I'm trying to show that given vector spaces $V_1,\dots,V_k$ (not necessarily finite dimensional) over the same field $F$ then if $v_i\in V_i$ we have $v_1\otimes\cdots\otimes v_k = 0$ if and only if at least one $v_i = 0$. Showing that if at least one $v_i = 0$ implies the product zero is a simple fact that comes from the multilinearity of the tensor product. My doubt know is about the other way around.
My thought was: we know that if $\mathcal{L}(V_1,\dots,V_k;W)$ means the space of multilinear mappings from $V_1\times\cdots\times V_k$ into another space $W$ then for each $\psi \in \mathcal{L}(V_1,\dots,V_k;W)$ there corresponds uniquely a linear $f\in \mathcal{L}(V_1\otimes\cdots\otimes V_k;W)$ with the property $$\psi(v_1,\dots,v_k)=f(v_1\otimes\cdots\otimes v_k)$$
Then, since by hypothesis $v_1\otimes\cdots\otimes v_k = 0$ this means that for each space $W$ and each $\psi \in \mathcal{L}(V_1,\dots,V_k;W)$ we have $\psi(v_1,\dots,v_k)=0$. Now I need to show that this implies that there must be one $v_i = 0$ and this makes sense, it seems to be a question of taking the convenient $\psi$, but I'm stuck.
Can someone give just a hint on how to proceed? I don't want the full proof, just a hint to go on.