# Axiom of choice and dual of a tensor product

EDIT : This question (and other related questions) was also asked on mathoverflow : here.

Let $$V$$, and $$W$$ be vector spaces. By the universal property of the tensor product, there is a canonical map from $$V^*\otimes W^*$$ into $$(V\otimes W)^*$$ (since the map $$(\omega_1,\omega_2)\mapsto \omega_1\otimes\omega_2$$ is bilinear from $$V^*\times W^*$$ into $$(V\otimes W)^*$$.

I have read that this map is actually injective by using some basis on $$V$$ and $$W$$.

Since the existence of basis for any arbitrary vector space relies on the axiom of choice, my question is : is the axiom of choice necessary to prove the injectivity of the canonical map $$V^*\otimes W^*\to(V\otimes W)^*$$ ?

• For future reference, when you cross-post (and you've done it more-or-less by the book, so it's fine), you should always add a disclaimer and a link here as well. This lets people who don't follow the [axiom-of-choice] tag on both sites know that it's there, and easy access to see if someone might have already answered it. Mar 14, 2019 at 19:54
Suppose an element $$\omega\in V^*\otimes W^*$$ maps to $$0$$ in $$(V\otimes W)^*$$. Using the description of $$V^*\otimes W^*$$ as a quotient of the free vector space generated by symbols $$\varphi\otimes \psi$$, we can write $$\omega=\sum_{i=1}^n \varphi_i\otimes \psi_i$$, with $$\varphi_i\in V^*$$, $$\psi_i\in W^*$$. Let $$V^f$$, $$W^f$$ be the subspaces of $$V^*$$, $$W^*$$ spanned by $$\varphi_i$$, $$\psi_i$$ respectively. Because $$V^f$$ and $$W^f$$ are finite-dimensional, we can find bases $$\alpha_1,\ldots,\alpha_n$$ of $$V^f$$ and $$\beta_1,\ldots,\beta_m$$ of $$W^f$$. Let $$\alpha:V\to K^n$$ and $$\beta:W\to K^m$$ be the linear maps whose coordinate functions are the $$\alpha_i$$ and $$\beta_i$$, respectively. The maps $$\alpha$$ and $$\beta$$ are surjective because the $$\varphi_i$$ and $$\psi_i$$ are linearly independent.
Now consider the commutative diagram $$\begin{array}{ccc} (K^n)^*\otimes (K^m)^*&\rightarrow&(K^n\otimes K^m)^*\\ \downarrow&&\downarrow\\ V^*\otimes W^*&\rightarrow&(V\otimes W)^*\\ \end{array}$$ The element $$\omega\in V^*\otimes W^*$$ lifts to $$(K^n)^*\otimes (K^n)^*$$ by construction, the top horizontal map is an isomorphism because the spaces are finite dimensional, and map on the right is injective because $$V\otimes W\to K^n\otimes K^m$$ is surjective. So if $$\omega$$ maps to $$0$$ in $$(V\otimes W)^*$$, then the lift of $$\omega$$ in $$(K^n)^*\otimes (K^m)^*$$ must have been $$0$$, which implies $$\omega=0$$.