Something of this form has already been answered here:
Why is $\text{Hom}(V,W)$ the same thing as $V^* \otimes W$?
I'm starting introductory category theory stuff, and I'm looking for some help.
I know that $W\otimes V^{\ast}$ and $\text{Hom}(V,W)$ are naturally isomorphic in the finite dimensional case using the isomorphism $v^{\ast}\otimes w(x)\mapsto v(x)w$. My question is constructing the isomorphism in the other direction, namely $\text{Hom}(V,W)\to W\otimes V^{\ast}$. I'm curious if this can be done without using:
1) Bases, so in particular it should be a natural transformation of finite dimensional spaces, using finiteness only to prove that the given map is indeed an isomorphism.
2) Hom-tensor adjointness. I have read this online and in Rotman, and see how it gives a natural isomorphism for these using a series of equivalences like $\text{Hom}(V\otimes W, U)\to \text{Hom}(V,\text{Hom}(W,U))$.
The basic idea is to say we want a natural transformation $\tau:\text{Hom}(V,-)\to -\otimes V^{\ast}$, so define $\tau_W:\text{Hom}(V,W)\to W\otimes V^{\ast}$ and define $\tau_W(f)=$. It's this latter part I'm having trouble with.
Any help would be greatly appreciated.
