The diagonal and graph classes contract to the alternating trace

For disconnected $M$, apply [F1] on each component: the diagonal has support only in $C\times C$, and the product components $C\times D$ with $C\mathrm{\ne }D$ have zero diagonal class. In component-adapted bases its expansion is the sum of the component expansions; it is independent of the basis since a basis change and its inverse dual change cancel in the tensor sum. Components mapped to a different component have zero diagonal trace block and no diagonal intersection. Thus the graph-pullback trace identity also sums over the components, including the empty case. Trivialize the orientation system by the given orientation of $M$. The cap-normalized diagonal class of [F1] becomes $\mathrm{P}\mathrm{D}[{\mathrm{\Delta }}_M]$ by [F2]. Its expansion is exactly the displayed formula, and [F1]'s graph pullback gives $⟨{\gamma }_f^{\ast }\mathrm{P}\mathrm{D}[{\mathrm{\Delta }}_M],[M]⟩=L(f)$.

Since the graph is oriented by ${\gamma }_f$, its fundamental homology class is $({\gamma }_f{)}_{\ast }[M]$. For every degree-$n$ class $\phi$, [F2] gives $⟨\mathrm{P}\mathrm{D}[{\mathrm{\Gamma }}_f]⌣\phi ,[M\times M]⟩=⟨\phi ,\mathrm{P}\mathrm{D}[{\mathrm{\Gamma }}_f]\cap [M\times M]⟩=⟨\phi ,({\gamma }_f{)}_{\ast }[M]⟩=⟨{\gamma }_f^{\ast }\phi ,[M]⟩$. Taking $\phi =\mathrm{P}\mathrm{D}[{\mathrm{\Delta }}_M]$ proves the cup contraction identity from step 1.1.

If $n\ge 1$ and the fixed points are nondegenerate, [F1] evaluates this graph pullback as the sum of the local signs $\mathrm{sign}\det (I-D{f}_x)$. By [F3] this is both $I(f)$ and the stated graph-diagonal intersection number. AC is inherited from [F1] and Poincare duality.