Compatibility complexes and first homology

We now construct a different finite object. Its vertices are indices of selected tests, not the valuations themselves. It records which groups of tests can all be true at one valuation. Label the carrier type before drawing a triangle. Select three indexed tests. Test one is X. Test two is Y. Test three is the negation of the whole conjunction X and Y. It is not a new variable Z. Their regions are A and B; B and C; and A, C and D. Each test is satisfiable, so all three indices become vertices. The first two tests can be jointly true at B. Draw the edge between their test indices and retain B as its witness. This edge concerns two available formulas, not a physical connection between state cards. X is compatible with the third test at A, and Y with it at C. Each pair therefore has an edge. The three edges form the boundary of a triangle, with a different valuation witnessing each pair. Requiring all three tests together is contradictory. No valuation witnesses their triple intersection. We therefore do not fill the triangle. Three pairwise witnesses do not supply a single witness satisfying all three formulas at once. A simplicial complex contains every subset of any included simplex. Joint satisfiability has this property: any subset of a jointly satisfiable test group is also satisfiable. This construction is the observation-side relation nerve of the satisfaction table. Following the three edges gives a loop, with no triangular face filling it. In this elementary example the first homology detects one independent one-dimensional hole. The companion note gives a small algebraic check; no general homology calculation is required here. The nerve is connected, but the valuation observation space for the same menu has components B and A, C and D. These objects even disagree about connectedness. Do not infer the topology or homology of one from the other's drawing. The observation topology is generated from selected true-row regions by unions and finite intersections. Symmetric distance adds weights at differing slots. Directed loss adds weights only at true-to-false changes from the source. We can check each construction separately when a new test is added. A tautology is true at every valuation. Its region was already the whole open carrier, so adding it changes no generated topology. Its constant coordinate adds no disagreement or directed loss, so both signature distances also remain unchanged. The tautology can accompany every previously satisfiable test group. In the nerve it adds a vertex and a cone over the old complex, filling the old loop through new faces. The resulting cone has no reduced homology, unlike the triangle boundary. Does preserving the observation topology and both distance functions force the indexed test nerve to preserve its hole? State what the tautology changes and what it leaves fixed. The topology and distances remain the same, while the nerve becomes a cone and loses its reduced homology. Therefore the nerve's homology is not determined by the observation topology. Any invariant claim needs additional restrictions and a stated presentation policy. The nerve usefully records higher-order compatibility of indexed tests. Its value does not require pretending it is a topological invariant of the valuation space. State its vertices, simplices, presentation and intended comparison before interpreting a hole.