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Can someone help me with the Tensor-Product? We defined a Tensor-Product $V \otimes W$ ($V,W$ Vectorspaces) as a Quotient space $K^{V\times W} / R(V,W)$, where $R(V,W)$ is generated by the vectors $(v,\lambda w) - \lambda (v,w)$

$(v,w+w')-(v,w)-(v,w')$

$(\lambda v,w) - \lambda(v,w)$

$(v+v',w)-(v,w)-(v',w)$

and $K^{V\times W} := \{f: V\times W \rightarrow\ K\}$.

How can I imagine this Vectorspace? Each vector should be of the form $(v,w) + R(V,W)$ ? And what is the universal property?

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  • $\begingroup$ There are numerous answers on MSE about tensor products. A quick search will probably help. However to help you to go in the right direction, suppose $V$ has basis $v_1,...,v_n$ and $W$ has basis $w_1,...,w_m$. Then a basis for $V\otimes W$ is given by the $mn$ vectors $v_i \otimes w_j$. Quotienting out by $R(V,W)$ ensures you have 'bilinearity' of sorts i.e. so that $(v_1 + av_2) \otimes w = v_1 \otimes w + a v_2 \otimes w$ and similarly in the other factor. This gives us a universal property i.e. any time we have a bilinear map from $V \times W$, we have a unique linear map from... $\endgroup$ – Osama Ghani Jul 11 '18 at 11:08
  • $\begingroup$ ...$V \otimes W$ so that the bilinear map factors through the linear one. $\endgroup$ – Osama Ghani Jul 11 '18 at 11:11
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Do not try to imagine it concretely with cosets. That is essentially useless. Tensor products are hard the first (and second?) time you see them. The only way to learn how to work with a tensor product of vector spaces is through proving basic theorems about them, so you see how you can extract properties by that universal mapping property (universal for turning bilinear maps into linear maps).

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