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For two vector spaces, $V$ and $W$, and a map $f: V \to W$, it is clear that: $ \ker(f) \otimes V + V \otimes \ker(f) \subseteq \ker(f \otimes f). $ Does the opposite inclusion hold? If so, I'd like a proof, and if not, a counterexample.

Bascially, given an element an element $\sum_i a_i \otimes b_i \in V \otimes V$, for which it holds that $ \sum_i f(a_i) \otimes f(b_i) = 0 $ can we show that $\sum_i a_i \otimes b_i \in \ker(f) \otimes V + V \otimes \ker(f)$?

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    I think th$a$t does it! Write up your comment as an answer.2012-06-01

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Take a basis $\{x_i\}_{i \in I_0}$ for the kernel of $f$, and extend it to a basis $\{x_i\}_{i \in I}$ for $V$, where $I_0 \subset I$. Thus $\{x_i \otimes x_j : i,j \in I\}$ is a basis for $V \otimes V$. Write out a general element of $\ker(f\otimes f)$ in terms of this basis, and note that the set $\{f(x_i) \otimes f(x_j) : i,j\in I\setminus I_0\}$ is linearly independent.

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    @MikhailMatrix Only after some fumbling. I think it's true over a commutative ring if you just assume that $V$ is flat, but you were right—no need for that language here. Might add it later just for fun. Cheers,2012-06-01