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For vector spaces $V,W$, probe that $\Lambda (V)\otimes\Lambda (W) \cong \Lambda (V\oplus W)$.

I have tried to use the universal property, but I can not create the necessary linear transformation.

SirWeigel
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    $\Lambda$ is the left adjoint to the forgetful functor from graded-commutative algebras to vector spaces given by taking the degree $1$ component, left adjoints preserve coproducts, and the coproduct of graded-commutative algebras is the graded tensor product. – Qiaochu Yuan Jul 01 '14 at 06:52
  • The idea of ​​the exercise is to use universal properties. – SirWeigel Jul 01 '14 at 06:54
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    This can be shown degreewise by showing that $\Lambda^i(V\oplus W)=\bigoplus \Lambda^j(V)\otimes \Lambda^{i-j}(W)$ and this is so by writing a basis. If you need universal properties, you can get linear maps $\Lambda(V)\to \Lambda(V\oplus W)$ (by the universal property of $\Lambda(V)$) and likewise for $W$ and then by linearlity (since tensorproduct is the coproduct) to a map from $\Lambda(V)\otimes \Lambda(W)\to \Lambda(V\oplus W)$. Now use the universal property to get a map backwards, and use uniqueness to get that they are inverses. Drawing all the diagrams is a good exercise. – Pax Jul 01 '14 at 06:59
  • @SirWeigel: the only thing I use in that argument is universal properties. – Qiaochu Yuan Jul 01 '14 at 07:29

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