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 $\sum_i a_i \otimes b_i \in \ker(f) \otimes V + V \otimes \ker(f)$?
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.