# Is the Tensor product in the category of finite vector spaces, enriched over itself strong monoidal?

The category $\mathcal{V}_k$ of vector spaces, over a field $k$, can be considered as a symmetric, closed monoidal category with respect to the tensor product of vector spaces, the vector space of linear maps as internal hom $[V,W]$ and the symmetry $\tau: x \otimes y = y \otimes x$.

In addition we can see $\mathcal{V}_k$ as enriched over itself and from the tensor $\leftrightarrow$ hom adjunction we get the following natural inclusion

$[V_1,W_1]\otimes [V_2,W_2] \to [V_1 \otimes V_2, W_1 \otimes W_2]$ $\quad\quad(A)$

which corresponds under the tensor - hom ajunction to a preimage of the identity in

$hom([V_1,W_1]\otimes [V_2,W_2], [V_1 \otimes V_2, W_1\otimes W_2])\simeq\\ hom([V_1,W_1]\otimes [V_2,W_2]\otimes V_1\otimes V_2, W_1\otimes W_2)\simeq\\ hom(([V_1,W_1]\otimes V_1) \otimes ([V_2,W_2]\otimes V_2), W_1\otimes W_2)\to\\ hom(W_1\otimes W_2, W_1 \otimes W_2)$

where we used the symmetry and the natural evaluation $[V,W]\otimes V\to W$. In the most right hom set, there is of course the natural choice $id_{W_1\otimes W_2}$.

Now the question: In case we restrict to finite dimensional vector spaces. Is $(A)$ an isomorphism? If yes, how can we compute the inverse?

## 1 Answer

The definition of "finite dimensional" is "has a finite basis". So to answer this problem we will definitely have to pick bases. Let $(e^1_i)$, $(e^2_i)$, $(g^1_i)$ and $(g^2_i)$ be bases of $V_1$, $V_2$, $W_1$ and $W_2$, and let the dual bases be $(f^1_i)$, $(f^2_i)$, $(h^1_i)$ and $(h^2_i)$. Then $[V_1,W_1]\otimes[V_2,W_2]$ has a basis given by $(f^1_i\otimes g^1_j\otimes f^2_k\otimes g^2_l)$ and $[V_1\otimes V_2,W_1\otimes W_2]$ has a bais given by $(f^1_i\otimes f^2_j\otimes g^1_k\otimes g^2_l)$. The inverse isomorphism is the one that takes $f^1_i\otimes f^2_j\otimes g^1_k\otimes g^2_l$ to $f^1_i\otimes g^1_k\otimes f^2_j\otimes g^2_l$.