Let \(X\) be a compact orientable surface (without boundary) and let \(f:X\to X\) be a continuous map. Assume that \(f\) is homotopic to the identity map but that \(f\) has no fixed points. What is the genus of \(X\)? Prove your answer.
Solution
1. If \(X\) has genus \(g\), then\[ H_k(X;\QQ) \cong \begin{cases} \QQ,&k=0,2,\\ \QQ^{2g},&k=1,\\ 0,&\text{otherwise}. \end{cases} \]
A closed connected orientable surface of genus \(g\) has a CW structure with one \(0\)-cell, \(2g\) \(1\)-cells, and one \(2\)-cell, with both cellular differentials zero. Thus its rational homology has the displayed form.
2. Since \(f\) is homotopic to the identity, its Lefschetz number is\[ L(f)=2-2g. \]
Homotopic maps induce the same maps on homology, so \[ f_*=\operatorname{id} \] on every \(H_k(X;\QQ)\). Therefore \[ \begin{aligned} L(f) &=\sum_{k\ge0}(-1)^k \operatorname{tr}\bigl(f_*:H_k(X;\QQ)\to H_k(X;\QQ)\bigr)\\ &=1-2g+1\\ &=2-2g. \end{aligned} \] Equivalently, \(L(f)=L(\operatorname{id}_X)=\chi(X)\).
3. The assumption that \(f\) has no fixed points forces\[ L(f)=0. \]
The Lefschetz fixed-point theorem states that a continuous self-map of a compact manifold with nonzero Lefschetz number has a fixed point. Taking the contrapositive, a fixed-point-free self-map must have Lefschetz number zero.
4. Hence\[ g=1. \]
Combining 2 and 3 gives \[ 2-2g=0, \] so \(g=1\). Thus \(X\) is a torus.
5. The genus-\(1\) conclusion is sharp: the torus admits fixed-point-free self-maps homotopic to the identity.
Write \[ T^2=\RR^2/\ZZ^2 \] and choose \[ v=(1/2,0). \] The translation \[ \tau_v([x])=[x+v] \] has no fixed point, since a fixed point would imply \(v\in\ZZ^2\). The map \[ H:T^2\times[0,1]\longrightarrow T^2, \qquad H([x],t)=[x+tv], \] is a homotopy from the identity to \(\tau_v\). Hence genus \(1\) actually occurs under the hypotheses.