# Is the canonical map $(\prod_{i \in I} V_i)\otimes (\prod_{j \in J} W_j)\to \prod_{(i,j)\in I\times J} V_i\otimes W_j$ injective?

Let $$k$$ be a field and $$\{V_i\}_{i \in I}$$ and $$\{W_j\}_{j \in J}$$ be a collection of $$k$$-vector spaces. We have a canonical map $$\left(\prod_{i \in I} V_i\right)\otimes \left(\prod_{j \in J} W_j\right)\to \prod_{(i,j)\in I\times J} V_i\otimes W_j: (v_i)_{i \in I}\otimes (w_j)_{j \in J}\mapsto (v_i\otimes w_j)_{(i,j)\in I\times J}.$$ Is this map injective in general? I can prove this for the direct sum, because there I can simply use a basis, however the direct product doesn't have a nice basis so I don't know how to proceed here. I tried tricks with linear independency, linear functionals etc.

Thanks in advance for any help!

• @WilliamM. How is that a counterexample? Apr 4, 2022 at 21:33
• I confused the left-hand tensor product with a Cartesian product. This made me wonder, what definition of tensor product are you using? I often work with $V \otimes W$ as the set of all bilinear forms on $V \times W$ of the form $(a,b) \mapsto x(a) y(b)$ with $x$ and $y$ linear forms on $V$ and $W,$ respectively. When applied to a Cartesian product, I reach a bilinear form $((a_i), (b_j)) \mapsto x((a_i)) y((b_j)),$ with $x$ and $y$ linear on $\prod V_i$ and $\prod W_j,$ but I fail to identify them with $(v_i)$ and $(w_j).$ So, our definitions of tensor product probably differ. Apr 4, 2022 at 21:52
• See also this closely related question: math.stackexchange.com/questions/4232175/… (which is essentially just the special case of your question where each $V_i$ and $W_j$ is 1-dimensional). Apr 4, 2022 at 22:13
• @WilliamM It doesn't matter what definition of the tensor product one uses, as long as it satisfies the universal property of the tensor product. Apr 5, 2022 at 7:59

Yes, this map is always injective. It suffices to show the canonical map $$\left(\prod_{i \in I} V_i\right)\otimes W\to \prod_{i\in I}(V_i\otimes W)$$ is always injective, since you can use this twice (first with $$W=\prod_{j\in j}W_j$$ and then on each $$V_i\otimes W$$ swapping the roles of the two sides of the tensor) to deduce injectivity of your map. But now note that any element of $$\left(\prod_{i \in I} V_i\right)\otimes W$$ actually comes from $$\left(\prod_{i \in I} V_i\right)\otimes W_0$$ for some finite-dimensional subspace $$W_0\subseteq W$$, and there is a commutative square
$$\require{AMScd} \begin{CD} \left(\prod_{i \in I} V_i\right)\otimes W_0 @>{}>> \prod_{i\in I}(V_i\otimes W_0)\\ @V{}VV @V{}VV \\ \left(\prod_{i \in I} V_i\right)\otimes W @>{}>> \prod_{i\in I}(V_i\otimes W) \end{CD}$$ in which the vertical maps are injective, so it suffices to show the top map is injective. In other words, we may assume $$W$$ is finite-dimensional. But in that case it is easy to see that the map is in fact an isomorphism, since each side preserves finite direct sums in the $$W$$ variable and the map is an isomorphism when $$W=k$$.