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!

  • 1
    $\begingroup$ @WilliamM. How is that a counterexample? $\endgroup$
    – Andromeda
    Apr 4, 2022 at 21:33
  • $\begingroup$ 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. $\endgroup$
    – William M.
    Apr 4, 2022 at 21:52
  • $\begingroup$ 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). $\endgroup$ Apr 4, 2022 at 22:13
  • $\begingroup$ @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. $\endgroup$
    – Andromeda
    Apr 5, 2022 at 7:59

1 Answer 1


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 $W=k$.

  • $\begingroup$ Lovely argument! Thanks for the answer! $\endgroup$
    – Andromeda
    Apr 5, 2022 at 8:31

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