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Let X and Y be k-vector spaces, if they are finite dimensional and $D=Hom_k(-,k)$ we have a natural (in X and Y) isomorphism:

$$ D Hom_k(X,Y) \cong Hom_k(DY,DX) $$

Can anyone help me prove this? I already tried to give the explicit isomorphism, but failed.

Another question, can we replace the statement for X and Y finite dimensional A-modules for a finite dimensional k-algebra A? Namely:

$$ D Hom_A(X,Y) \cong Hom_{A^{op}}(DY,DX) $$

Is it still natural in X and Y?

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    Altough, the first isomorphism is obvious by a dimension argument, but I would really like the explicit isomorphism. – 2017-01-05
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    Why do you think that such an isomorphism exists? If you take $Y = k$ then you are seeking for a natural isomorphism between $X^{**}$ and $X^{*}$... – 2017-01-06
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    Well, first of all because both vector spaces $DHom_k(X,Y)$ and $Hom_k(DY,DX)$ have the same dimension over k, and by the classification theorem of vector spaces, they are isomorphic. Also I do not see why $X^{**}$ is not isomorphic to $X^*$, since they have the same dimension. Maybe I am not understanding your confusion, or I didn't explain properly on the post. – 2017-01-06
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    Of course they are isomorphic, but they are not naturally isomorphic! (This is the same as asking whether $X$ and $X^{*}$ are naturally isomorphic as $X^{**}$ is naturally isomorphic to $X$). – 2017-01-06
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    I do not see why $X$ and $X^*$ are not naturally isomorphic. – 2017-01-06

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I think that this cannot be true. Via passing to the dual map, $L(X,Y)$ is naturally isomorphic to $L(Y^*,X^*)$ (for finite dimensional $X$ and $Y$). If what you claim were true, then $L(X,Y)$ would be naturally isomorphic to $L(X,Y)^*$ which for $X=k$ would give you a natural isomorphism between $Y$ and $Y^*$.

In your orginal setup things alos do not fit together since one of the two functors you are looking at is covariant in $X$ and contravariant in $Y$ while the other is contravariant in $X$ and covaraint in $Y$.