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Every finitely presented group is the fundamental group of a closed connected smooth oriented four-manifold. Build a compact four-dimensional handle decomposition with one zero-handle, a one-handle for each generator of a group and a two-handle for each relator. Turning the handle decomposition upside down shows that its boundary surjects on the fundamental group. Its double of a manifold has the same fundamental group by the Seifert-van Kampen theorem.

A chain complex consists of abelian groups $C_q$ and homomorphisms ${\partial }_q:C_q\to C_{q-1}$ with ${\partial }_{q-1}{\partial }_q=0$. Its groups of chain cycles and chain boundaries are $Z_q=\ker {\partial }_q$ and $B_q=\mathrm{im}{\partial }_{q+1}$; the identity ensures $B_q\subseteq Z_q$. Its homology groups areFor a nonnegative complex, take $C_q=0$ for $q<0$.

The singular chain group $C_q(X)$ is the free abelian group on continuous maps $\sigma :{\Delta }^q\to X$, called singular simplices. With ${\delta }_i$ the inclusion of the face opposite vertex $i$, defineDeleting any two vertices in the two possible orders gives the same face with opposite signs, proving ${\partial }^2=0$. This defines the singular chain complex and its singular homology. A continuous map $f$ induces the chain map $f_{\#}(\sigma )=f\circ \sigma$; it commutes with the boundary because composition commutes with restriction to faces.

To prove homotopy invariance of homology, let $F:X\times I\to Y$ have endpoints $f,g$. Triangulate ${\Delta }^q\times I$ by the $q+1$ simplices with ordered verticesLet $p_i:{\Delta }^{q+1}\to {\Delta }^q\times I$ be their affine parameterizations. The singular prism operator isOn taking boundaries, shared internal faces of consecutive prism simplices cancel. The top and bottom faces give $g_{\#}\sigma -f_{\#}\sigma$, and the remaining faces are $-P_{q-1}(\partial \sigma )$ with their induced signs. ThusFor a chain cycle $z$, $g_{\#}z-f_{\#}z=\partial Pz$ is a chain boundary, so $f_{\ast }=g_{\ast }$ on homology. If $f$ is a homotopy equivalence, its homotopy inverse therefore induces an inverse to $f_{\ast }$, proving that singular homology depends only on homotopy type.

Use the standard CW complex for an orientable genus-three surface: one vertex, six edges $a_1,b_1,a_2,b_2,a_3,b_3$, and a two-cell whose attaching word is ${\prod }_{j=1}^3[a_j,b_j]$. The cellular homology theorem identifies this cellular calculation with singular homology. The boundary of every edge is zero, and each edge's total exponent in the attaching word is zero, so the two-cell boundary is zero. HenceChoose $C_1$ to be the nonseparating curve represented by $a_1$, and choose $C_2$ as a small circle bounding a disc in a surface patch. Then $[C_1]$ is a nonzero primitive homology class, whereas the inclusion of $C_2$ induces zero on $H_1$. If a homotopy equivalence carried $C_1$ onto $C_2$, its restriction would factor through $C_2$ and kill $[C_1]$, contradicting injectivity on first homology. This is the homology obstruction to carrying surface curves onto one another.

There is nevertheless such a continuous map. Map the edge $a_1$ homeomorphically onto $S^1$ and all other edges to its base point. The two-cell attaching word then has winding number zero in $S^1$, hence is null-homotopic and the map extends to $r:{\Sigma }_3\to S^1$. Compose $r$ with a homeomorphism $S^1\to C_2$ and the inclusion of $C_2$ in the surface. This gives $g(C_1)=C_2$. Thus an arbitrary map can carry the first curve onto the second, but a homotopy equivalence cannot.

For an unambiguous drawing, realize ${\Sigma }_3$ as the double of a manifold formed from a disc with three holes. Its two copies are glued along all four boundary circles. The red curve below follows an arc from the outer boundary to the first hole on the front copy and returns along the matching arc on the back copy; it is a nonseparating curve. The blue circle is entirely in a disc patch on the front copy and is contractible.

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.

All homology and cohomology in this solution use ${\mathbb{F}}_2$ coefficients, so no orientation hypothesis is necessary. The Poincare-Lefschetz duality degrees are complementary:The starred notation in the question must be understood with this degree reversal.

Turn the handle decomposition upside down. An absolute $k$-handle becomes a relative $(n-k)$-handle based on $\partial W$. The absolute cellular boundary coefficient counts intersections of an attaching sphere with a belt sphere. In the reversed handle decomposition, the same intersections are counted in the opposite order. Over ${\mathbb{F}}_2$ the two incidence matrices are transposes, with no sign ambiguity. Thus the absolute chain complex is the complementary-degree dual of the relative chain complex. Taking homology and using the universal coefficient theorem for cohomology over a field proves the displayed isomorphism. This is mod-two handle duality.

To obtain a perfect pairing from the assumed closed case, form the double of a manifold $D(W)$ from two copies of $W$. The fold map $D(W)\to W$ is a retraction, so the inclusion of either copy induces an injection on homology. If $a\ne 0$ in $H_k(W)$, its image in $H_k(D(W))$ is nonzero. Closed-manifold mod-two Poincare duality supplies a class $z\in H_{n-k}(D(W))$ with intersection $1$ against that image. Cut a representative of $z$ along the boundary and retain its part in the first copy; it defines $b\in H_{n-k}(W,\partial W)$. Equivalently, apply the quotient map $D(W)\to D(W)/W_{-}\cong W/\partial W$ and excision. Intersections with a representative of $a$ in the interior are unchanged, so $a\cdot b=1$. This proves nondegeneracy in the first variable. Mod-two handle duality gives equal finite dimensions for the two spaces, hence nondegeneracy in the second variable as well.

For the three-dimensional conclusion, setThe long exact sequence in relative homology says $L=\mathrm{im}\partial$, where $\partial :H_2(W,\partial W)\to V$. The closed-surface intersection pairing on $V$ is nondegenerate. A collar calculation gives the adjoint identityIndeed, push $x$ slightly into the collar: intersections with the relative surface representing $b$ correspond to its boundary intersections with $x$. By the absolute-relative perfect pairing already proved, $x$ is orthogonal to every $\partial b$ precisely when $i_{\ast }x=0$. Therefore $L^{\perp }=L$.

For any finite-dimensional space with a perfect pairing, $\dim L+\dim L^{\perp }=\dim V$. Hence the half-lives-half-dies theorem givesBut $H_1({\mathbb{RP}}^2;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ has dimension one. It cannot be the entire boundary of a compact three-manifold, because the displayed dimension would be $1/2$. This excludes nonorientable three-manifolds as well as orientable ones.

A closed connected three-manifold has a handle decomposition with one zero-handle and one three-handle. To arrange this, cancel the zero-handles along a spanning tree of connecting one-handles; perform the corresponding construction in the dual handle decomposition to consolidate the three-handles. This does not require the three-manifold to be orientable.

If there are $r$ one-handles and $s$ two-handles, the Euler characteristic isMod-two Poincare duality pairs the Betti numbers of a closed odd-dimensional manifold, giving $\chi (Y)=0$. Thus $r=s$. The one-handles give $r$ generators of a group of the fundamental group, the attaching circles of the two-handles give $r$ relators, and the three-handle does not change the fundamental group. Therefore ${\pi }_1(Y)$ admits a balanced presentation.

Now letbe any finite group presentation. Start with a four-dimensional zero-handle and $r$ one-handles. Its fundamental group is the free group on the $x_i$. Represent the finitely many relators by disjoint embedded circles in its three-dimensional boundary; a small perturbation makes the circles disjoint. Choose framings of an embedded sphere and attach $s$ two-handles. The resulting compact connected oriented four-manifold $W$ has ${\pi }_1(W)\cong G$ by the Seifert-van Kampen theorem.

The map ${\pi }_1(\partial W)\to {\pi }_1(W)$ is surjective. One way to see this is to turn the handle decomposition upside down: the relative handles based on $\partial W$ have indices $2,3,4$, and so add no fundamental group generators. Also $\partial W$ is connected, since surgery on embedded circles in a connected three-manifold preserves connectedness.

Take the double of a manifoldIt is a closed connected smooth oriented four-manifold. The Seifert-van Kampen theorem givesBoth maps from ${\pi }_1(\partial W)$ are the same surjection, under the natural identification of the two copies. The pushout identifies the two copies of every element of $G$ and imposes no new relations: the fold homomorphism is inverse to either inclusion. ThusThis is four-manifold realization of finitely presented groups.