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A nonzero Lefschetz coincidence number forces a coincidence of the two maps. If their paired map misses the diagonal, the diagonal Thom class restricts to zero on its image and its pullback vanishes, making the coincidence number zero. The argument works for continuous maps and requires neither isolated coincidences nor transversality.

Let $m=\dim M$ and $q=m-\dim Y$. Here a closed submanifold means compact and without boundary, which is the condition needed for a compactly supported dual class. The orientations of $M$ and $Y$ orient the normal bundle ${\nu }_Y$. Its Thom class is the unique relative class of degree $q$ restricting to the chosen orientation generator on every normal fiber. A tubular neighborhood identifies this class with one near $Y$ in $M$. Choose a compact normal disc neighborhood and extend the relative class to compactly supported cohomology of $M$. Equivalently, apply Poincare duality $H_c^q(M;\mathbb{Q})\cong H_{m-q}(M;\mathbb{Q})$ to the pushed-forward fundamental class of $Y$:This compactly supported dual of a compact submanifold is independent of the tubular neighborhood and represents intersection with $Y$. Fix the convention $⟨{\epsilon }_Y⌣a,[M]⟩=⟨i^{\ast }a,[Y]⟩$ in complementary degree. Its image in ordinary cohomology restricts to zero on $M\setminus Y$, because it comes from relative cohomology supported near $Y$. If a submanifold is only closed as a subset but is noncompact, its usual dual lies in ordinary cohomology instead; compact support is not asserted in that situation.

Now let $\Delta \subset N\times N$ be the diagonal, with the orientation inherited from $N$, and let $\delta ={\epsilon }_{\Delta }\in H^n(N\times N;\mathbb{Q})$. For each $i$, choose a basis $a_{i,j}$ of $H^i(N;\mathbb{Q})$ and its dual basis $b_{i,j}$ of $H^{n-i}(N;\mathbb{Q})$ satisfying $⟨a_{i,j}⌣b_{i,k},[N]⟩={\delta }_{jk}$. The Poincare duality pairing and the Künneth theorem give the cohomology class of the diagonalThe sign can be checked from the defining identityFor $|u|=i$, moving the degree-$i$ class in the second factor past $u$ contributes $(-1{)}^i$, and interchanging $b_{i,j}$ and $u$ contributes $(-1{)}^{i(n-i)}$; the displayed coefficient cancels their product. Equivalently its exponent is $ni$ modulo two.

The cohomological pushforward between closed oriented manifolds is characterized byfor complementary degrees. This follows immediately from its definition by Poincare duality and naturality of evaluation on pushed-forward homology. In degree $i$, the matrix trace of $f^{!}{g}^{\ast }$ on $H^i(N;\mathbb{Q})$ is consequentlyFor finite-dimensional maps between two possibly different vector spaces, $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. Thus this matrix trace equals that of $g^{\ast }{f}^{!}$ on $H^i(M;\mathbb{Q})$. Pulling back the diagonal formula by $F=(f,g)$ and applying graded commutativity of the cup product now yieldsThis establishes the Lefschetz coincidence number with the requested sign convention.

If $f(m)\ne g(m)$ for every $m$, the map $(f,g)$ factors through $(N\times N)\setminus \Delta$. The diagonal class restricts to zero there, so its pullback is zero and $L(f,g)=0$. Taking the contrapositive proves the Lefschetz coincidence theorem:No transversality or isolated-coincidence assumption is needed. This cohomological argument also applies to continuous maps, which covers the last request.

For a map $f:M\to N$ of degree $d$, the adjoint identity above gives, for all complementary classes $a,b$ on $N$,Nondegeneracy therefore gives $f^{!}{f}^{\ast }=d\mathrm{id}$. Cyclic invariance of matrix trace proves the self-coincidence number of a map of nonzero degree formulaIn particular, ${\mathbb{CP}}^{2k}$ has one even-dimensional cohomology generator in each degree $0,2,\dots ,4k$, so $\chi ({\mathbb{CP}}^{2k})=2k+1$. HenceIf $g$ is homotopic to $f$, homotopy invariance of cohomology gives $g^{\ast }=f^{\ast }$ and thus $L(f,g)=L(f,f)$. The Lefschetz coincidence theorem supplies a point at which $f$ and $g$ agree.