@Steve, agree, we don't lose any information while doing tensor product and always can recover the original vector spaces using projections.
If I rewrite the previous example using the dual basis \\(e^j\\), such that \\(e^j(e_i) = e_i(e^j) = 1\ iff\ i = j,\ otherwise\ 0\\), I get a tensor product of \\(v= a \cdot e_1 + b \cdot e_2\\) and \\(v^* = a' \cdot e^1 + b' \cdot e^2\\) to look as \\(v^* \otimes v = aa' \cdot (e^1 \otimes e_1) + a'b \cdot (e^1 \otimes e_2) + ba' \cdot (e^2 \otimes e_1) + b'b \cdot (e^2 \otimes e_2)\\).
A cup \\(\cup: v^* \otimes v \to k\\) then just a dual map to this space.
**EDIT** following John's [comment 42](https://forum.azimuthproject.org/discussion/comment/20851/#Comment_20851), what goes below doesn't make any sense, because we should treat \\(e^i \otimes e_j\\) as a single abstract symbol.
If we assume that it has basis vectors of the form \\((e^j \otimes e_i)^* = e_i \otimes e^j\\), then, (using interchange law), it seems that \\((e^j \otimes e_i) \circ (e_i \otimes e^j) = (e^j \circ e_i) \otimes (e_i \circ e^j) = (e^j(e_i)) \otimes (e_i(e^j))\\). For example
\\((e^1 \otimes e_1) \circ (e_1 \otimes e^1) = (e^1 \circ e_1) \otimes (e_1 \circ e^1) = 1 \otimes 1 \to 1 \\)
\\((e^1 \otimes e_2) \circ (e_2 \otimes e^1) = (e^1 \circ e_2) \otimes (e_2 \circ e^1) = 0 \otimes 0 \to 0 \\)
using the definition of the dual basis.