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An oriented normal bundle of a compact embedded submanifold has a Thom class. A tubular neighborhood and excision identify it with a relative class in $H^d(X,X\setminus Y)$. Its image in absolute cohomology is ${\epsilon }_Y$. It restricts to zero off Y and, in an oriented closed ambient manifold, represents the Poincare dual of Y with the normal-first induced orientation.

Let $M$ be a closed oriented $n$-manifold, let $e_{p,i}$ be a homogeneous basis of $H^p(M;\mathbb{Q})$, and choose its Poincare-dual basis $e_{p,i}^{\ast }\in H^{n-p}(M;\mathbb{Q})$ by $⟨e_{p,i}{e}_{p,j}^{\ast },[M]⟩={\delta }_{ij}$. With the product-orientation convention, the Poincare dual of the diagonal is

The cycle class of an algebraic cycle is its image under the cycle class map. For a codimension-$r$ cycle on a smooth complex projective variety, it belongs to $H^{2r}(X,\mathbb{Z})$ and is the Poincare dual of its fundamental homology class. The map factors through the Chow group, but a Chow class retains finer information than its cohomological cycle class.

For a smooth complex projective algebraic variety, the cycle class map sends an algebraic cycle of codimension $r$ to the Poincare dual of its homology class. Rational equivalence gives the same cohomology class, so the map descends to the Chow group. It respects intersection products, pullback and proper pushforward; a cohomological obstruction therefore obstructs the corresponding proposed identity of Chow classes.

Give Complex projective space its complex orientation. In the homology cross product basis of $H_{2n}({\mathbb{CP}}^n\times {\mathbb{CP}}^n;\mathbb{Z})$, the diagonal has the displayed class. If $a,b$ are the two normalized degree-two hyperplane classes, its Poincare dual is ${\sum }_{j=0}^n{a}^j{b}^{n-j}$. Indeed every degree-$2n$ monomial $a^r{b}^{n-r}$ restricts on the diagonal to the top power of the hyperplane class, whose evaluation on the fundamental class is one. These evaluations determine all coefficients.

The diagonal map sends $x$ to $(x,x)$. It is an embedding of a topological space onto its diagonal subset, because either coordinate projection is its inverse on the image. For a closed oriented manifold, that subset has a fundamental class, whose Poincare dual is the cohomology class of the diagonal.

The cohomology ring of the product of two real projective planes is ${\mathbb{F}}_2[x,y]/(x^3,y^3)$, with $|x|=|y|=1$. Pairing against $x^2$, $xy$ and $y^2$ identifies the Poincare dual of the diagonal as $x^2+xy+y^2$. For a self-map $f$ with $f^{\ast }x=\epsilon x$, the graph pullback evaluates on the mod-two fundamental class as $1+\epsilon +{\epsilon }^2=1$. Thus no graph of a map on the whole projective plane can avoid the diagonal. The argument uses a relative Thom class and does not assume transverse intersections.

The standard CW complex structure on Complex projective space ${\mathbb{CP}}^3$ has one cell in each dimension $0,2,4,6$ and no other cells. Its cellular chain complex therefore has zero differentials: each boundary has an odd-dimensional target group, which is zero. The product of the usual two-cell structures on $S^2$ and $S^4$ likewise has exactly one cell in dimensions $0,2,4,6$. Thus cellular homology gives, for both spaces,The cohomology rings detect more than these additive groups. Let $c\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ be the hyperplane class. The cohomology ring of complex projective space isFor the normalization, a hyperplane is Poincare dual to $c$; intersecting three generic hyperplanes gives one point with positive complex orientation, so $⟨c^3,[{\mathbb{CP}}^3]⟩=1$. Together with the computed additive groups this makes $c,c^2,c^3$ generators in their respective degrees.

Let $u$ and $v$ be the pullbacks of the orientation classes of $S^2$ and $S^4$. The Künneth theorem and naturality of the cup product givewhere $uv$ generates degree six. In particular, every degree-two class on the product has square zero, whereas $c^2\ne 0$. Any graded ring isomorphism would send $c$ to $\pm u$, contradicting this product relation. A homotopy equivalence induces an isomorphism of cohomology rings, so the spaces have isomorphic integral homology but are not homotopy equivalent.

For an oriented real vector bundle $\pi :E\to X$ of rank $d$, the Thom isomorphism theorem gives a unique Thom class $u_E\in H^d(D(E),S(E);\mathbb{Z})$ restricting to the chosen generator in each fibre, and isomorphismsEquivalently use the pair $(E,E\setminus X)$, identifying $X$ with the zero section. After forgetting relative supports, pull the Thom class back along the zero section $s$ to define the Euler class:In this expression the relative class is first mapped to absolute cohomology.

A compact embedded submanifold equipped with a coorientation $Y\subset X$ has an oriented normal bundle ${\nu }_Y$ of rank $d$. A tubular neighborhood and excision identify its Thom class with a class $u_Y\in H^d(X,X\setminus Y)$. Its image in $H^d(X)$ is the cohomology class of a cooriented submanifold ${\epsilon }_Y$. In particular it restricts to zero on $X\setminus Y$. If the ambient manifold is oriented, orient $Y$ by the normal-first convention. The class is its Poincare dual, characterized byfor the appropriate complementary degree. This is the stated integration identity in cohomological notation.

First prove the Lefschetz fixed-point theorem for an oriented closed n-manifold $X$. Orient its diagonal by the identification $\Delta \cong X$, and choose the normal orientation so this agrees with the normal-first convention in the product. This produces the integral class ${\epsilon }_{\Delta }$. To compute the part that contributes to the trace, tensor with $\mathbb{Q}$: the trace on a finitely generated integral cohomology group means its trace on the free quotient, equivalently on rational cohomology; torsion contributes nothing.

Choose a basis $e_{p,j}$ of $H^p(X;\mathbb{Q})$ and its Poincare duality pairing dual basis $e_{p,j}^{\vee }\in H^{n-p}(X;\mathbb{Q})$, normalized by $⟨e_{p,i}⌣e_{p,j}^{\vee },[X]⟩={\delta }_{ij}$. The rational image of the cohomology class of the diagonal isIndeed, pairing this expression against $\alpha \times \beta$ with degrees $p,n-p$ picks out the p-summands. The cross-product multiplication contributes $(-1{)}^{p^2}$ and moving $\alpha$ past $e_{p,j}^{\vee }$ contributes $(-1{)}^{p(n-p)}$. Together with $(-1{)}^{np}$ their product is $+1$. The result is $⟨\alpha ⌣\beta ,[X]⟩$, exactly the diagonal's defining identity. The rational Künneth theorem and nondegeneracy of the duality pairing determine the class. An integral torsion component, if present, is not determined by this rational expression and does not affect any evaluation below.

For the graph map ${\Gamma }_f=(1,f):X\to X\times X$, write $f^{\ast }{e}_{p,j}={\sum }_i{A}_{ij}{e}_{p,i}$. Pulling back the expression and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph lies in $(X\times X)\setminus \Delta$. The relative Thom class restricts to zero there, so ${\Gamma }_f^{\ast }{\epsilon }_{\Delta }=0$ and $L(f)=0$. The contrapositive proves the required fixed-point assertion, without requiring smoothness of $f$ or transversality of its graph.

The printed hypothesis does not require $X$ to be orientable. For a nonorientable $X$, the diagonal's normal bundle is $TX$, so an ordinary integral coorientation cannot be assumed. We extend the proved theorem as follows, rather than apply the untwisted diagonal computation incorrectly. We first construct a smooth embedding in Euclidean space using finitely many charts ${\chi }_i:U_i\to {\mathbb{R}}^n$ and smooth functions ${\rho }_i$ supported in those charts, with at least one ${\rho }_i(x)>0$ at every point. The map $\Psi (x)=({\rho }_i(x),{\rho }_i(x){\chi }_i(x){)}_i$, with zero extensions, is injective: a positive first coordinate identifies a chart in which its remaining coordinates recover $x$. Its differential is injective as well, since $d{\rho }_i(v)=0$ and $d({\rho }_i{\chi }_i)(v)=0$ force $d{\chi }_i(v)=0$ in a chart with ${\rho }_i>0$. Compactness makes this injective immersion a homeomorphism onto its image. This finite-chart Euclidean embedding of a compact smooth manifold, enlarged by zero coordinates if necessary, has positive codimension. A compact tubular disc neighbourhood $W$ is an oriented N-manifold with boundary, inheriting its orientation from the ambient Euclidean space. Its double of a manifold $M$ is closed and oriented. Projection of each copy of $W$ to $X$ agrees on their common boundary, giving $r:M\to X$; inclusion into the first copy gives $i:X\to M$ with $ri=1_X$.

Set $F=ifr$. A fixed point of $F$ must lie in $i(X)$, and then is exactly the image of a fixed point of $f$. Moreover $F^{\ast }=r^{\ast }{f}^{\ast }{i}^{\ast }$, and cyclicity of trace for maps between finite-dimensional rational cohomology groups givesin each degree, because $i^{\ast }{r}^{\ast }=1$. Thus $L(F)=L(f)$. The oriented theorem for $M$ now gives a fixed point of $F$, hence of $f$, whenever $L(f)\ne 0$. This oriented manifold retract for the Lefschetz theorem proves the full printed result. The double construction also works componentwise if necessary.

Let $\pi :E\to B$ be an oriented real vector bundle of rank $r$ over a paracompact base. A Thom class is a class $U\in H^r(E,E\setminus B;\mathbb{Z})$ restricting on every oriented fiber pair $({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ to its positive generator. The Thom isomorphism theorem states that this class exists uniquely with those fiber restrictions and thatHere $B$ is identified with the zero section. Equivalently use the disk/sphere bundle pair. The compact-support version is $H_c^q(B;\mathbb{Z})\cong H_c^{q+r}(E;\mathbb{Z})$, with fiberwise compact support in the Thom construction. If the bundle is not oriented, the integral assertion requires the corresponding orientation local system.

For a closed co-oriented embedded submanifold $Y$ of codimension $r$, co-orientation means an orientation of its normal bundle. A tubular neighborhood identifies a neighborhood of $Y$ with that bundle. Its Thom class, transferred by excision, is a class in $H^r(M,M\setminus Y;\mathbb{Z})$. For compact $Y$ this maps into the direct limit defining compactly supported cohomology, giving $\boxed{{\epsilon }_Y\in H_c^r(M;\mathbb{Z})}$. The construction is independent of the tubular neighborhood and uses the chosen co-orientation. In an oriented ambient manifold it is the Poincare dual of the oriented submanifold. Here “closed submanifold” has the usual compact-without-boundary meaning; if it meant merely a closed subset and $Y$ were noncompact, the supported relative class would still exist but need not have compact support.

We next prove the fixed-point implication without imposing an unprinted orientability assumption on $M$. SetUse a finite triangulation of the closed smooth manifold. The elementary Hopf trace identity says that the alternating trace of a chain endomorphism equals its alternating trace on homology. To verify it, split each finite-dimensional chain space as boundaries, a complement representing homology, and a complement mapping isomorphically onto the preceding boundaries. The traces on these two boundary summands cancel in adjacent degrees, leaving exactly the homology traces. The cohomology traces equal the homology traces because the induced maps are dual over $\mathbb{Q}$.

If $f$ has no fixed points, compactness gives $\delta ={\min }_{x}d(x,f(x))>0$. Choose a triangulation $K$ of mesh less than $\delta /4$ and a sufficiently fine subdivision $K^{\prime }$ admitting a simplicial approximation $F:K^{\prime }\to K$ uniformly within $\delta /4$ of $f$. The subdivision chain map $s:C_{\ast }(K)\to C_{\ast }(K^{\prime })$ sends each simplex to simplices inside it. For a vertex $v$ of such a small simplex, $d(v,F(v))>3\delta /4$, so $F(v)$ cannot lie in the original coarse simplex. Consequently $F_{\#}s$ has zero diagonal coefficient on every coarse simplex. Its trace vanishes in every degree, while it induces $f_{\ast }$ on homology. The trace identity yields $L(f)=0$. Taking the contrapositive provesThis is the Lefschetz fixed-point theorem, established here rather than invoked. For oriented $M$ the same identity is the graph-diagonal formula for the Lefschetz number: $L(f)=⟨(1,f{)}^{\ast }{\epsilon }_{\Delta },[M]⟩$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same cohomology maps, so their Lefschetz numbers agree.

Let the integer degree of a map between oriented manifolds be $d\ne 0$, so $f_{\ast }[X]=d[Y]$. For any nonzero $\alpha \in H^k(Y;\mathbb{Q})$, nondegeneracy of the Poincare duality pairing supplies $\gamma \in H^{n-k}(Y;\mathbb{Q})$ with $⟨\alpha ⌣\gamma ,[Y]⟩\ne 0$. Naturality of the cup product and its evaluation givesThus $f^{\ast }\alpha \ne 0$, proving injectivity of pullback on rational cohomology. By the universal coefficient theorem for cohomology, these finite-dimensional rational cohomology groups are the duals of the corresponding homology groups, and the dual of $f_{\ast }$ is $f^{\ast }$. Injectivity of the dual map therefore provesOne can see the surjection constructively. Given $u\in H_k(Y;\mathbb{Q})$, choose its Poincare dual $\eta \in H^{n-k}(Y;\mathbb{Q})$ with $\eta \cap [Y]=u$. Then $v=d^{-1}(f^{\ast }\eta \cap [X])$ satisfies $f_{\ast }v=d^{-1}(\eta \cap f_{\ast }[X])=u$ by naturality of the cap product. Division by $d$ is the reason rational coefficients suffice.

The cellular homology of complex projective space gives one integral generator $[{\mathbb{CP}}^j]$ in each degree $2j$, for $0\le j\le 2$, and zero in odd degrees. Since these groups are free, the Künneth theorem gives the homology cross product basisThus the ranks in degrees $0,2,4,6,8$ are $1,2,3,2,1$, respectively, and every other integral homology group vanishes.

Let $h$ be the normalized hyperplane class, with $⟨h^2,[{\mathbb{CP}}^2]⟩=1$, and write $a=p_1^{\ast }h$, $b=p_2^{\ast }h$. The cohomology ring of complex projective space and the Künneth theorem give $H^{\ast }({\mathbb{CP}}^2\times {\mathbb{CP}}^2;\mathbb{Z})=\mathbb{Z}[a,b]/(a^3,b^3)$. If $d$ is the diagonal embedding, then $d^{\ast }a=d^{\ast }b=h$. Hence each of $a^2,ab,b^2$ evaluates as one on the diagonal's fundamental class. These three classes are dual to the three degree-four product generators, so the diagonal class of complex projective space givesEquivalently its Poincare dual is $a^2+ab+b^2$. The signs are positive with the complex and product orientations.

For finite-dimensional Complex projective space ${\mathbb{CP}}^r$, use the filtration by its linear projective subspaces. The difference ${\mathbb{CP}}^j\setminus {\mathbb{CP}}^{j-1}$ is ${\mathbb{C}}^j$, so there is one topological cell in every even dimension $0,2,\dots ,2r$, and none in odd dimensions. Every cellular boundary vanishes for degree reasons. The integral cohomology is therefore $\mathbb{Z}$ in those even degrees and zero elsewhere.

For the cohomology ring, let $x\in H^2({\mathbb{CP}}^r;\mathbb{Z})$ be the Poincare dual of a projective hyperplane, using the complex orientation. Generic hyperplanes meet transversely, and a collection of $j$ of them intersects in ${\mathbb{CP}}^{r-j}$. The geometric interpretation of the cup product identifies $x^j$ with the dual of that intersection. Restricting to a linear ${\mathbb{CP}}^j$, the $j$ hyperplanes meet in one point, with positive sign because the transverse normal directions are complex lines. HenceThus $x^j$ is a primitive generator in degree $2j$, not merely a nonzero multiple. There are no classes beyond degree $2r$, soThe point case $r=0$ has ring $\mathbb{Z}$. Taking the union of the finite projective spaces gives $H^{\ast }({\mathbb{CP}}^{\infty };\mathbb{Z})=\mathbb{Z}[x]$, since each fixed cohomological degree stabilizes. This proves the cohomology ring of complex projective space computation.

For the rank bound for a nonsingular complex bilinear map, write $a={\dim }_{\mathbb{C}}U$, $b={\dim }_{\mathbb{C}}V$, $R=\mathrm{im}\varphi$, and $s={\dim }_{\mathbb{C}}R$. Assume $a,b\ge 1$. Injectivity with either nonzero factor fixed implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Thereforeis well defined and continuous. Scaling either vector only scales the output, and in local projective charts the formula is given by bilinear coordinates with no simultaneous zero. Let $h$ be the target's degree-two generator and $x,y$ the pullbacks of the two factor generators. The Künneth theorem givesFixing $[v]$ or $[u]$ makes $F$ a projective linear embedding induced by an injective linear map. Such an embedding pulls back the hyperplane class to the factor's hyperplane class. Since the two restrictions determine $H^2$ of the product, $F^{\ast }h=x+y$, with the generator of a point factor interpreted as zero.

Put $d=a+b-2$. In the source cohomology ring, every term in $(x+y{)}^d$ vanishes except the one with exponents $a-1,b-1$, andThe binomial coefficient is a positive integer and the final monomial generates the top integral cohomology. If $s\le d$, the class $h^d$ in ${\mathbb{CP}}^{s-1}$ would be zero. This contradiction provesWhen $a=b=1$, the nonzero pure-tensor image already gives $s\ge 1$. A positivity convention is genuinely necessary: if $U=0$, $V={\mathbb{C}}^2$, and $W=0$, all restrictions to the zero tensor subspaces are injective, but the literal bound would read $0\ge 1$. The displayed theorem is therefore the nonzero-factor version. If a nonzero factor is infinite-dimensional, either factor restriction already bounds the image dimension below by that factor dimension; the sum of two positive cardinal dimensions, with at least one infinite, is their maximum, giving the corresponding cardinal bound.

Use cohomology with coefficients in ${\mathbb{F}}_2={\mathbb{Z}}_2$. The cellular homology of real projective space has one cell in degrees zero, one and two, and its boundary coefficients become zero modulo two. Hence there is a one-dimensional cohomology group in each of those degrees. Let $x\in H^1(P;{\mathbb{F}}_2)$ be the Poincare dual of a projective line. Two distinct projective lines intersect transversely in one point, so $⟨x^2,[P{]}_2⟩=1$. ThusThis is the mod-two cohomology ring of real projective space in dimension two. The Künneth theorem over a field makes the cross product an isomorphism of graded rings. Writing $x,y$ for the factor pullbacks givesThere are no additional signs over ${\mathbb{F}}_2$.

The obstruction to a graph avoiding the diagonal is the mod-two diagonal class of the real projective plane. Its dual class $D\in H^2(P\times P;{\mathbb{F}}_2)$ is determined bywhere $i:P\to P\times P$ is the diagonal embedding. Test $z=x^2,xy,y^2$. Each diagonal pullback is $x^2$, with evaluation one. Since $x^2{y}^2$ is the product's top generator, comparison of coefficients givesIf the submanifold were the graph of a smooth map $f:P\to P$, its parametrization would be $j=(\mathrm{id},f)$. For some $\epsilon \in {\mathbb{F}}_2$, $f^{\ast }x=\epsilon x$, and thereforeOn the other hand, the Poincare dual of the diagonal comes from its relative Thom class in $H^2(P\times P,(P\times P)\setminus \Delta ;{\mathbb{F}}_2)$. It restricts to zero on the complement. A graph disjoint from the diagonal would factor through that complement and force $j^{\ast }D=0$, a contradiction. Such a submanifold cannot be the graph of a smooth map $P\to P$. The relative Thom-class argument also applies to continuous self-maps.

Cells and attaching maps. Regard ${\mathbb{RP}}^k$ as the lines in ${\mathbb{R}}^{k+1}$ and include ${\mathbb{RP}}^{k-1}$ as the lines whose last coordinate is zero. Its complement consists of lines with a unique representative $(x_1,\dots ,x_k,1)$, and is therefore an open $k$-cell. Inductively this makes Real projective space a CW complex with one cell in each dimension from zero to $n$.

A characteristic map is obtained from the northern closed hemisphere of $S^k$: send a unit vector to the line it spans. Its interior maps homeomorphically onto the open cell, while its equator has the antipodal identification. Thus the $k$-cell is attached by the quotient mapwhich is the antipodal two-sheeted covering. This includes the two endpoints of the one-cell attaching to the zero-cell.

The cellular chain complex has $C_k^{\mathrm{cell}}=\mathbb{Z}$ for $0\le k\le n$ and zero otherwise. To compute its differential, follow the attaching map by collapse of the $(k-2)$-skeleton. The resulting map to ${\mathbb{RP}}^{k-1}/{\mathbb{RP}}^{k-2}\cong S^{k-1}$ has two local contributions. They differ by the mapping degree $(-1{)}^k$ of the antipodal map on $S^{k-1}$. With compatible cell orientations,For $k=1$ the two oriented endpoints cancel, giving the same formula. Consecutive differentials compose to zero, as required.

The mod-two cup products. Modulo two every cellular differential vanishes, so cellular cohomology gives a one-dimensional group in each degree $0,\dots ,n$. Let $a\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$ be the Poincare dual of a projective hyperplane. This class is nonzero: a projective line transverse to that hyperplane meets it once. Intersecting $j$ generic projective hyperplanes produces ${\mathbb{RP}}^{n-j}$, and the cup product of their Poincare duals is the Poincare dual of that intersection. In particular, $a^n$ evaluates to one on the mod-two fundamental class. Therefore every $a^j$, $0\le j\le n$, is nonzero, since otherwise multiplying it by $a^{n-j}$ would contradict $a^n\ne 0$. Dimension makes $a^{n+1}=0$. The mod-two cohomology ring of real projective space isFor $n=0$ this simply means the cohomology of a point.

The product with integral coefficients. The final product's coefficients are unstated; take $\mathbb{Z}$ as the default. Dualizing the cellular chain complex above giveswith all unlisted groups zero. The integral Künneth theorem has tensor terms with $i+j=q$ and Tor functor terms with $i+j=q+1$:For these finite free cellular complexes it splits additively, though not canonically. Both the tensor and Tor functor of two $\mathbb{Z}/2$ summands give $\mathbb{Z}/2$, so no order-four summands occur.

For an efficient count, write $F_A(t)$ for the free-rank polynomial and $T_A(t)$ for the number of $\mathbb{Z}/2$ summands in each degree. The integral Künneth torsion polynomial rule isThe last two factors record respectively the tensor contribution in summed degree and the Tor functor contribution one degree lower. The three factors haveThe first two give $F_{23}=1+t^3$ and $T_{23}=2t^2+t^3+t^4+t^5$. Multiplying by the third givesConsequently the integral cohomology of a product of finite real projective spaces in this case isIf the intended coefficients were instead ${\mathbb{F}}_2$, the Künneth theorem over a field gives the dimension polynomialThus the mod-two groups in degrees zero through nine are respectivelyand all other degrees vanish.

The assertion is true. The standard filtration ${\mathbb{CP}}^0\subset \cdots \subset {\mathbb{CP}}^n$ gives one cell in each dimension $0,2,\dots ,2n$. There are no odd-dimensional cells, so all cellular differentials vanish. The cellular homology theorem and the universal coefficient theorem for cohomology give one copy of $\mathbb{Z}$ in each even degree from zero to $2n$, and zero in every other degree.

Let $x$ be the Poincare dual of a projective hyperplane, with the complex orientation. It has degree two and evaluates to $+1$ on a complex projective line, so it is the positive generator of $H^2$. We use the intersection interpretation of the cup product: the product of duals of oriented submanifolds in transverse position is the dual of their oriented intersection. Distinct transverse complex hyperplanes intersect in ${\mathbb{CP}}^{n-j}$ after $j$ intersections, with positive complex orientation. Thus $x^j$ is the dual of that linear subspace.

Pairing $x^j$ with a transverse linear ${\mathbb{CP}}^j$ gives one positively oriented intersection point. Therefore $x^j$ is a primitive generator of $H^{2j}$, for every $0\le j\le n$. There is no cohomology above dimension $2n$, so $x^{n+1}=0$. These facts show that the surjective graded ring map from $\mathbb{Z}[x]$ has exactly the indicated kernel:This proves the cohomology ring of complex projective space, rather than only its additive groups. For $n=0$ it is $\mathbb{Z}$, with $x=0$.

Let $m=\dim M$. Choose a homogeneous basis $e_{q,i}$ of $H^q(M;\mathbb{Q})$ and its Poincare dual basis $e_{q,i}^{\ast }\in H^{m-q}(M;\mathbb{Q})$, normalized byWith the product orientation, the cohomology class of the diagonal isIndeed, multiplying this class by $\alpha \times \gamma$ and evaluating on $[M\times M]$ gives $⟨\alpha ⌣\gamma ,[M]⟩$, which characterizes the Poincare dual of $\Delta$.

Pulling back along the graph map $(1,f):M\to M\times M$ and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph is disjoint from $\Delta$. Represent ${\epsilon }_{\Delta }$ with support in a tubular neighbourhood disjoint from the graph; its pullback is zero, so $L(f)=0$. Contrapositively, $L(f)\ne 0$ implies that $f$ has a fixed point, the Lefschetz fixed-point theorem.

Now let $M$ be three disjoint circles. A homeomorphism permutes their three components. Its action on $H^1(M;\mathbb{Z})\cong {\mathbb{Z}}^3$ is a signed permutation matrix. A component fixed setwise contributes to the trace by the degree of the corresponding circle homeomorphism. An orientation-reversing circle homeomorphism has a fixed point, so fixed-point-freeness forces that degree to be $+1$. Nonfixed components contribute zero. The trace is therefore the number of fixed points of a permutation of three objects, and

For a compact manifold $N$ with boundary, let $D(N)=N_{+}{\cup }_{\partial N}{N}_{-}$ be its double and define $F$ by applying $f$ on both copies. The Mayer–Vietoris sequence for this decomposition is natural under $F$. Alternating traces in a finite-dimensional exact sequence sum to zero, so the two copies of $N$ contribute twice and their intersection contributes with the opposite sign:This is the Lefschetz number of a doubled map.

Finally let $N$ be a pair of pants. If $f$ is fixed-point-free, so are its double $F$ and its boundary restriction. The Lefschetz fixed-point theorem and the displayed identity give $L(F)=L(f{|}_{\partial N})=0$, hence $L(f)=0$. Since $N$ is connected and $H^i(N;\mathbb{Q})$ is nonzero only for $i=0,1$,Suppose the boundary permutation $\sigma$ had a fixed component. The restriction there is a fixed-point-free circle homeomorphism and thus has degree $+1$, forcing $f$ to preserve the surface orientation. The homology action of a pair-of-pants homeomorphism would then have trace $\#\mathrm{Fix}(\sigma )-1$, which is $2$ for the identity permutation and $0$ for a transposition. Neither is $1$. Therefore $\sigma$ has no fixed component; a permutation of three objects with no fixed point is a three-cycle. Thus