The first part of the question was asked on Math-stackexchange.

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 questions are

1. is the axiom of choice necessary to prove the injectivity of the canonical map $V^*\otimes W^*\to(V\otimes W)^*$ ?

2. A (possibly) connected question is the following : is it necessary to use the axiom of choice to prove that if $(v_i)$ is a linearly independant family in $V$, and $(w_j)$ is a linearly independant family in $W$, then $(v_i\otimes w_j)_{i,j}$ is linearly independant in $V\otimes W$.

The first part of the question was asked on Math-stackexchange.

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 questions are

1. is the axiom of choice necessary to prove the injectivity of the canonical map $V^*\otimes W^*\to(V\otimes W)^*$ ?

2. A (possibly) connected question is the following : is it necessary to use the axiom of choice to prove that if $(v_i)$ is a linearly independent family in $V$, and $(w_j)$ is a linearly independent family in $W$, then $(v_i\otimes w_j)_{i,j}$ is linearly independent in $V\otimes W$.