The geometric intersection pairing on a closed oriented manifold

Let $M$ be a closed oriented smooth $n$-manifold and let $A^a,B^b\subseteq M$ be closed oriented embedded submanifolds with $a+b=n$. Using the oriented intersection number of The oriented intersection number (whose source is the compact submanifold and whose target is the closed submanifold, first factor first), set $⟨A,B{⟩}_M:=I(A,B)=I(i_A,B)\in \mathbb{Z},$ where $i_A:A\hookrightarrow M$ is the inclusion and $I(i_A,B)$ is evaluated on any smooth map homotopic to $i_A$ and transverse to $B$; when $A$ and $B$ are transverse this is the finite signed count ${\sum }_{p\in A\cap B}\epsilon (p)$ of the local signs of The local oriented intersection sign. For non-transverse $A,B$ the value is the common value on all such transverse representatives, by The oriented intersection number is homotopy invariant. With ${\mathbb{F}}_2$ coefficients the same construction with the mod 2 intersection number The mod 2 intersection number defines $⟨A,B{⟩}_2\in {\mathbb{F}}_2$ without orientability of $A,B$ or $M$; there The mod 2 intersection number is homotopy invariant supplies the same independence. The number depends only on the homotopy classes of the two inclusions, so replacing a factor by a homotopic submanifold, or by a homotopic embedding of the same manifold, leaves it unchanged. The factor order is part of the definition: Intersection number under factor interchange gives $⟨B,A⟩=(-1{)}^{ab}⟨A,B⟩$ and $⟨B,A{⟩}_2=⟨A,B{⟩}_2$. No claim is made yet that the number depends only on the homology classes of $A$ and $B$ (homology invariance is proved in The geometric intersection number is the Poincare-dual cup pairing; Bordant cycles have equal intersection numbers separately proves bordism invariance), nor that every homology class has an embedded representative (see Not every integral homology class is represented by an embedded submanifold). Countable Choice is inherited from the transverse-representative selection in The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$); the finite signed counts themselves are choice-free.