TITLE: Tensor product of duals
QUESTION [5 upvotes]: Can we show $V^* \otimes W^* \simeq (V \otimes W)^*$, when $V$ or $W$ is finite-dimensional without referring to the basis?
I can inject $V^* \otimes W^*$ into $(V \otimes W)^*$ using the obvious way, the problem is other way around.

REPLY [4 votes]: There is a canonical linear map $V^* \otimes W^* \to (V \otimes W)^*$. It is an isomorphism when $W=K$ (both sides identify with $V^*$ and the linear map becomes the identity). The class of $W$s for which it is an isomorphism is closed under finite direct sums - this is because both sides are additive (contravariant) functors in $W$. It follows that it is an isomorphism for every finite-dimensional vector space $W$. Here, $V$ is arbitrary.