Rational oriented bordism is detected by Pontryagin numbers

The case $4\mid n$. Write $n=4k$ and expand a rational bordism class uniquely in the basis of [F1]: $z={\sum }_J{q}_J[P_J]$ with $q_J\in \mathbb{Q}$. Applying the linear functional $p_I\otimes \mathbb{Q}$ [F2] gives $p_I[z]={\sum }_J{q}_J{A}_{I,J}$ with the invertible matrix $A$ of [F3]; hence the map $z\mapsto (p_I[z]{)}_I$ is a $\mathbb{Q}$-linear isomorphism from ${\mathrm{\Omega }}_{4k}^{SO}\otimes \mathbb{Q}$ onto ${\prod }_I\mathbb{Q}$ (finitely many coordinates). Applying this to $[M]-[N]=[M]+[-N]$ shows that $M$ and $N$ agree in ${\mathrm{\Omega }}_{4k}^{SO}\otimes \mathbb{Q}$ exactly when all their Pontryagin numbers agree, since $A$ is invertible and the vector of differences $(p_I[M]-p_I[N]{)}_I$ vanishes if and only if the differences of coefficients vanish.

The case $4\nmid n$. By [F1] the group ${\mathrm{\Omega }}_n^{SO}\otimes \mathbb{Q}$ is zero, there is no degree-$n$ monomial in the Pontryagin classes, and the product over the empty index set is the zero vector space; the asserted equivalence and isomorphism hold vacuously. For $n=0$ the empty partition gives the single monomial $1$, [F4] identifies ${\mathrm{\Omega }}_0^{SO}\otimes \mathbb{Q}\cong \mathbb{Q}$ through the signed count, and the value $p_{\mathrm{\varnothing }}$ of the positively oriented point is $1$, so the one-by-one map is the identity.

Multiple of a boundary. Suppose all Pontryagin numbers of a closed oriented $n$-manifold $M$ vanish. By steps 1.1 and 1.2 the class $[M]\otimes 1$ is zero in ${\mathrm{\Omega }}_n^{SO}\otimes \mathbb{Q}$. By the fraction criterion [F4] there is a positive integer $s$ with $s[M]=0$ in ${\mathrm{\Omega }}_n^{SO}$, that is, the sum of $s$ copies of the class is zero. Since the group operation is disjoint union [F2], $s[M]=[M\bigsqcup \cdots \bigsqcup M]$, so some positive multiple of $M$ is null-cobordant, i.e. an oriented boundary. This is the rational multiple-boundary consequence; the theorem makes no integral single-copy assertion, and the integral refinement is recorded separately.

Conventions and boundaries. The product on the right of the displayed map runs over the finitely many partitions of $n\mathrm{/}4$ when $4\mid n$ and is the empty product otherwise; the target is a finite-dimensional rational vector space and no completion occurs. For the empty manifold all numbers are zero and its class is zero. The statement uses only rational coefficients, and the cited suppliers carry their own choice declarations, so no further choice is made.