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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 $V^*\otimes W^*\to(V\otimes W)^*$ ?

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  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Mar 14 '19 at 19:54
  • $\begingroup$ Indeed, I added the link. $\endgroup$ – Phil-W Mar 14 '19 at 23:34
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Here is an argument the avoids choice.

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 $\omega=0$.

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