Pointed spherical recharting preserves the integral odd and Bar–Natan complexes

The zero-face cyclic-order test, with both outside placements. Name the inside arc $a$ and the outside arc $b$, and write their directed endpoints as $a_{-},a_{+},b_{-},b_{+}$. Give their common circle the positive boundary orientation of its bounded disk. The endpoints alternate, so the two possible cyclic words starting at $a_{-}$ are $W_{+}=(a_{-},b_{-},a_{+},b_{+})$ and $W_{-}=(a_{-},b_{+},a_{+},b_{-})$. Put $\chi (W_{+})=1$ and $\chi (W_{-})=-1$. The precise calibration must include both possible placements of the unbounded face on the outside-arc side: they need not be related by a planar isotopy fixing all endpoint labels. Straighten the circle locally at its four endpoints, with inside $a$ directed from South to North, and put the endpoints of $b$ at West and East. Read the outside arrow at an endpoint germ, as in the stipulated standard picture, rather than at a midpoint of its exterior route. There are four cases. If $b$ runs from West to East above the circle, it enters its head at East leftwards, the cyclic word is $W_{-}$, and $\kappa =1$. If it runs from West to East below the circle, the head germ is still leftwards, its word is still $W_{-}$, and $\kappa =1$. If $b$ runs from East to West above the circle, it enters West rightwards, its word is $W_{+}$, and $\kappa =-1$. If it runs from East to West below the circle, its head germ remains rightwards, its word remains $W_{+}$, and $\kappa =-1$. The same direction is obtained at the tail germ: outgoing at West is leftwards and outgoing at East rightwards. Thus all four placements give the single formula $\kappa =-\chi$. This table also treats the two possible faces on each side of the circle cut by its surgery arc. Straightening an endpoint germ uses only its transverse attachment on the indicated side and does not change which directed endpoint it is; no isotopy exchanging the two outside-face placements is assumed.

Changing the omitted point. Consider the same zero-face circle on the oriented sphere. Its two complementary sides contain its two surgery arcs. If the two omitted points are on the same side, both planar boundary orientations agree and both presentations call the same arc inside. Their cyclic word and $\kappa$ agree by step 1.1. If the omitted points are on opposite sides, the inside/outside names are exchanged and the positive bounded-side boundary orientation is reversed: the boundary orientations of the two sides of an oriented surface are opposite. Both changes are necessary. For example, an old word $W_{+}=(a_{-},b_{-},a_{+},b_{+})$, read in the reverse orientation starting at $b_{-}$, is $(b_{-},a_{-},b_{+},a_{+})$, which is the new word $W_{+}$ when $b$ is the new inside arc and $a$ the new outside arc. An old word $W_{-}=(a_{-},b_{+},a_{+},b_{-})$ becomes $(b_{-},a_{+},b_{+},a_{-})$, which is the new word $W_{-}$. Hence $\chi$, and therefore $\kappa$, is unchanged in this case too. This is the full two-case calculation; exchanging only the names without reversing the boundary orientation would give an incorrect sign. The same argument applies to an arbitrary orientation-preserving spherical change of coordinates by transporting the circle and its two sides first.

Naturality of every unsigned saddle. At state $v$ let $P_v$ send each circle generator to that same spherical circle in the other chart. It is a bijection of bases. For a merge quotient $q$, both $P_{v+e}q$ and $q^{\mathrm{\prime }}{P}_v$ send the two merging generators to the same merged generator and fix every spectator. Thus their exterior extensions agree. For a split, transport the arrow and its ordered output circles. Then $P_{v+e}$ carries the difference $C_1-C_2$ to $C_1^{\mathrm{\prime }}-C_2^{\mathrm{\prime }}$. Choose the transported lift $j^{\mathrm{\prime }}$ of the split input. On every wedge $z$ one has $\mathrm{\Lambda }{P}_{v+e}((C_1-C_2)\wedge \mathrm{\Lambda }j(z))=(C_1^{\mathrm{\prime }}-C_2^{\mathrm{\prime }})\wedge \mathrm{\Lambda }{j}^{\mathrm{\prime }}(\mathrm{\Lambda }{P}_v(z))$. The alternative lift gives the same expression by [F1]. Circle order changes are absorbed by the exterior relabelling, including their actual wedge signs. Thus the two unsigned composites on every face are conjugate by invertible relabellings. On a nonzero face their relative commutation sign $\kappa$ is consequently unchanged. On a zero face it is unchanged by step 2.1. All face products prescribed by [F2] therefore agree in the two charts.

Odd chain isomorphisms and their actual inverse. Choose a full admissible assignment in the first chart using [F2] and give each corresponding edge of the second chart the same sign. By step 3.1 these are full admissible signs in the second chart, and $P_o={⨁}_v\mathrm{\Lambda }{P}_v$ intertwines their signed differentials. Its inverse is ${⨁}_v\mathrm{\Lambda }(P_v^{-1})$, so both composites are exactly the identity; the chain homotopies for an inverse are zero. If the second chart instead has independently chosen full admissible signs, divide those signs by the transported signs. The quotient has product one on every face, including the zero faces. Fix the lowest vertex sign to be one and assign each other vertex the product of the edge quotients along a monotone path. The square equations make this independent of the path, since adjacent coordinate exchanges connect all monotone paths. Multiplication by the resulting scalar at each vertex conjugates the transported differential to the chosen one. Compose this diagonal map with $P_o$. Its inverse is the inverse relabelling followed by the same scalar signs in the inverse order. This proves the asserted isomorphism for the independently chosen assignments, without claiming independence for arbitrary weak square-compatible assignments.

Marked linearity and gradings. Relabelling carries the marked generator $Z_p$ to the marked generator $Z_p^{\mathrm{\prime }}$. Hence $P_o(z\wedge Z_p)=P_o(z)\wedge Z_p^{\mathrm{\prime }}$. Vertexwise scalar signs commute with this right action too. Both the odd map and its inverse are thus $\mathbb{A}$-linear. An orientation-preserving surface chart preserves the signed crossing, its designated 0/1 smoothing and its over/under information. It consequently preserves $n_{+},n_{-},\mathrm{\mid }v\mathrm{\mid },k(v)$; relabelling preserves exterior degree. The formula in [F1] then proves preservation of both normalized gradings. Vertexwise signs have degree zero. There is no semilinear substitution $X\mapsto -X$ in these maps.

The integral Bar–Natan map. On each state relabel the tensor factors by their spherical circles and keep the labels $1,X$ and the coefficient $h$ unchanged. Commutativity of multiplication and cocommutativity of comultiplication make this independent of a tensor-factor ordering. For a merge, split, birth or death the relabelling intertwines the operations because their input and output circles and all coefficients in the formulas of [F1] agree. In particular it intertwines every cube edge. The crossing order has been kept, so the ordinary cube edge signs agree as well. This yields a chain isomorphism $P_h$ with inverse the inverse tensor-factor relabelling. It takes multiplication by the marked $X$ to multiplication by the target marked $X$, and fixes $h$, so is $\mathbb{R}$-linear. The grading check is the same state-shift calculation as in step 5.1, with unchanged circle labels. Both inverse composites are identities with zero homotopies.

The common reduction and the local maps. Modulo two all vertex signs and exterior reorder signs are one. The reduced odd map sends a selected subset of circles to exactly the same spherical subset in the other chart. After setting $(2,h)=0$, the Bar–Natan map sends the tensor state with those $X$-labels to that same subset. Under the comparison of [F1] these maps are identical on every basis state, including its marked action. Thus $P_{h}f=f^{\mathrm{\prime }}{P}_o$ holds exactly after the specified quotients, not merely up to homotopy. The inverse maps satisfy the same calculation. Since $P_h$ is already an integral marked-module chain isomorphism, localizing at $h$ retains its explicit inverse and zero homotopies. This verifies all local-map conditions, with the full asserted degrees and module structures, whenever the endpoint triples satisfy the LEO-object hypotheses; it does not assert rank-one localization for arbitrary multi-component links. The only sign assignment and comparison constructions used finite cube paths and finite multiplications, so require no choice axiom.