Cap product on the oriented circle

The two endpoints of $\sigma$ equal $x=(1,0)$, so $\mathrm{\partial }z=x-x=0$. The normalization says $a(z)=1$. This does not depend on the representative: replacing $a$ by $a+\delta u$ changes the evaluation by $u(\mathrm{\partial }z)=0$.

The front face for degree one is all of $\sigma$, and its back face is its last vertex $x$. Hence $a\cap z=a(\sigma )x=x\mathrm{.}$ By [F2] this identity passes to $\alpha \cap [z]=[x]$.

Every point of $S^1$ can be joined to $x$ by an arc $t\mapsto (\cos (t\theta ),\sin (t\theta ))$, using any angle $\theta$ for that particular point. Thus there is one path component, and [F3] sends $[x]$ to $1$, not $-1$. This is the asserted positive generator. The coincident endpoints cause no cancellation of the retained vertex: only the boundary subtracts them. The input and coefficient ring are fixed and nonzero; neither an empty space nor a constant loop can satisfy this evaluation normalization. No simultaneous choice of angles or representatives is required, and no AC is used.