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A closed manifold is a compact manifold with boundary whose boundary is empty. Here “closed” includes compactness.

Attaching an index-$k$ handle to an $n$-dimensional manifold with boundary glues $D^k\times D^{n-k}$ along $\partial D^k\times D^{n-k}$. The complementary $D^k\times \partial D^{n-k}$ replaces the attaching region in the boundary.

An index-$\lambda$ handle in dimension $d$ is $D^{\lambda }\times D^{d-\lambda }$. A handle attachment glues its attaching region $S^{\lambda -1}\times D^{d-\lambda }$ to a corresponding neighborhood in the boundary of the preceding manifold with boundary. Besides the embedding of the attaching sphere, one specifies a framing of an embedded sphere, namely the normal-disk directions used in this gluing. Corners are then smoothed.

In dimension four, a two-handle is $D^2\times D^2$ and attaches along $S^1\times D^2$. Its attaching circle is a knot in the boundary of the zero-handle $D^4$. Identify that boundary with $S^3$ and remove a point away from the circles to draw them in ${\mathbb{R}}^3$. Disjoint attaching circles give a framed link. Each integer records the twist of its normal framing relative to the zero Seifert framing; the integer is also the linking number of the circle and its framed push-off. The resulting boundary change is Dehn surgery on a framed link. This is the two-handle part of a Kirby diagram.

For $S^2\times S^2$, use the following explicit handle decomposition. Write each factor as two disks, $S^2=D_{-}^2\cup D_{+}^2$, glued along their boundary circles. The four product pieces giveThe first is a four-dimensional zero-handle after rounding its corners. Its boundary is the union of the solid tori $S^1\times D^2$ and $D^2\times S^1$. The two middle product pieces attach along these tori. Their attaching circles $S^1\times \{0\}$ and $\{0\}\times S^1$ form the standard Hopf link, with linking number one after choosing orientations.

Both framings are zero. For example, the normal disk of $S^1\times \{0\}$ is the second disk factor. Its constant normal push-off has linking number zero with the original circle: the circle bounds $D^2\times \{0\}$ in the four-ball, and a parallel push-off bounds a disjoint translated disk. Thus the framed self-intersection is zero; the same argument applies with the factors interchanged. These product framings are the zero Seifert framings.

After these two two-handles, the remaining product piece $D_{+}^2\times D_{+}^2$ is a four-ball. Its boundary is exactly the boundary of the assembled first three pieces, so it attaches as a four-handle. Equivalently, zero surgery on the Hopf link gives $S^3$, which this handle caps. We have therefore constructed the actual product, provingThere are no one- or three-handles. As a check, the two factor spheres have self-intersection zero and meet once, giving the intersection form $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The explicit product construction, rather than the intersection form alone, identifies the smooth four-manifold. This is the zero-framed Hopf-link handle decomposition of the sphere product.

Let $M$ be an oriented smooth $m$-dimensional manifold with boundary and $\alpha$ a smooth compactly supported $(m-1)$-differential form. Give $\partial M$ the outward-normal-first orientation. A partition of unity subordinate to oriented interior and boundary charts reduces the proof to a compactly supported form on ${\mathbb{R}}^m$ or on a half-space. Indeed $\alpha ={\sum }_j{\chi }_j\alpha$, so $d\alpha ={\sum }_{j}d({\chi }_j\alpha )$; the sum of terms involving $d{\chi }_j$ is zero. Exterior derivative commutes with chart pullbacks, so each summand can be calculated in coordinates.

Write the local form asFor an interior chart, integrate each derivative using the fundamental theorem of calculus: its integral is zero because $f_j$ has compact support. For a boundary chart modeled on $x^m\ge 0$, all tangential derivatives again integrate to zero, while ${\int }_0^{\infty }{\partial }_m{f}_{m}d{x}^m=-f_m(x^{\prime },0)$. The outward normal is $-{\partial }_m$; its contraction with the volume form gives exactly this sign in the induced boundary integral of $\alpha$. Summing the chart identities proves the Generalized Stokes theorem:Compact support can be replaced by compactness of $M$; on a noncompact manifold, sufficient support or convergence conditions are necessary.

An embedded smooth $n$-dimensional manifold $X\subseteq {\mathbb{R}}^N$ has, around every point, a parametrization by an open subset of ${\mathbb{R}}^n$ which is a homeomorphism onto a relative neighbourhood, has rank $n$, and has smooth coordinate changes. Equivalently, after smooth ambient coordinates it is locally a coordinate $n$-plane. For a manifold with boundary use relatively open subsets of ${\mathbb{H}}^n=\{u_n\ge 0\}$ instead, with parametrizations and transitions smoothly extendible across the boundary. Points corresponding to $u_n=0$ form the boundary. Interior charts are ordinary manifold charts; a boundary point is not a corner.

The integral identity requires compactness of $M$, compact support of $X$, or suitable conditions eliminating flux at infinity. The printed question does not state such a hypothesis. We prove the Riemannian divergence theorem for a compact manifold, and more generally for a compactly supported smooth vector field on an oriented Riemannian manifold with boundary.

Write $j:\partial M\hookrightarrow M$ for the inclusion. Give $\partial M$ the outward-normal-first boundary orientation, and let $N$ be the outward unit normal vector. The Riemannian volume form of the induced Riemannian metric isIndeed, if $(e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis tangent to the boundary, $(N,e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis in $M$. Both sides therefore evaluate to one on the boundary basis.

At a boundary point decompose $X=g(X,N)N+X^{\top }$, where $X^{\top }$ is tangent to $\partial M$. Evaluating on $n-1$ tangent vectors, the term with $X^{\top }$ vanishes: all $n$ arguments of the alternating volume form then lie in the $(n-1)$-dimensional boundary tangent space. Thus the interior product of a differential form satisfiesThe general Stokes theorem for an oriented manifold with boundary states ${\int }_{M}d\eta ={\int }_{\partial M}{j}^{\ast }\eta$ for any smooth differential form $\eta$ of degree $n-1$ with compact support; compactness of $M$ makes the support condition automatic. Applying it to $\eta ={\iota }_X{\omega }_g$ and using the defining equation for the divergence of a Riemannian vector field proves

For a concrete failure without the extra hypothesis, take $M=[0,\infty )$ with Riemannian metric $d{x}^2$ and $X=(\arctan x){\partial }_x$. The divergence of a Riemannian vector field is $\mathrm{div}X=(1+x^2{)}^{-1}$, sobecause $X$ vanishes at the only boundary point. Both integrals are finite. The missing $\pi /2$ is the limiting flux at infinity, so mere integrability of the divergence cannot repair the unrestricted assertion.

Subtract the sphere equation from the last equality to obtain $x_2^2=1/2$. Thus the equality locus is the disjoint union of two sphereseach a smooth two-dimensional manifold. The two equality gradients are independent there: $x_2\ne 0$ and at least one of $x_1,x_3,x_4$ is nonzero.

On this locus the inequality becomes $1-2x_4^2\le a$, orSince $0<a<1$, this cuts out two closed caps on each sphere. At the cap boundary, $x_1^2+x_3^2=a/2>0$, so the restricted differential of $x_4$ is nonzero. The regular level set theorem for a sublevel set therefore gives a manifold with boundary. Explicitly it consists of four disks, with four boundary circles, and

A smooth $n$-dimensional manifold with boundary has charts onto relatively open subsets of the half-space ${\mathbb{H}}^n=\{u_n\ge 0\}$, with smooth transition maps extending smoothly locally across the boundary. A point is a boundary point when its last coordinate in such a chart is zero; this condition is independent of chart. One way to see the independence is that a local diffeomorphism between half-spaces preserves their interior and hence their topological boundary.

Restrict each boundary chart to $u_n=0$ and drop the last coordinate. It maps the boundary onto an open subset of ${\mathbb{R}}^{n-1}$, and the restricted transitions are smooth diffeomorphisms. These charts, with the inherited Hausdorff and countability properties, make $\partial X$ a smooth manifold without boundary. Therefore

An orientation of a smooth $n$-dimensional manifold can be specified by a nowhere-vanishing smooth $n$-form $\omega$: an ordered tangent basis is positive precisely when $\omega$ evaluates positively on it. Multiplication by a positive smooth function preserves the orientation. This is the differential-form formulation of the equivalent formulations of orientability of a smooth manifold.

Use the standard convention that the compact connected manifold is nonempty and has no boundary. Choose a positive volume form $\omega$ for its orientation. It is a closed differential form, because there are no nonzero $(n+1)$-forms. For $n>0$, compactness makes its integral finite, and positivity on a coordinate neighborhood together with nonnegativity elsewhere givesIf $\omega =d\eta$, the Generalized Stokes theorem would give ${\int }_M\omega ={\int }_{\partial M}\eta =0$, a contradiction. ThereforeFor $n=0$, a nonempty connected manifold is a point, and $H_{\mathrm{dR}}^0(M)=\mathbb{R}$ directly. The boundary hypothesis matters: a closed interval is an oriented compact connected manifold with boundary with $H_{\mathrm{dR}}^1=0$. Thus the result is about ordinary boundaryless manifolds, not an unrestricted version including boundary.

All groups in the first calculation have integral coefficients. Let $U=S^n\setminus j(W)$ and $V=S^n\setminus \mathrm{int}j(W)$. A two-sided tubular neighborhood of $j(\partial W)$ identifies $V$ as a compact manifold with boundary, with interior $U$. Pushing its boundary inward in a collar neighborhood gives a homotopy equivalence $U\simeq V$. After thickening across the same collar, excision identifies the relative groupsThe orientation of $S^n$ gives $W$ an orientation. The Poincare-Lefschetz duality isomorphism therefore givesEach component of $W$ has boundary: a closed component would have open image by the local embedding condition and closed image by compactness, hence would occupy the whole connected sphere, leaving no room for the components with boundary. In particular $U$ is nonempty, and $H^n(W)=0$.

Apply the long exact sequence in relative homology of $(S^n,V)$, using reduced homology for the absolute terms. Away from the top-dimensional homology of the sphere, it identifiesThe exceptional map iswhere $r$ is the number of components of $W$. Local compatibility of the orientations sends $1$ to $(1,\dots ,1)$, the relative fundamental classes of all the components. Its kernel is zero and its cokernel is ${\tilde{H}}^0(W)$. Thus the exceptional term has exactly the reduced form required, and all higher groups vanish. We obtain the Alexander duality formulawith negative-index cohomology zero. Ordinary degree zero is recovered byBoth formulas depend on the abstract manifold $W$, so they establish the requested independence up to group isomorphism. They make no claim that the complements themselves are homeomorphic or have isomorphic fundamental groups. The argument also covers $n=1$, when the diagonal map handles the degree-zero exception. This is the complement homology of a compact codimension-zero submanifold.

For the second calculation write $F={\mathbb{F}}_2$, $C=S^n\setminus i(M)$ and $c=n-k$. First suppose $c>0$, with $n\ge 1$. The rank-$c$ normal bundle has a mod-two Thom class, regardless of orientability. A tubular neighborhood, excision and the homological Thom isomorphism theorem identifyThe long exact sequence in relative homology becomes the Gysin sequence of an embedding:The map $\gamma$ takes the mod-two fundamental class of $S^n$ to that of $M$: restricting to each normal fiber evaluates the Thom class as $1$. Since $M$ is closed and connected, the degree-$n$ mapis an isomorphism. This cancels the exceptional top term and gives $H_i(C;F)=0$ for $i\ge n-1$ when $i>0$. In all the intervening degrees the sphere groups vanish. In degree zero, the remaining $F$ is removed by the augmentation. Consequently the mod-two homology of a submanifold complement iswhere negative-index homology of $M$ is zero. To recover ordinary homology, add one copy of $F$ in degree zero and change nothing in positive degrees. For $n=1$ the first range is empty, so the complement of a connected zero-manifold has the homology of a point. For example, codimension at least two gives $H_0(C;F)=F$, while codimension one gives $H_0(C;F)=F^2$ when $n\ge 2$.

If $k=n\ge 1$, an embedding of the closed connected manifold has image both open and closed in $S^n$, hence is onto; its complement is empty and all its ordinary homology groups vanish. The positive-codimension formula is not asserted in that case. The standard sphere calculations here assume $n\ge 1$; in ambient dimension zero the complement of the embedded connected point in $S^0$ is the other point.

A surface with boundary is a two-dimensional manifold with boundary: it is locally homeomorphic to the plane or to the closed upper half-plane. Points represented on the edge of a half-plane chart form its one-dimensional boundary.

For $n\ge 1$, a topological $n$-manifold with boundary is a Hausdorff second-countable space locally homeomorphic to relatively open sets in the half-space ${\mathbb{H}}^n$. The openness here is in the subspace topology of ${\mathbb{H}}^n$, not necessarily in ${\mathbb{R}}^n$. Its intrinsic boundary $\partial M$ consists of points sent to $x_n=0$; the remaining points have ordinary coordinate charts with images open in Euclidean space. The intrinsic boundary is independent of the chosen chart.^[ref]

The closed interval $[0,1]$ illustrates the distinction: its interior points have open interval charts, while $0$ and $1$ have charts modeled on $[0,\epsilon )\subseteq {\mathbb{H}}^1$, using $x$ and $1-x$. Its intrinsic boundary is $\{0,1\}$. The circle has empty intrinsic boundary even though its boundary of a set as a subset of the Euclidean plane is the whole circle. A topological manifold without boundary is therefore a special case with $\partial M=\varnothing$, and a smooth atlas on the half-space charts gives a manifold with boundary in the smooth sense.