In the proof of proposition 1.1 found in this paper, what is the rationale for the map $$ w^{\prime}\otimes (x\otimes w)\mapsto w^{\prime} \otimes v(x,w) $$ found at the bottome of page 7? Is it simply that you can always think of the tensor product as a bilinear map (i.e. v(x,w))?
Tensor Product and Non-degenerate Pairing
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1 Answers
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Thinking of a tensor product as a bilinear form isn't quite right. The tensor product $V \otimes W$ isn't the space of bilinear forms $V\times W \to k$, but rather its dual is. This is the defining universal property of tensor products.
If you read a couple lines up you have a map $ev_M: Z(M)\otimes Z(\bar{M}) \to k$, and $v: Z(M) \times Z(\bar{M})$ is the corresponding bilinear form. Explicitly $v$ is the unique bilinear form on $Z(M)\times Z(\bar{M})$ sending $(x,w)$ to $ev_M(x \otimes w)$ for all $x \in Z(M)$ and $w \in Z(\bar{M})$.