishes. Induction givesThe separate $S^0$ calculation avoids conflating its two degree-zero generators.

Choose an orientation of $S^n$, with $n\ge 1$, and let $[S^n]$ be its fundamental class. The mapping degree is the integer determined by $f_{\ast }[S^n]=\mathrm{deg}(f)[S^n]$. To define the local degree at $x_0$, set $y_0=f(x_0)$ and require $x_0$ to be isolated in $f^{-1}(y_0)$. Choose an open neighborhood $B$ containing no other point of that fiber. Then $f$ defines a map of pairsThe degree-$n$ relative homology groups on both sides are $\mathbb{Z}$, by a manifold chart and excision. The local orientation of a manifold chooses their generators. The induced map multiplies these generators by an integer, denoted ${\mathrm{deg}}_{x_0}f$. Excision and naturality show independence of the choice of $B$. Differentiability is unnecessary; if $f$ is smooth and $D{f}_{x_0}$ is invertible, the inverse function theorem gives ${\mathrm{deg}}_{x_0}f=\mathrm{sign}\det D{f}_{x_0}$ relative to the orientations. Without isolation in the fiber this pointwise definition need not apply.

For the degree as a sum of local degrees, assume $y\in S^n$ has finite fiber $F=\{x_1,\dots ,x_s\}$. Disjoint coordinate neighborhoods and excision identifyThe absolute fundamental class maps to the tuple of local orientation generators. Under the relative map induced by $f$, the $j$th generator goes to ${\mathrm{deg}}_{x_j}f$ times the generator at $y$, and the resulting map from the direct sum adds these contributions. On the other hand, naturality says that first applying the absolute map $f_{\ast }$ and then passing to the local group at $y$ gives $\mathrm{deg}(f)$ times that generator. ThusIf the fiber is empty the sum is zero: the map factors through the contractible punctured sphere and has degree zero. If every point of a fiber is isolated, compactness makes that closed fiber finite, so the theorem applies. In particular, a regular value of a smooth map gives the signed count of its inverse images. For $n=0$, define a mapping degree on the reduced generator $[p_{+}]-[p_{-}]$: the identity has mapping degree one, the swap of the two points minus one, and either constant map zero. With local orientation signs positive at $p_{+}$ and negative at $p_{-}$, the same sum formula holds. The connected-sphere convention above is the standard positive-dimensional one.

Collapsing the boundary identifies the top relative homology of a disk pair with the top reduced homology of its quotient sphere. Together with the boundary connecting homomorphism, this shows that the induced quotient map and the boundary map have the same mapping degree, using the corresponding orientations.