Intersection number under factor interchange

By [F1] the coincidence set is finite, so both sums $I(f,g)$ and $I(g,f)$ are finite sums over the same index set; this exhibits $I(f,g)$ for two maps as a finite sum of the local signs of [F2] and provides the map--map intersection numbers used in the statement.

At a coincidence point $(a,b)$ with $f(a)=g(b)=y$, transversality and $x+z=n$ make $L=(d{f}_a,d{g}_b):T_{a}X\oplus T_{b}Z\to T_{y}M$ an isomorphism. The corresponding isomorphism for $(g,f)$ is $L\circ s^{-1}$, where $s$ swaps the source factors and has sign $(-1{)}^{xz}$ by [F3]. Hence ${\epsilon }_{(g,f)}(b,a)=(-1{)}^{xz}{\epsilon }_{(f,g)}(a,b)$, including signed determinant rays in dimension zero.

For any transverse pair $u:A^a\to M$, $v:B^b\to M$, the product map $u\times v:A\times B\to M\times M$ is transverse to ${\mathrm{\Delta }}_M$: the quotient $(r,s)\mapsto r-s$ has kernel $T{\mathrm{\Delta }}_M$ and sends its derivative onto $du(TA)+dv(TB)=TM$. Its local intersection sign is $(-1{)}^b$ times that of $(u,v)$. Indeed in oriented determinant frames put $C=[dudv]$; the derivative matrix for $(u\times v,i_{\mathrm{\Delta }})$ has columns $(du,0),(0,dv),(I,I)$. Subtract the bottom row block from the top. The resulting block lower triangular matrix has diagonal blocks $[du-dv]$ and $I$. Its determinant sign is therefore $(-1{)}^b\mathrm{sgn}\det C$. In dimension zero the supplied point rays multiply the same comparisons, with ${\mathrm{\Delta }}_M$ carrying the ambient ray. Summing gives $I(u\times v,{\mathrm{\Delta }}_M)=(-1{)}^{b}I(u,v)$.

The swap $(a,b)\mapsto (b,a)$ identifies the two finite coincidence sets. Summing 1.2 gives $I(g,f)=(-1{)}^{xz}I(f,g)$. In particular this proves the submanifold identity when $A,B$ are transverse, by taking their inclusions.

Now let $A,B$ be arbitrary compact oriented complementary-dimensional submanifolds without boundary. Under [F5] choose $u:A\to M$ homotopic to $i_A$ and transverse to $B$, and $v:B\to M$ homotopic to $i_B$ and transverse to $A$. By definition $I(A,B)=I(u,i_B)$ and $I(B,A)=I(v,i_A)$. The transverse maps $u\times i_B$ and $i_A\times v$ are homotopic to $i_A\times i_B$, so [F5] applied to the closed oriented diagonal gives equality of their intersection numbers. By 1.3 this reads $(-1{)}^{b}I(u,i_B)=(-1{)}^{b}I(i_A,v)$. Applying 2.1 to the transverse pair $(i_A,v)$ gives $I(i_A,v)=(-1{)}^{ab}I(v,i_A)$, hence $I(B,A)=(-1{)}^{ab}I(A,B)$. Compactness of the source products and closedness of the diagonal supply finite counts; no transversality of the original inclusions is required. Countable Choice is used only for this homotopy-class extension; the finite transverse swap calculation is choice-free and includes $a=0$ or $b=0$ with sign $+1$.