 Solution

Its homology and chain formulas areFor 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, thenwould 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.