= Solution
Its homology and chain formulas are
$$
L(f)=\sum_i(-1)^i\operatorname{tr}(f_*|H_i(K;\mathbb Q))
=\sum_i(-1)^i\operatorname{tr}(f_\#|C_i(K;\mathbb Q)).
$$
For <Lefschetz fixed-point theorem>, argue contrapositively: after barycentric subdivision, a fixed-point-free map has a simplicial approximation such that no simplex meets its image. Every diagonal coefficient on chains is then zero, so $L(f)=0$.
The antipodal map on $S^n$ has degree $(-1)^{n+1}$. For even $n$ this is $-1$, whereas the identity has degree $1$, so they are not homotopic.
If the tangent field $v$ were nowhere zero, then
$$
H(x,t)=\cos(\pi t)x+\sin(\pi t)\frac{v(x)}{|v(x)|}
$$
would stay on the sphere because $x\cdot v(x)=0$, and would homotope the identity to the antipodal map. This is impossible for even $n$, so $v$ has a zero.
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