Orientation local system on a manifold with boundary

Let $M$ be a compact $n$-manifold with boundary $A=\mathrm{\partial }M$, let $N=M\setminus A$, and let $R$ be a commutative unital ring. The pointwise local-homology formula for a boundaryless manifold is not used at points of $A$: its degree-$n$ group there would be zero. Instead choose a collar push-in $r:M\to N$ supplied by Compact topological manifold boundaries admit collars and define the orientation local system of $M$ by ${\mathcal{O}}_M^R:=r^{\ast }{\mathcal{O}}_N^R,$ where ${\mathcal{O}}_N^R$ is the boundaryless orientation system from The orientation system is a local system. Thus transport along a path $\gamma$ in $M$ is orientation transport along $r\gamma$ in the interior.

This definition is independent of the push-in up to a specified natural isomorphism. If $r_0,r_1:M\to N$ are homotopy inverses to the inclusion $i:N\hookrightarrow M$, then $r_0\simeq r_{0}i{r}_1\simeq r_1$. For a chosen such homotopy $H$, transport along the track $t\mapsto H(x,t)$ gives an isomorphism $(r_0^{\ast }{\mathcal{O}}_N^R{)}_x\to (r_1^{\ast }{\mathcal{O}}_N^R{)}_x$. The boundary of the square $(s,t)\mapsto H(\gamma (s),t)$ shows that these stalk maps commute with transport along every $\gamma$; hence they form a natural isomorphism. No claim is made that different homotopies give literally the same isomorphism.

The restriction to $N$ is naturally isomorphic to ${\mathcal{O}}_N^R$ because $ri\simeq 1_N$. On the boundary, collar product charts identify ${\mathcal{O}}_M^R{\mathrm{\mid }}_A$ with ${\mathcal{O}}_A^R$: cross a local $(n-1)$-orientation class of $A$ with the collar interval oriented from positive height toward the boundary, placing that outward direction first, and transport the resulting ambient class to positive collar height. This fixes the outward-normal-first sign. Reversing a boundary loop reverses the ambient local orientation exactly when it reverses the boundary local orientation, so these stalk identifications commute with transport.

When $A=\mathrm{\varnothing }$, take $r=1_M$ and recover the published boundaryless system literally. A compact zero-manifold has empty boundary, so no negative-dimensional boundary system occurs. Empty and disconnected manifolds are treated componentwise, and the zero ring gives the corresponding zero stalks. A particular collar push-in is finite geometric data, not a simultaneous choice from a family; no AC is required.