Two-map intersection as a diagonal preimage

The identity $(f\times g{)}^{-1}({\mathrm{\Delta }}_M)=\{(a,b):f(a)=g(b)\}=X{\times }_{M}Z$ is immediate from the definitions. At $(a,b)$ with $y=f(a)=g(b)$, put $U:=d{f}_a(T_{a}X)$ and $W:=d{g}_b(T_{b}Z)$; the differential of $f\times g$ has image $U\times W$, and [F2] gives $T_{(y,y)}{\mathrm{\Delta }}_M={\mathrm{\Delta }}_{T_{y}M}$. Then $d(f\times g{)}_{(a,b)}$ is transverse to ${\mathrm{\Delta }}_M$ exactly when $U\times W+{\mathrm{\Delta }}_{T_{y}M}=T_{y}M\times T_{y}M$, the quotient map $(v,w)\mapsto v-w$ has kernel ${\mathrm{\Delta }}_{T_{y}M}$ and sends $U\times W$ onto $U+W$, so this is equivalent to $U+W=T_{y}M$, that is, to transversality of $f$ and $g$ at $(a,b)$.

Assume the equivalent transversality conditions hold. At a coincidence write $A=d{f}_a$ and $B=d{g}_b$ in supplied oriented determinant frames, and $C=[AB]$. The local sign of $(f,g)$ is $\mathrm{sgn}\det C$. For $(f\times g,{\mathrm{\Delta }}_M)$ the ordered derivative matrix in the ambient product frame has block columns $(A,0)$, $(0,B)$, $(I,I)$. Subtracting its top row block from the bottom and moving the final $n$ diagonal columns to the front gives the sign factor $(-1{)}^{n^2}=(-1{)}^n$ and bottom block $[-AB]$, of determinant $(-1{)}^x\det C$. Thus the original determinant has sign $(-1{)}^{n+x}\mathrm{sgn}\det C=(-1{)}^z\mathrm{sgn}\det C$. The determinant-ray calculation includes zero-dimensional source factors: their point signs multiply both comparisons equally, while the diagonal carries the ambient ray.

Summing 2.1 over the finite coincidence set gives $I(f,g)=(-1{)}^{z}I(f\times g,{\mathrm{\Delta }}_M)$: the two sums are over the same finite index set, and $(-1{)}^z$ is a common factor. Exchanging the roles of the two maps and using the factor-interchange sign of [F5] gives $I(f,g)=(-1{)}^{x}I(i_{\mathrm{\Delta }},f\times g)$: the map--submanifold number $I(f\times g,{\mathrm{\Delta }}_M)$ equals the map--map number $I(f\times g,i_{\mathrm{\Delta }})$ because the differential of the inclusion is the inclusion of $T{\mathrm{\Delta }}_M$, and the same pointwise block-swap calculation as [F5], of dimensions $n,n$, gives $I(i_{\mathrm{\Delta }},f\times g)=(-1{)}^{n\cdot n}I(f\times g,i_{\mathrm{\Delta }})=(-1{)}^{n}I(f\times g,{\mathrm{\Delta }}_M)$. This finite transverse sum is defined even if the diagonal is noncompact, since its coincidence set is exactly the finite preimage in the compact $X\times Z$ with $n=x+z$.