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Hello :) i think the tensor product is very difficulty, and i have some questions (let $U$ be any $k$-vectorspace and $k$ a field)

  • How to construct a natural bijection (thus without choosing a basis) between bilinear maps $V\times W\rightarrow U$ and linear maps $V\otimes W\rightarrow U$?
  • How to construct a natural isomorphism $V^*\otimes W\rightarrow Hom(V,W)$ (with V finite dimensional and $V^*$ the dual space of $V$.

I don't see how to produce such bijections and isomorphims. I think thats not difficult but i can't imagine such a mapping. Someone who can help me? Thanks.

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For the first one, given a bilinear map, and using the universal property of tensor products, don't you think there is an obvious candidate? – rschwieb Jan 17 at 15:30
i thought that you make the following map $\psi(f(x\times y))=f(x\otimes y)$ – Lie-algebra Jan 17 at 15:31

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For the first one: given a bilinear map $B:V\times W\rightarrow U$, use the universal property of tensor products to produce a linear map from $\hat B: V\otimes W\rightarrow U$. Work to show that the assigment $B\mapsto \hat B$ is a bijection.

For the second one: First, given $(f,w)\in V^*\times W$, we try to produce an element of $Hom(V,W)$. The choice that "jumps out" is to try: $(f,w)(v):=f(v)\cdot w$. Notice this assigment is a bilinear map $V^*\times W\rightarrow Hom(V,W)$. Lift this with the universal property and work to show it's a natural isomorphism.

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For the first one: if you choose such a $B$ then you know from the universal property that there exists a bilinear map from $i:V\times W\rightarrow V\otimes W$. We get also a unique map $L:V\otimes W\rightarrow U$ such that $L\circ i=B$, but then $L$ is also linear or not? – Lie-algebra Jan 17 at 15:46
sorry we can make the following map: $v\otimes w\mapsto B(v,w)$ – Lie-algebra Jan 17 at 15:50
@Lie-algebra Yes, $L$ is linear: that is part of the universal property. – rschwieb Jan 17 at 15:57

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