When do exterior and tensor algebras commute with dual spaces? Suppose $V$ is a vector space, and $V^*$ is its dual space. Furthermore, let $\Lambda(V)$ be the exterior algebra of $V$, and let $T(V)$ be the tensor algebra.
When do the following two statements hold for some natural, canonical isomorphism?


*

*$\Lambda(V^*) \simeq \Lambda(V)^*$

*$T(V^*) \simeq T(V)^*$


It's clear they hold for finite-dimensional real vector spaces. Do they hold in the infinite-dimensional case? What about over arbitrary fields?
 A: First, it's pretty clear that the field $K$ does not matter in any way (if you know how to prove things for $\mathbb{R}$, then just check that you don't use any special property of this field).
Then, be careful : the statement for tensor algebras is already false for finite-dimensional vector spaces : if $V$ has finite dimension, $T(V^*)$ has countable dimension, but $T(V)^*$ has uncountable dimension. So canonical or not, there does not exist any isomorphism.
On the other hand, the result is true for exterior algebra if $V$ has finite dimension.
Let's treat the general case : since $T(V^*) = \bigoplus_n (V^*)^{\otimes n}$ and $T(V)^* = \left(\bigoplus_n V^{\otimes n}\right)^* = \prod_n \left(V^{\otimes n}\right)^*$, defining a linear map $\varphi: T(V^*)\to T(V)^*$ is canonically the same as defining linear maps $\varphi_{n,m}: (V^*)^{\otimes n}\to \left( V^{\otimes m}\right)^*$ for all $n,m\in \mathbb{N}$, which amounts to a $(n+m)$-linear form $$\widetilde{\varphi}_{n,m}: V\times\cdots \times V\times V^*\times \cdots \times V^*\to K.$$
Likewise, defining $\psi: \Lambda(V^*)\to \Lambda(V)^*$ is the same as defining $\psi_{n,m}: \Lambda^n(V^*)\to \left( \Lambda^m(V)\right)^*$, which amounts to a $(n+m)$-linear alternating form $$\widetilde{\psi}_{n,m}: V\times\cdots \times V\times V^*\times \cdots \times V^*\to K.$$
Then the canonical choice is to take $\varphi_{n,m}=0$ and $\psi_{n,m}=0$ if $n\neq m$, and $$\widetilde{\varphi}_{n,n}(v_1,\dots,v_n,f_1,\dots,f_n) = \prod_i f_i(v_i)$$ and $$\widetilde{\psi}_{n,n}(v_1,\dots,v_n,f_1,\dots,f_n) = Det(f_i(v_j))_{i,j}.$$
If $V$ is finite-dimensional, all $\psi_{n,n}$ are linear isomorphisms, and thus $\psi$ is an isomorphism because the direct product in $\Lambda(V)^*$ is finite, so it's a direct sum. On the other hand, if $V$ is finite-dimensional, even though the $\varphi_{n,n}$ are linear isomorphisms, $\varphi$ is not an isomorphism.
If $V$ is infinite-dimensional, then it's worse : the $\varphi_{n,n}$ and $\psi_{n,n}$ are not isomorphisms, and a fortiori neither are $\varphi$ and $\psi$.
