$\hom(U, V)=U^*\otimes V$ [closed]

We know that $\hom(U, V)=U^*\otimes V$. Is the following true?

(1) $\hom(U, V)=U\otimes V^*$;

(2) $\hom(U, V)=V\otimes U^*$;

(3) $\hom(U, V)=V^*\otimes U$.

Thank you very much.

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What are your assumptions? Do you know the notions of covariant and contravariant functor? In some contexts the tensor product is commutative. –  Pierre-Yves Gaillard Aug 15 '11 at 14:45
It depends on what you mean by the equality sign. 1) and 3) are contravariant in $V$ while $\text{hom}(U, V)$ is covariant, so they can't be naturally isomorphic as functors. 2) holds in some situations but not others. –  Qiaochu Yuan Aug 15 '11 at 14:51
If you are taking your tensor products over a commutative ring, you need that $U$ is finitely generated and projective, and over a field, you need $U$ is finite dimensional, or else the original isomorphism does not hold. If you are commutative (and the module structures are symmetric, so $am=ma$), then tensor products are commutative. If you're working over vector spaces, you have a non-cannonical isomorphism for all the options (because the dimensions are the same). If you want a natural isomorphism, you need the star to be on the $U$. –  Aaron Aug 15 '11 at 17:37