@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.