Let's use the notation of `[A=>B]` for `Hom(A, B)`. Take a 1-dimensional algebraic torus `G`<sub>`m`</sub> and higher-dimensional torus `T` and let's live in the category of commutative algebraic groups over `k`.

Out of  four expressions like `[G`<sub>`m`</sub>` => [G`<sub>`m`</sub>`=>T]]` etc. half give back `T` `(*)`, others the *dual torus* `T`<sup>`V`</sup>, in the sense that `X`<sub>`*`</sub>`(T) := ` [`G`<sub>`m`</sub> ` => T] = [T`<sup>`V`</sup> ` => G`<sub>`m`</sub>`] =: X`<sup>`*`</sup>`(T`<sup>`V`</sup>`)`.

To prove equality  `(*)` use for `B = G`<sub>`m`</sub>
       
       (**)              A \otimes [B=>B] ==== [[A=>B] => B].


> **Question:**  Is there an example of commutative algebraic group `B`, other than `G`<sub>`m`</sub>, for which the identity `(**)` is also true or perhaps true in some other sense?

(One thing I specifically have in mind is that if we could write `[X => Y] = X`<sup>`*`</sup> `\otimes Y` whenever `X` and `Y` are groups, as if it was a with any rigid tensor category, the formula would hold for all `A` and `B`)