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For a smooth map of disk pairs whose graph of a function is transverse to the zero slice, orient the graph by its domain. With the tangent space of the zero slice ordered before that of the graph, the local smooth intersection number at a zero is $\mathrm{sgn}\det DF$. Summing these signs gives the relative mapping degree and hence the boundary mapping degree. Reversing the order of these two $m$-dimensional tangent spaces changes the sign by $(-1{)}^m$.

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.

An index-$h$ handle is $D^h\times D^{n-h}$, attached along $S^{h-1}\times D^{n-h}$. Its attaching sphere is $S^{h-1}\times \{0\}$ with its attaching framing of an embedded sphere, and its belt sphere in the new boundary is $\{0\}\times S^{n-h-1}$. A handle decomposition is an ordered sequence of such handle attachments.

A handle slide of one $h$-handle over another replaces its framed attaching sphere by a band sum with a parallel framed copy of the second attaching sphere, along a band in the boundary of the previously attached handles. The second handle stays fixed. The band transports the framing of an embedded sphere. The union of the two handles and their attaching collars can be reidentified by a diffeomorphism, so this changes the decomposition, rather than the underlying manifold. A sequence of these moves, allowing the usual isotopies of attaching data, gives handle decompositions related by slides. Reading the decomposition backwards interchanges attaching spheres and belt spheres; a slide of dual handles is a slide of the corresponding original handles.

For the standard handle-calculus assertion, take $1\le k\le n-2$, so both the relevant attaching and belt spheres are connected. We need the local geometric move behind handle-slide isolation of a cancelling pair. Write $p=A_1\cap B_1$. If $q$ is an intersection of some $A_j$, $j>1$, with $B_1$, choose an arc on $B_1$ from $q$ to $p$, and take a narrow parallel band to the unique sheet of $A_1$ at $p$. Slide $A_j$ over an appropriate parallel copy of $A_1$. Choose the copy's orientation so that the new sheet at $p$ has the opposite local smooth intersection number to the sheet at $q$.

Within a thin neighbourhood of the arc, the old sheet at $q$ and the added sheet at $p$ are the two ends of one band. An isotopy pushes that band through the neighbourhood of $B_1$ and removes both ends. Outside this neighbourhood the original $A_j$ is unchanged, apart from the added copy of the parts of $A_1$ away from $B_1$. Since $A_1$ meets $B_1$ only at $p$, this copy contributes no further intersections with $B_1$. Thus one slide followed by an isotopy removes one original geometric intersection with $B_1$. This is a local band move, not an appeal to handle cancellation or to an algebraic cancellation of signs.

The arcs and bands may be chosen one at a time; in the one-dimensional sphere case, start with an intersection adjacent to $p$, so that the arc interior contains no other intersection. Perturb to make all intersections transverse. The relevant spheres are compact, so there are only finitely many intersections. Repeating the move for every $j>1$ leaves $A_1$ and $B_1$ fixed and givesThe bands can avoid the remaining attaching data; any consequent change to intersections with other belt spheres is allowed at this stage.

For the second stage, view the same interface from the dual decomposition. Now $B_i^{\prime }$ are attaching spheres and $A_j^{\prime }$ are belt spheres. Apply the same local move to each intersection of $B_i^{\prime }$, $i>1$, with $A_1^{\prime }$, sliding that dual handle over the one with attaching sphere $B_1^{\prime }$. This changes the original $k$-handles. It keeps $A_1^{\prime }$ and $B_1^{\prime }$ fixed, while removing the unwanted intersections with $A_1^{\prime }$. The parallel copies and bands may be taken disjoint from all $A_j^{\prime }$, $j>1$, because those spheres are already disjoint from $B_1^{\prime }$ and from $A_1^{\prime }$. Hence their intersections with $B_1^{\prime }$ remain empty. We obtainNo handle has been removed.

There is an implicit index-range qualification in this geometric assertion. If arbitrary zero-handles are included, the second conclusion need not hold: two zero-handles joined by one one-handle have an attaching zero-sphere with one endpoint on each zero-handle belt sphere. Slides preserving this handle inventory cannot make its second endpoint avoid both belts. The dual obstruction occurs for top-index handles. Thus the argument uses the customary intermediate-index setting of the handle-slide lemma; the unqualified assertion at the extreme indices would require additional hypotheses or additional moves.

For the numerical formula, fix coherent orientations for the integer smooth intersection numbers, and letKeep the orientation at the surviving point, so its number remains $\delta$. A slide of a $(k+1)$-handle adds or subtracts the first row of the intersection matrix. Removing all geometric intersections in the first column has net row operationEven if opposite-sign intersections require extra slides, their net signed coefficient is the displayed one. In particular,The dual slides then perform the column operationsSince $C_1^{\prime }=(\delta ,0,\dots ,0{)}^T$, these operations leave the lower-right block unchanged. Thus the final answer, including both stages of slides, isThe remaining block is the Schur complement of the unit pivot $\delta$. If the original orientation is chosen so that $\delta =1$, it is simply $m_{lk}-m_{l1}{m}_{1k}$. For unoriented handles use intersection numbers modulo $2$, with the same formula interpreted in ${\mathbb{F}}_2$.