A vanishing Lefschetz number with canceling fixed points

The fixed points. A point $\theta$ is fixed by $g$ exactly when $\epsilon \sin \theta =0$ in $\mathbb{R}\mathrm{/}2\pi \mathbb{Z}$, i.e. exactly when $\theta \in \{0,\pi \}$. Both solutions are nondegenerate because $g^{\mathrm{\prime }}(\theta )=1+\epsilon \cos \theta$ is $1+\epsilon$ at $0$ and $1-\epsilon$ at $\pi$, neither equal to $1$ for $0<\epsilon <1$; their indices are $\mathrm{sign}(1-(1+\epsilon ))=-1$ at $0$ and $\mathrm{sign}(1-(1-\epsilon ))=+1$ at $\pi$. So $f$ has fixed points and its index sum is $(-1)+(+1)=0$.

The periodic function $\epsilon \sin \theta$ makes $g_t(\theta )=\theta +t\epsilon \sin \theta$ a well-defined homotopy of circle maps from the identity to $f$. By [L1], $L(f)=L(\mathrm{i}\mathrm{d})$, and [F1] gives the two identity traces one in degrees zero and one, so $L(f)=1-1=0$. Thus $L(f)=0$ while $f$ has the two fixed points found in step 1.1.

The refutation. The displayed statement asserts that a vanishing Lefschetz number forces the map to be fixed-point-free; $f$ has $L(f)=0$ and the fixed points $0,\pi$, so the implication fails. The example is the curved version of the standard caution that $L$ is only a signed count: the nonvanishing of $L$ guarantees a fixed point, but its vanishing merely allows fixed points to cancel, as here with indices $-1$ and $+1$.