Axiom of choice and algebraic tensor product 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 $V\otimes W$. 
 A: I think both can be proved without choice, essentially because, in both cases, whenever you're tempted to choose a basis, you can manage with a little care to get by with a basis of a finite dimensional subspace.
For (2), if there's a linear dependence between the $v_i\otimes w_j$ then it involves only finitely many $v_i$ and $w_j$. Also, the linear dependence must be a (finite) linear combination of the usual relations such as $(u+u')\otimes v-u\otimes v-u'\otimes v$ for the tensor product, so there are finite dimensional subspaces $V'\leq V$ and $W'\leq W$ so that you have the same linear dependence in $V'\otimes W'$. And now you can use bases without invoking choice.
For (1), an element of the kernel is a finite sum of simple tensors $\varphi\otimes\psi$. By choosing a basis of the finite-dimensional subspaces of $V^*$ and $W^*$ spanned by the $\varphi$ and $\psi$ that occur, we can write the element of the kernel as a linear combination of $\{\varphi_i\otimes\psi_j\}_{i,j}$, where $\{\varphi_i\}_i$ and $\{\psi_j\}_j$ are finite linearly independent subsets of $V^*$ and $W^*$.
Now, again without choice, we can find finite dimensional subspaces $V'\leq V$ and $W'\leq W$ together with (finite) bases $\{v_i\}_i$ and $\{w_j\}_j$ that are dual bases to the restrictions of $\{\varphi_i\}_i$ and $\{\psi_j\}_j$ to $V'$ and $W'$, and prove that the kernel element is zero using these bases.
[To add a bit more detail to the last step, suppose that $U$ is a vector space over $k$, and $\alpha_1,\dots,\alpha_d$ a finite linearly independent list of elements of $U^*$. Then the subspace $S=\left\{\left(\alpha_1(u),\dots,\alpha_d(u)\right)\mid u\in U\right\}$ of $k^d$ must be the whole of $k^d$, or else there would be a nonzero linear functional $k^d\to k$ vanishing on $S$, and hence a linear dependence between the $\alpha_i$.
Hence there are elements $u_1,\dots,u_d\in U$ with $\alpha_i(u_j)=\delta_{ij}$ and so $U$ has a finite dimensional subspace $U'=\langle u_1,\dots,u_d\rangle$ with the $u_i$ forming a basis dual to the basis of $(U')^*$ consisting of the restrictions of the $\alpha_i$ to $U'$.]
In fact, answering a question asked in comments, there is a common generalization of (1) and (2). For any vector spaces $V$, $V'$, $W$, $W'$, the fact that the natural map
$$\text{Hom}(V,V')\otimes\text{Hom}(W,W')\to\text{Hom}(V\otimes W,V'\otimes W')$$
is injective can be proved without the axiom of choice.
To see this, note that an element of the kernel can be written in terms of $\alpha_i\otimes\beta_j$, where $\alpha_1,\dots,\alpha_s$ are finitely many linearly independent elements of $\text{Hom}(V,V')$ and $\beta_1,\dots,\beta_t$ finitely many linearly independent elements of $\text{Hom}(W,W')$. 
I claim that there is a finite dimensional subspace $V''\leq V$ such that the restrictions of $\alpha_1,\dots,\alpha_s$ to $V''$ are still linearly independent (and similarly for $W$ and the $\beta_j$). This follows by induction on $s$. Suppose there is a finitely generated subspace $U$ so that the restrictions of $\alpha_1,\dots,\alpha_k$ to $U$ are linearly independent. Then either the restrictions of $\alpha_1,\dots,\alpha_{k+1}$ to $U$ are linearly independent, or there is a linear dependence $\sum_{i=1}^{k+1}\lambda_i(\alpha_i|_U)=0$, which is unique up to a scalar multiple. But since $\alpha_1,\dots,\alpha_{k+1}$ are linearly independent, there is some $v\in V$ such that $\sum_{i=1}^{k+1}\lambda_i\alpha_i(v)\neq0$, and then the restrictions of $\alpha_1,\dots,\alpha_{k+1}$ to $U+\langle v\rangle$ are linearly independent.
Replacing $V$ and $W$ by $V''$ and $W''$, and $V'$ and $W'$ by $\sum_i\alpha_i(V'')$ and $\sum_j\beta_j(W'')$ reduces the original question to one about finite dimensional vector spaces, which can easily be solved without invoking choice.
