Verify that the Lefschetz fixed point theorem holds also when \(\tau(f)\) is defined using homology with coefficients in a field \(F\).
Solution
Let \(F\) be a field and define \[ \tau_F(f)=\sum_n(-1)^n\operatorname{tr}\bigl(f_*:H_n(X;F)\to H_n(X;F)\bigr). \]
1. For a finite chain complex of finite-dimensional \(F\)-vector spaces and a chain endomorphism \(T\), one has\[ \sum_n(-1)^n\operatorname{tr}(T|C_n) = \sum_n(-1)^n\operatorname{tr}(T_*|H_n). \]
Use the invariant subspaces \(B_n\subset Z_n\subset C_n\). Additivity of trace in short exact sequences gives \[ \operatorname{tr}(T|C_n)=\operatorname{tr}(T|Z_n)+\operatorname{tr}(T|B_{n-1}) \] and \[ \operatorname{tr}(T|Z_n)=\operatorname{tr}(T|B_n)+\operatorname{tr}(T_*|H_n). \] Summing with alternating signs cancels the boundary terms.
2. If \(f:X\to X\) has no fixed points, then after the subdivisions and simplicial approximation used in Hatcher’s proof of the Lefschetz theorem there is a simplicial map \(g\) homotopic to \(f\) such that\[ g(\sigma)\cap\sigma=\varnothing \]for every simplex \(\sigma\) of the domain subdivision.
This is the geometric part of Hatcher’s proof and is independent of the coefficient ring.
3. The induced map on each simplicial chain group has trace zero over \(F\).
The condition in 2 says that no oriented simplex maps with a nonzero coefficient to itself, so the matrix of the chain map has zero diagonal. This statement is valid over every field.
4. Hence a fixed-point-free map has\[ \tau_F(f)=0. \]
By 1 and 3 the alternating trace on homology equals the zero alternating trace on chains. Homotopy invariance identifies the homology maps of \(f\) and \(g\).
Taking the contrapositive proves the field-coefficient Lefschetz theorem: \[ \boxed{\tau_F(f)\ne0\Longrightarrow f\text{ has a fixed point}.} \]