Zero Lefschetz number does not imply absence of fixed points

The map $f$ is continuous, since the inverse image of every open set is itself. Its map on every singular chain is the identity: composing a singular simplex with $f$ changes nothing. Consequently its map on each homology group is the identity. By [F2], the only nonzero rational homology groups are the two one-dimensional groups in degrees zero and one. Their identity matrices are each $(1)$, with trace $1$. All other homology endomorphisms are on zero spaces and have trace zero.

Substitution into [F1] gives $L(f)=(-1{)}^0\cdot 1+(-1{)}^1\cdot 1=1-1=0.$ On the other hand $f(x)=x$ for every $x\in S^1$, and $(1,0)\in S^1$ is an explicit fixed point. Thus the premise of the proposed implication holds and its conclusion fails. The fixed-point set is the whole circle.

The witness is nonempty and connected, so the failure is not due to an empty space or a disconnected-component convention. The zero value in step 2.1 is cancellation of two traces equal to one, not the vanishing of the homology groups. The sole one-cell's two endpoints are attached to the same vertex, giving a valid nonregular CW structure; the calculation uses singular homology and includes degenerate singular simplices. No limiting, relative, orientation or homotopy-endpoint choice is involved, and no AC is used. The fixed-point theorem asserts the different implication from nonzero Lefschetz number to a fixed point, so this example does not contradict it.